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Hardy type inequalities for one weight function and their applications R. G. Nasibullin Institute of Mathematics and Mechanics, Kazan (Volga Region) Federal University
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     § 1. IntroductionLet $\Omega$ be an open connected proper subset of the Euclidean space $\mathbb{R}^n$, $n\geqslant 2$, and let $C_{\mathrm{c}}^1(\Omega)$ be the family of continuously differentiable functions $g\colon \Omega\to\mathbb{R}$ with compact supports in $\Omega$. In a domain $\Omega$, the distance function $\delta$ from a point $x$ of this domain to its boundary $\partial\Omega$ is defined by
$$
\begin{equation*}
\delta(x)=\operatorname{dist}(x, \partial\Omega)=\inf_{y\in \partial\Omega}|x-y|.
\end{equation*}
\notag
$$
More information on properties and applications of the distance function associated with an open subset of the Euclidean space can be found in the recent papers [1], [2]. Applications of the distance function include the theory of Hardy type variational inequalities. It is well known that if $\Omega$ is a convex domain, then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, the following Hardy inequality holds
$$
\begin{equation}
\frac{1}{4}\int_\Omega\frac{|g(x)|^2}{(\delta(x))^2}\, dx\leqslant \int_\Omega |\nabla g(x)|^2\, dx
\end{equation}
\tag{1.1}
$$
with sharp (but non-attainable) constant $1/4$ (see, for instance, [3]–[7]). This non-attainability of inequality (1.1), which is now considered classical, is one features of this inequality. The second feature of this inequality, also closely related, to the sharp constant, is the possibility of strengthening the inequality by adding an additional term. In this regard, we mention the following result of Brezis and Marcus [3]: if $\Omega\subset\mathbb{R}^n$ is a bounded domain of finite diameter $D (\Omega)$, then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$,
$$
\begin{equation}
\frac{1}{4}\int_\Omega\frac{|g(x)|^2}{(\delta(x))^2}\, dx+\frac{1}{4 (D(\Omega))^2}\int_\Omega|g(x)|^2\, dx\leqslant \int_\Omega |\nabla g(x)|^2\, dx.
\end{equation}
\tag{1.2}
$$
The problem of adding an additional term to (1.1) is related to classical estimates for the first eigenvalue $\lambda_1(\Omega)$ of the Laplacian under Dirichlet boundary conditions and to the following Poincaré inequality:
$$
\begin{equation*}
\lambda_1(\Omega)\int_\Omega|g(x)|^2\, dx\leqslant \int_\Omega |\nabla g(x)|^2\, dx \quad\forall\, g\in C_{\mathrm{c}}^1(\Omega).
\end{equation*}
\notag
$$
In this regard, the Poincaré estimate $\lambda_1(\Omega)>\pi^2/(D(\Omega))^2$ and the famous Rayleigh–Faber–Krahn isoperimetric inequality
$$
\begin{equation*}
\lambda_1(\Omega) >\frac{\omega_n^{2/n}}{|\Omega|^{2/n}}(j_{n/2-1})^2,
\end{equation*}
\notag
$$
are well known, where $j_\nu$ is the first zero of the Bessel function $J_\nu$ of order $\nu$, and $\omega_n$ is the volume of the $n$-dimensional unit ball (see [8]). In the same paper [3], H. Brezis and M. Marcus have asked whether it is possible to replace the constant in the additional term by a value of the form $K'(n) |\Omega|^{-2/n} $, where $|\Omega|$ is the volume of the domain, and $K'(n)$ is some positive universal constant. Hoffmann-Ostenhof, Hoffmann-Ostenhof, and Laptev [9] gave a positive answer to this question. Namely, they showed that
$$
\begin{equation}
\frac{1}{4}\int_\Omega\frac{|g(x)|^2}{(\delta(x))^2}\, dx +\frac{1}{4}\, \frac{K(n)}{|\Omega|^{2/n}}\int_\Omega |g(x)|^2\, dx\leqslant \int_\Omega |\nabla g(x)|^2\, dx \quad\forall\, g\in C_{\mathrm{c}}^1(\Omega),
\end{equation}
\tag{1.3}
$$
where $K(n)=n[|\mathbb{S}^{n-1}|/n]^{2/n}$ and $|\mathbb{S}^{n-1}|$ is the surface area of the $(n-1)$-dimensional unit sphere. Evans and Lewis [10] proved an analogue of (1.3) with constant $3/2K(n)$ instead of $1/4K(n)$, but, as their proof shows, this inequality is valid on a more restrictive class of functions than the family $C_{\mathrm{c}}^1(\Omega)$. In the case of convex domains of fixed volume, as a corollary to the result of M. Hoffmann-Ostenhof, T. Hoffmann-Ostenhof, and A. Laptev, the first eigenvalue is estimated as
$$
\begin{equation*}
\lambda_1(\Omega)\geqslant \frac{1}{4}\, \frac{K(n)}{|\Omega|^{2/n}}.
\end{equation*}
\notag
$$
The question of finding sharp constants in the above isoperimetric inequalities and the question of the non-improvability of the constant in the additional term in Hardy-type inequalities remain open, and even the problem of strengthening these inequalities is quite involved and relevant. Despite the external similarity of inequalities (1.2) and (1.3), they are quite different if considered as variational problems. The first extremal problem is considered in domains of finite diameter, and the second one, in domains of finite volume. The third kind of extremal problems involves inequalities with additional terms in domains of finite inner radius There are domains with finite inner radius, but with infinite volume and diameter (for example, a strip), and hence the third class of problems is wider (see [11]–[16] for details).In this direction, the most complete solution of the “Brezis–Marcus problem” is the result of Avkhadiev and Wirths [14], which states that
$$
\begin{equation}
\frac{1-\nu^2q^2}{4}\int_\Omega\frac{|g(x)|^2}{(\delta(x))^2} \, dx +\frac{q^2|\lambda_q|^2}{4(\delta_0(\Omega))^2}\int_\Omega\frac{|g(x)|^2}{(\delta(x))^{2-q}} \, dx \leqslant \int_\Omega |\nabla g (x)|^2 \, dx
\end{equation}
\tag{1.4}
$$
for any continuously differentiable function $ g$ with compact support in a convex domain $\Omega$ with finite inner radius $\delta_0(\Omega)$, where $q>0$, $\nu \in [0,1/q]$, $\delta(x)$ is the distance function to the boundary of the domain, and the constant $\lambda_q$ is the solution of the Lamb-type equation for the Bessel function $J_\nu$ of order $\nu$. The constants $(1-\nu^2q^2)/4$ and $q^2(\lambda_q)^2/4$ in this inequality are sharp. We only note that, for $\nu>0$, there is an extremal function for which this equality is attained, and, for $\nu=0$, Avkhadiev and Wirths, constructed a minimizing sequence verifying sharpness of the inequality and unattainability of the constant (see also [15]). In the case of $n$-dimensional convex domains, the result of F. G. Avkhadiev and K.-J. Wirths provides a second method of the proof of the known estimates (see [17])
$$
\begin{equation*}
\lambda_1(\Omega)\geqslant \frac{\pi^2}{4(\delta_0(\Omega))^2}\geqslant \frac{\pi^2}{(D(\Omega))^2}.
\end{equation*}
\notag
$$
In what follows, we will obtain some estimates for $\lambda_1(\Omega)$ in terms of the diameter of the domain for other classes of domains. The proofs of all the above inequalities are original and beautiful, and are based on one-dimensional inequalities. The one-dimensional inequalities do not involve difficulties associated with the geometry of the domain. V. I. Levin can be considered the first to obtain a one-dimensional Hardy-type inequality with additional terms. In [18], for all continuously differentiable functions $y$ such that $y(0)=0$ and $y\not\equiv 0$, V. I. Levin obtained the sharp inequality
$$
\begin{equation}
\int_0^{1}\frac{(y(x))^2}{x^2(2-x)^2} \, dx< \int_0^{1}(y'(x))^2\, dx,
\end{equation}
\tag{1.5}
$$
which can be rewritten in the form
$$
\begin{equation*}
\frac{1}{4}\int_0^{1}\frac{(y(x))^2}{x^2} \, dx +\frac{1}{2}\int_0^{1}\frac{(y(x))^2}{x(2-x)}\, dx +\frac{1}{4}\int_0^{1}\frac{(y(x))^2}{(2-x)^2} \, dx < \int_0^{1}(y'(x))^2\, dx.
\end{equation*}
\notag
$$
Multivariate Hardy-type inequalities with additional terms were first obtained by Maz’ya [19] in the case where $\Omega$ is the upper half-plane and the weight functions depend on the distance to the origin. In the present paper, we will prove inequalities on a finite interval that strengthen inequality (1.5) and lead us to spatial inequalities with a greater constant in front of $K(n)$. For example, we will show that, for any absolutely continuous function $y\colon [0,1]\to \mathbb{R}$ such that $y(0)=0$ and $y'\in L_2([0,\rho])$, for $q\in (0,\infty)$ and $\nu\in [0,1/q]$, the sharp inequality
$$
\begin{equation*}
(1\,{-}\,\nu^2q^2)\int_0^\rho \!\frac{(y(t))^2}{t^2(2\,{-}\,t)^2}\, dt \,{+}\,q^2(C_\nu(q))^2\frac{(2\,{-}\rho)^q}{\rho^q}\int_0^\rho \!\frac{(y(t))^2}{(2\,{-}\,t)^{2+q}t^{2-q}}\, dt\,{\leqslant}\! \int_0^\rho \!(y'(t))^2\, dt
\end{equation*}
\notag
$$
holds, and if $q\in [1,2]$, then
$$
\begin{equation*}
\int_{0}^\rho \frac{(y(t))^2}{t^{2-q}(2-t)^{2-q}}\, dt\leqslant \kappa'(q,\rho)\int_0^\rho (y'(t))^2\, dt,
\end{equation*}
\notag
$$
where $\rho\in (0,1]$, $C_\nu(q)$ and $\kappa'(q,\rho)$ are some known constants. Below, we will also obtain some applications of these one-dimensional inequalities. Using the first inequality, we will establish multivariate inequalities in arbitrary domains in terms of the mean distance, and, in addition, in regular domains, in domains satisfying the cone condition, in domains $\lambda$-close to convex ones, and in convex domains, we will prove the inequalities in terms of the distance to the boundary of the domain. As a consequence of multivariate inequalities, we will obtain new estimates for the first eigenvalue $\lambda_1(\Omega)$ in various domains. For example, in a convex domain $\Omega$ of fixed volume or diameter, respectively, we have
$$
\begin{equation*}
\lambda_1(\Omega)\geqslant \frac{(j'_1)^2}{2}\, \frac{K(n)}{|\Omega|^{2/n}} \quad\text{and}\quad \lambda_1(\Omega)\geqslant \frac{(j'_1)^2}{2}\, \frac{3n}{(D(\Omega))^2},
\end{equation*}
\notag
$$
where $j'_1\approx 1.84118$ is the first positive root of the derivative $J_1'$ of the Bessel function $J_1$, and in the case of a domain $\lambda$-close to a convex domain, we get
$$
\begin{equation*}
\lambda_1(\Omega)\geqslant \frac{2.75972n}{(D(\Omega))^2}.
\end{equation*}
\notag
$$
The proof of multivariate inequalities depends on upper estimates of the mean distance in terms of the distance to the boundary of the domain. In addition, we establish new estimates for the mean distance in terms of the distance to the boundary of a domain $\lambda$-close to a convex domain (see § 3.4). According to [13]–[15], the constant in the additional term in inequalities with inner radius does not depend on the dimension of the space, but this is not the case in (1.3) and in its generalizations (see [20]). In this regard, we show, for the first time, that, in inequalities with diameter, the constant also depends on the dimension. As an application of the second one-dimensional inequality, we establish new sufficient Nehari–Pokornii type conditions for univalence, which continues the study of sufficient conditions for univalence from [21]–[26]. In order not to overburden the introduction, a comparison of the known and our results is put off until § 4. § 2. One-dimensional inequalities on a finite interval2.1. Basic notation and auxiliary resultsThis section is devoted to new one-dimensional inequalities with power-law singularities in weight functions. For further purposes, we need some facts about the Bessel function $J_\nu$ of order $\nu\geqslant 0$. Recall that the Bessel function can be written as the convergent power series
$$
\begin{equation*}
J_\nu(t)=\sum_{k=0}^\infty\frac{(-1)^kt^{2k+\nu}}{2^{2k+\nu}k!\,\Gamma(k+1+\nu)}, \qquad t\in[0,1],
\end{equation*}
\notag
$$
where $\Gamma$ is the Euler gamma function. Let $j_\nu$ be the first zero of the Bessel function $J_\nu$. For more on the Bessel function and its zeros, see the book [27]. We will need the following of the Bessel function. For example, it is known that a) $J_\nu$ is the canonical solution of the Bessel differential equation (see [27]) b) for sufficiently small $t$, the asymptotic formula
$$
\begin{equation*}
J_\nu(t)= \frac{t^\nu}{2^\nu\Gamma(1+\nu)} \biggl(1-\frac{\Gamma(1+\nu)}{\Gamma(2+\nu)}\, \frac{t^2}{2^2}+\frac{\Gamma(1+\nu)}{\Gamma(3+\nu)}\, \frac{t^{4}}{2^4}-\cdots\biggr) \sim \frac{1}{\Gamma(\nu+1)}\biggl(\frac{t}{2}\biggr)^{\nu}
\end{equation*}
\notag
$$
holds. We will also need the following result, which we give with a proof. Lemma 1. The continuous function $h(t)=J_1(t)/(tJ_0(t))$ is increasing on $[0,2]$, and $\inf_{t\in [0,2]} h(t)=1/2$. Proof. Let us show that the derivative $h'(t)$ is non-negative on $[0,2]$. Using the well-known equalities for the Bessel function and its derivative (see, for example, [27], §§ 3.2 and 5.51),
$$
\begin{equation*}
\begin{gathered} \, J_0'(t)= -J_1(t), \qquad tJ'_1(t)-J_1(t)=-tJ_2(t), \\ (J_1(t))^2-J_0(t)J_2(t)=\frac{4}{t^2}\sum_{j=0}^\infty(2+2j)(J_{2+2j}(t))^2, \end{gathered}
\end{equation*}
\notag
$$
we get that
$$
\begin{equation*}
\begin{aligned} \, h'(t) &= \frac{tJ_1'(t)J_0(t)-J_1(t)J_0(t)-tJ'_0(t)J_1(t)}{t^2(J_0(t))^2} =\frac{J_0(t)(tJ_1'(t)-J_1(t))+t(J_1(t))^2}{t^2(J_0(t))^2} \\ &=\frac{(J_1(t))^2-J_0(t)J_2(t)}{t (J_0(t))^2} =\frac{4}{t^3}\sum_{j=0}^\infty(2+2j)(J_{2+2j}(t))^2 \geqslant 0. \end{aligned}
\end{equation*}
\notag
$$
Therefore, $h(t)$ is increasing, and $\inf_{t\in [0,2]}h(t)=\lim_{t\to 0}h(t)= 1/2$ by property b). The lemma is proved. 2.2. Inequalities required for multivariate analoguesThe following theorem holds. Theorem 1. Let $q\in (0,\infty)$, $\nu\in [0,1/q]$, and $\rho\in (0,1]$. Then, for any absolutely constant function $y\colon [0,1]\to \mathbb{R}$ such that $y(0)=0$ and $y'\in L_2([0,\rho])$, the sharp inequality Proof. Consider the function where $\lambda_\nu(q)$ is some constant which will be chosen later.
Note that the choice (selection) of the form of the function $\varphi(t)$ is the key and novel part of the proof. Using properties a) and b), it can be shown that
$$
\begin{equation}
\begin{aligned} \, -\frac{\varphi''(t)}{\varphi(t)} &= \frac{1-\nu^2q^2}{(z(t))^2} +\frac{q^2|\lambda_\nu(q)|^2}{(2-t)^{2+q}t^{2-q}}, \\ \frac{\varphi'(t)}{\varphi(t)} &= \frac{1-t}{z(t)}+\frac{q\lambda_\nu(q) t^{q/2-1}}{(2-t)^{q/2+1}}\, \frac{J'_\nu(\lambda_\nu(q) t^{q/2}/(2-t)^{q/2})}{J_\nu(\lambda_\nu(q) t^{q/2}/(2-t)^{q/2})}, \end{aligned}
\end{equation}
\tag{2.1}
$$
and
$$
\begin{equation*}
\frac{\varphi'(t)}{\varphi(t)}= \frac{1+\nu q}{z(t)}+o(1)\quad \text{as}\quad t\to 0.
\end{equation*}
\notag
$$
Next, using the series expansion of the Bessel function and the well-known equality (see [27], § 3.2), we have
$$
\begin{equation*}
\begin{aligned} \, \varphi'(t) &=\frac{1-t-\nu q}{\sqrt{z(t)}}J_\nu \biggl( \frac{\lambda_\nu(q)t^{q/2}}{(2-t)^{q/2}}\biggr) +\frac{q\lambda_\nu(q) t^{q/2}}{\sqrt{z(t)}\, (2-t)^{q/2}}J_{\nu-1}\biggl( \frac{\lambda_\nu(q) t^{q/2}}{(2-t)^{q/2}}\biggr) \\ &=\frac{1-t-\nu q}{\sqrt{z(t)}}\biggl(\frac{t}{2-t}\biggr)^{q\nu/2} \frac{|\lambda_\nu(q)|^\nu}{2^{\nu}\Gamma(\nu+1)}(1+\cdots) \\ &\qquad+ \frac{1}{\sqrt{z(t)}}\biggl(\frac{t}{2-t}\biggr)^{q\nu/2} \frac{q|\lambda_\nu(q)|^\nu}{2^{\nu-1}\Gamma(\nu)}(1+\cdots) \\ &=\frac{1-t+\nu q}{\sqrt{z(t)}}\biggl(\frac{t}{2-t}\biggr)^{q\nu/2} \frac{1}{\Gamma(\nu+1)}\, \frac{|\lambda_\nu(q)|^\nu}{2^\nu} \\ &\qquad +O\biggl(|\lambda_\nu(q)|^{2+\nu} \biggl(\frac{t}{2-t}\biggr)^{q(2+\nu)/2}\biggr),\qquad t\to 0^+. \end{aligned}
\end{equation*}
\notag
$$
Consequently, $\varphi'(t)\in L^2(0,1)$ if $\nu\neq 0$, and $\varphi'(t)\notin L^2(0,1)$ if $\nu=0$.
Next, integrating by parts, we obtain
$$
\begin{equation*}
\begin{aligned} \, 0&\leqslant P:=\int_0^\rho\biggl(y'(t)-y(t)\frac{\varphi'(t)}{\varphi(t)}\biggr)^2\, dt \\ &=\int_0^\rho (y'(t))^2\, dt -\int_0^\rho\frac{\varphi'(t)}{\varphi(t)}\, d(y(t))^2+\int_0^\rho (y(t))^2\frac{(\varphi'(t))^2}{(\varphi(t))^2}\, dt \\ &=\int_0^\rho (y'(t))^2\, dt- \lim_{t\to \rho} \biggl((y(t))^2 \frac{\varphi'(t)}{\varphi(t)}\biggr) +\lim_{t\to 0}\biggl((y(t))^2\frac{\varphi'(t)}{\varphi(t)}\biggr) + \int_0^\rho (y(t))^2\frac{\varphi''(t)}{\varphi(t)} \, dt. \end{aligned}
\end{equation*}
\notag
$$
For any absolutely continuous function $y\colon [0,1]\to \mathbb{R}$ such that $y(0)=0$ and the integral $\int_0^1(y'(\tau))^2\, d\tau$ converges, we have
$$
\begin{equation*}
(y(t))^2\leqslant \left(\int_0^t |y'(\tau)| \, d\tau\right)^2\leqslant t\int_0^t |y'(\tau)|^2 \, d\tau.
\end{equation*}
\notag
$$
Hence,
$$
\begin{equation*}
\lim_{t\to 0}\biggl((y(t))^2\frac{\varphi'(t)}{\varphi(t)}\biggr)=0.
\end{equation*}
\notag
$$
Let us further assume that $C_\nu(q)$ is the first positive root of the equation
$$
\begin{equation*}
1-\rho+ qz\frac{J'_\nu(z)}{J_\nu(z)}=0, \qquad z\in (0,j_\nu),
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\lambda_\nu(q)=C_\nu(q)\frac{(2-\rho)^{q/2}}{\rho^{q/2}}.
\end{equation*}
\notag
$$
Consequently, by (2.1) we get
$$
\begin{equation*}
\lim_{t\to \rho} \biggl((y(t))^2\frac{\varphi'(t)}{\varphi(t)}\biggr) =0.
\end{equation*}
\notag
$$
So, we have
$$
\begin{equation*}
(1-\nu^2q^2)\int_0^\rho \frac{(y(t))^2}{(z(t))^2}\, dt+q^2(C_\nu(q))^2\frac{(2-\rho)^q}{\rho^q}\int_0^\rho \frac{(y(t))^2}{(2-t)^{2+q}t^{2-q}}\, dt\leqslant \int_0^\rho (y'(t))^2\, dt.
\end{equation*}
\notag
$$
The equality $P=0$ is possible if and only if Consequently, the equality $P=0$ is attained on the function $y_0(t)=C \varphi(t)$, whose derivative, according to the above, lies in $L_2(0,1)$ only for $\nu>0 $. Theorem 1 is proved. The following result is a consequence of this theorem with $\nu> 0$ and $\rho=1$. Corollary 1. For each absolutely continuous function $y\colon [0,1]\to \mathbb{R}$ such that $y(0)=0$ and $y'\in L_2([0,1])$, where $j'_{\nu}$ is first positive root of the derivative $J'_{\nu}$ of the Bessel function $J_{\nu}$. The constants in the inequality of Theorem 1 are also sharp in the case of $\nu=0$. Namely, following result holds. Proposition 1. If $\nu=0$ and $\rho\in (0,1]$, then, for any $\varepsilon>0$, there exist functions $f_\varepsilon$ and $g_\varepsilon$ satisfying the conditions of Theorem 1, and such that Proof. If $\rho=1$, then $C_0(q)=0$, and we have the well known inequality (1.5) from [18] with sharp unattainable constant.
Now let $\rho\in(0,1)$. Consider $f_\varepsilon (t)= (t(2-t))^{(1+\varepsilon)/2}$. It is clear that
$$
\begin{equation*}
\begin{aligned} \, &\int_0^\rho(f'_\varepsilon (t))^2\, dt =(1+\varepsilon)^2\int_0^\rho(1-t)^2((2-t)t)^{\varepsilon-1}\, dt \\ &\qquad <(1+\varepsilon)^2 \int_0^\rho((2-t)t)^{\varepsilon-1}\, dt+q^2(C_0(q))^2\frac{(2-\rho)^q}{\rho^q}\int_0^\rho \frac{(f_\varepsilon(t))^2}{(2-t)^{2+q}t^{2-q}}\, dt \\ &\qquad =(1+\varepsilon)^2 \int_0^\rho\frac{(f_\varepsilon(t))^2}{(z(t))^2}\, dt+q^2(C_0(q))^2\frac{(2-\rho)^q}{\rho^q}\int_0^\rho \frac{(f_\varepsilon(t))^2}{(2-t)^{2+q}t^{2-q}}\, dt. \end{aligned}
\end{equation*}
\notag
$$
Therefore, the first constant is sharp.
Let $\alpha=\alpha(\varepsilon)\in (0, q)$, and let
$$
\begin{equation*}
g_\varepsilon(t)= z(t)^{\alpha/2}\varphi(t) =(z(t))^{(\alpha+1)/2}J_0\biggl(\lambda_\nu(q) \frac{t^{q/2}}{(2-t)^{q/2}}\biggr).
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{aligned} \, P_\varepsilon &:=\int_0^\rho \frac{(g_\varepsilon(t))^2}{(z(t))^2}\, dt+\left(q^2C^2\frac{(2-\rho)^q}{\rho^q}+\varepsilon\right)\int_0^\rho \frac{(g_\varepsilon(t))^2}{(2-t)^{2+q}t^{2-q}}\, dt-\int_0^\rho (g'_\varepsilon(t))^2\, dt \\ &=\varepsilon\int_0^\rho \frac{(g_\varepsilon(t))^2}{(2-t)^{2+q}t^{2-q}}\, dt-\int_0^\rho \biggl(g'_\varepsilon-g_\varepsilon(t)\frac{\varphi'(t)}{\varphi(t)}\biggr)^2\, dt. \end{aligned}
\end{equation*}
\notag
$$
It can be shown that
$$
\begin{equation*}
g'_\varepsilon-g_\varepsilon(t)\frac{\varphi'(t)}{\varphi(t)}=\alpha (1-t) (z(t))^{(\alpha-1)/2} J_0\biggl(\lambda_\nu(q) \frac{t^{q/2}}{(2-t)^{q/2}}\biggr)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, P_\varepsilon &=\varepsilon\int_0^\rho \frac{(z(t))^{\alpha+1}}{(2-t)^{2+q}t^{2-q}} \biggl(J_0\biggl( \frac{\lambda_\nu(q) t^{q/2}}{(2-t)^{q/2}}\biggr)\biggr)^2 \, dt \\ &\qquad- \alpha^2\int_0^\rho(z(t))^{\alpha-1} \biggl(J_0\biggl( \frac{\lambda_\nu(q) t^{q/2}}{(2-t)^{q/2}}\biggr)\biggr)^2(1-t)\, dt \\ &\geqslant\varepsilon\int_0^\rho \frac{(z(t))^{q+1}}{(2-t)^{2+q}t^{2-q}} \biggl(J_0\biggl( \frac{\lambda_\nu(q)t^{q/2}}{(2-t)^{q/2}}\biggr)\biggr)^2\, dt \\ &\qquad -\alpha\max_{t\in[0,\rho]}\biggl\{(2-t)^{\alpha-1}\biggl(J_0\biggl( \frac{\lambda_\nu(q) t^{q/2}}{(2-t)^{q/2}}\biggr)\biggr)^2\biggr\}. \end{aligned}
\end{equation*}
\notag
$$
Consequently, $P_\varepsilon>0$ for sufficiently small $\alpha$. Thus, we have shown that both constants are unimprovable also in the case $\nu=0$. The proposition is proved. We also have the following result. Theorem 2. Let $q\in (0,\infty)$ and $\rho\in (0,1]$. Then, for any absolutely continuous function $y\colon [0,1]\to \mathbb{R}$ such that $y(0)=0$ and $y'\in L_2([0,\rho])$, Proof. To prove this theorem we will use the function
$$
\begin{equation*}
\psi(t)=\sqrt{t}\, J_0\biggl(\lambda_0(q) \frac{t^{q/2}}{(2-t)^{q/2}}\biggr), \qquad t\in [0,1].
\end{equation*}
\notag
$$
A direct calculation shows that
$$
\begin{equation*}
\frac{\psi'(t)}{\psi(t)}=\frac{1}{2t}-\frac{q\lambda_0(q) t^{q/2-1}}{(2-t)^{q/2+1}}\, \frac{J_1(\lambda_\nu(q) t^{q/2}/(2-t)^{q/2})}{J_0(\lambda_\nu(q) t^{q/2}/(2-t)^{q/2})}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
-\frac{\psi''(t)}{\psi(t)}= \frac{1}{4t^2} +\frac{q^2(\lambda_0(q))^2}{t^{2-q}(2-t)^{2+q}} +\frac{q\lambda_0(q)}{t^{1-q/2}(2-t)^{2+q/2}}\frac{J_1(\lambda_0(q) t^{q/2}/(2-t)^{q/2})}{J_0(\lambda_0(q) t^{q/2}/(2-t)^{q/2})}.
\end{equation*}
\notag
$$
As the proof of Theorem 1, we have
$$
\begin{equation*}
\int_0^\rho (y'(t))^2\, dt=\lim_{t\to \rho} \biggl( (y(t))^2 \frac{\psi'(t)}{\psi(t)} \biggr) -\lim_{t\to 0} \biggl((y(t))^2\frac{\psi'(t)}{\psi(t)}\biggr) -\int_0^\rho (y(t))^2\frac{\psi''(t)}{\psi(t)} \, dt,
\end{equation*}
\notag
$$
where the constant $\lambda_0(q)$ is the root of the equation
$$
\begin{equation*}
\frac{\psi'(\rho)}{\psi(\rho)}= \frac{1}{2\rho}-\frac{q\lambda_0(q) \rho^{q/2-1}}{(2-\rho)^{q/2+1}}\, \frac{J_1(\lambda_0(q) \rho^{q/2}/(2-\rho)^{q/2})}{J_0(\lambda_0(q) \rho^{q/2}/(2-\rho)^{q/2})} =0, \qquad \lambda_0(q)\in (0,j_0).
\end{equation*}
\notag
$$
Next, arguing as in the proof of Theorem 1, we get that
In view of Lemma 1, we have
$$
\begin{equation*}
\begin{aligned} \, -& \frac{\psi''(t)}{\psi(t)} \geqslant\frac{1}{4t^2} +\frac{q^2(\lambda_0(q))^2}{t^{2-q}(2-t)^{2+q}} \\ &\ \, +q(\lambda_0(q))^2\frac{t^{-1+q}}{(2-t)^{2+q}}\min_{t\in[0,1]}\frac{J_1(\lambda_0(q) t^{q/2}/(2-t)^{q/2})}{(\lambda_0(q)t^{q/2}/(2-t)^{q/2}) J_0(\lambda_0(q)t^{q/2}/(2-t)^{q/2})} \\ &=\frac{1}{4t^2}+\frac{q^2(\lambda_0(q))^2}{t^{2-q}(2-t)^{2+q}} +\frac{q}{2}\, \frac{(\lambda_0(q))^2}{t^{1-q}(2-t)^{2+q}}, \end{aligned}
\end{equation*}
\notag
$$
proving the theorem. Theorem 3. Let $\nu\in(0,1]$, $\rho(t)=\min\{t,2b-t\}$ and $\mu(t)=2b-\rho(t)$. Then, for any function $y\in C_{\mathrm{c}}^1(0,2b)$, Proof. It is easy to see that
$$
\begin{equation*}
\frac{1}{(z(t))^2}=\frac{1}{t^2(2-t)^2}=\frac{1}{4}\biggl(\frac{1}{t^2} +\frac{2}{t(2-t)}+\frac{1}{(2-t)^2}\biggr)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\frac{4}{(2-t)^{2+q}t^{2-q}}= \frac{(2-t+t)^2}{(2-t)^{2+q}t^{2-q}} =\frac{1}{(2-t)^{q}t^{2-q}}+\frac{2}{(2-t)^{1+q}t^{1-q}}+\frac{t^{q}}{(2-t)^{2+q}}.
\end{equation*}
\notag
$$
Consequently, using Corollary 1 with $q=1$, we have
$$
\begin{equation*}
\begin{aligned} \, \int_0^1 (y'(t))^2\, dt &\geqslant \frac{1-\nu^2}{4}\int_0^1 \frac{(y(t))^2}{t^2}\, dt +\frac{2-2\nu^2+(j'_{\nu})^2}{4}\int_0^1\frac{(y(t))^2}{(2-t)t}\, dt \\ &\qquad+ \frac{1-\nu^2+2(j'_{\nu})^2}{4}\int_0^1\frac{(y(t))^2}{(2-t)^2}\, dt+\frac{(j'_{\nu})^2}{4}\int_0^1\frac{(y(t))^2t}{(2-t)^3}\, dt, \end{aligned}
\end{equation*}
\notag
$$
and now an application of Theorem 2 with $q=1$ and $\rho=1$ gives
$$
\begin{equation*}
\begin{aligned} \, \int_0^1 (y'(t))^2\, dt &\geqslant \frac{1}{4}\int_0^1 \frac{(y(t))^2}{t^2} \, dt+ \frac{(C_0(1))^2}{4}\int_0^1\frac{(y(t))^2}{(2-t)t}\, dt \\ &\qquad+ \frac{(C_0(1))^2}{2}\int_0^1\frac{(y(t))^2}{(2-t)^2}\, dt +\frac{(C_0(1))^2}{4}\int_0^1\frac{(y(t))^2}{(2-t)^{3}}(2+t)\, dt. \end{aligned}
\end{equation*}
\notag
$$
By changing the variable $t= \tau/b$ in the first and second inequalities, it is easy to show that
$$
\begin{equation*}
\begin{aligned} \, \int_0^b (y'(\tau))^2\, d\tau &\geqslant \frac{1-\nu^2}{4}\int_0^b \frac{(y(\tau))^2}{\tau^2}\, d\tau+\frac{2-2\nu^2+(j'_{\nu})^2}{4}\int_0^b\frac{(y(\tau))^2}{(2b-\tau)\tau}\, d\tau \\ &\qquad+ \frac{1-\nu^2+2(j'_{\nu})^2}{4}\int_0^b\frac{(y(\tau))^2}{(2b-\tau)^2}\, d\tau+\frac{(j'_{\nu})^2}{4}\int_0^b\frac{(y(\tau))^2\tau}{(2b-\tau)^3}\, d\tau \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \int_0^b (y'(\tau))^2\, d\tau &\geqslant \frac{1}{4}\int_0^b \frac{ (y(\tau))^2}{\tau^2} \, d\tau+ \frac{(C_0(1))^2}{4}\int_0^b\frac{(y(\tau))^2}{(2b-\tau)\tau}\, d\tau \\ &\quad +\frac{(C_0(1))^2}{2}\int_0^b\frac{(y(\tau))^2}{(2b-\tau)^2}\, d\tau +\frac{(C_0(1))^2}{4}\int_0^b\frac{(y(\tau))^2}{(2b-\tau)^{3}}(2b+\tau)\, d\tau. \end{aligned}
\end{equation*}
\notag
$$
Combining the last two inequalities with the following corresponding inequalities on the interval $[b,2b]$
$$
\begin{equation*}
\begin{aligned} \, \int_b^{2b} (y'(\tau))^2\, d\tau &\geqslant \frac{1-\nu^2}{4}\int_b^{2b} \frac{(y(\tau))^2}{(2b-\tau)^2}\, d\tau +\frac{2-2\nu^2+(j'_{\nu})^2}{4}\int_b^{2b}\frac{(y(\tau))^2}{(2b-\tau)\tau}\, d\tau \\ &\quad+ \frac{1-\nu^2+2(j'_{\nu})^2}{4}\int_0^b\frac{(y(\tau))^2}{\tau^2}\, d\tau+\frac{(j'_{\nu})^2}{4}\int_b^{2b}\frac{(y(\tau))^2(2b-\tau)}{\tau^3}\, dt \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \int_b^{2b} (y'(\tau))^2\, d\tau &\geqslant \frac{1}{4}\int_b^{2b} \frac{(y(\tau))^2}{(2b-\tau)^2} \, d\tau +\frac{(C_0(1))^2}{4}\int_b^{2b}\frac{(y(\tau))^2}{(2b-\tau)\tau}\, d\tau \\ &\quad+ \frac{(C_0(1))^2}{2}\int_b^{2b}\frac{(y(\tau))^2}{\tau^2}\, d\tau +\frac{(C_0(1))^2}{4}\int_{b}^{2b}\frac{(y(\tau))^2}{\tau^{3}}(4b-\tau)\, d\tau, \end{aligned}
\end{equation*}
\notag
$$
for any function $y\in C^1(b,2b)$ such that $y(2b)=0$, we arrive at the conclusion of the theorem. 2.3. Inequalities required for sufficient univalence conditionsNext, we consider inequalities for a weight function different from those considered in § 2.2. The following results will be required in the proof of sufficient univalence conditions. Lemma 2. Let $q\in [1,2]$ and $\rho\in (0,1]$. Then, for any absolutely continuous function $y$ such that $y(0)=0$, where Proof. Using the inequality $|y(x)|\leqslant \int_0^x|y'(t)|\, dt$ and changing the order of integration in iterated integrals, we have
$$
\begin{equation*}
\int_0^\rho \frac{|y(x)|}{x^{2-q}(2-x)^{2-q}}\, dx\leqslant \int_0^\rho |y'(t)|\int_t^\rho\frac{dx}{x^{2-q}(2-x)^{2-q}}\, dt.
\end{equation*}
\notag
$$
Applying the last estimate to the function $y(x)=|g(x)|^2$, we get that
$$
\begin{equation*}
\int_0^\rho \frac{|g(x)|^2}{x^{2-q}(2-x)^{2-q}}\, dx\leqslant 2\int_0^\rho |g(x)|\, |g'(x)|\int_x^\rho\frac{dt}{t^{2-q}(2-t)^{2-q}}\, dx.
\end{equation*}
\notag
$$
Now, from the well-known Opial inequality (see [28], [29])
$$
\begin{equation*}
\int_a^b s(x)|y(x)|\, |y'(x)|\, dx \leqslant \kappa\int_a^b |y'(x)|^2\, dx,
\end{equation*}
\notag
$$
which holds for any absolutely continuous function and non-negative measurable on $(0,\rho)$ weight function $s$ satisfying
$$
\begin{equation*}
\kappa=\frac{1}{\sqrt{2}}\, \biggl(\int_a^b (s(t))^2(t-a)\, dt\biggr)^{1/2}<\infty,
\end{equation*}
\notag
$$
we have Taking as above, we completes the proof of the lemma. We also need the following result. Theorem 4. Let $q\in [1,2]$ and $\rho\in (0,1)$. Then, for any absolutely continuous function $y$ such that $y(-\rho)=y(\rho)=0$ and $y\not\equiv 0$, where and $B_t(x,y)=\int_0^t\tau^{x-1}(1-\tau)^{y-1}\, d\tau$ is the incomplete Beta function. Proof. Since $\kappa'(q,\rho)$ is monotonic increasing with respect to $\rho$, we have, by changing of variables,
$$
\begin{equation*}
\kappa'(q,\rho)\leqslant \kappa'(q)=\frac{1}{2}\biggl(\int_0^1 \biggl(\int_0^t \tau^{-1/2}(1-\tau)^{q-2}\, d\tau\biggr)^2 \frac{1-\sqrt{t}}{\sqrt{t}}\, dt \biggr)^{1/2}.
\end{equation*}
\notag
$$
Indeed,
$$
\begin{equation*}
\begin{aligned} \, \kappa'(q,\rho) &=\biggl(2\int_0^\rho \biggl(\int_t^\rho\frac{dx}{x^{2-q}(2-x)^{2-q}}\biggr)^2t \, dt\biggr)^{1/2} \\ &\leqslant\biggl(2\int_0^1 \biggl(\int_t^1\frac{dx}{x^{2-q}(2-x)^{2-q}}\biggr)^2t \, dt\biggr)^{1/2}. \end{aligned}
\end{equation*}
\notag
$$
In the inner integral, we change of variable $x= 1-\tau$, and then put $\tau=\sqrt{y}$. As a result,
$$
\begin{equation*}
\begin{aligned} \, \kappa'(q,\rho) &\leqslant \biggl(2\int_0^1\biggl(\int_0^{1-t}\frac{d\tau}{(1-\tau^2)^{2-q}}\biggr)^2t \, dt\biggr)^{1/2} \\ &= \biggl(\frac{1}{2}\int_0^1\biggl(\int_0^{(1-t)^2}\frac{dy}{\sqrt{y}\, (1-y)^{2-q}}\biggr)^2t \, dt\biggr)^{1/2}. \end{aligned}
\end{equation*}
\notag
$$
By changing the variable $z=(1-t)^2$ in the outer integral, we get that
$$
\begin{equation*}
\kappa'(q,\rho) \leqslant \frac{1}{2}\biggl(\int_0^1\biggl(\int_0^{z}\frac{dy}{\sqrt{y}\, (1-y)^{2-q}}\biggr)^2\frac{1-\sqrt{z}}{\sqrt{z}} \, dz\biggr)^{1/2}.
\end{equation*}
\notag
$$
Let us apply the inequality proved in Lemma 2 to the function $g(x)=f(x-\rho)$ and $g(x)= f(\rho-x)$. Correspondingly, we have It is obvious that if $\rho\in (0,1)$ and $x\in [-\rho, 0]$, then and if $\rho\in (0,1)$ and $x\in [0,\rho]$, then Consequently, Now the theorem follows by summing the last two inequalities. Now let’s compare $\kappa'(q)$ with the constants in the already known inequalities. For example, F. G. Avkhadiev showed that in the inequality of Theorem 4, instead of $\kappa'(q)$, one can take the quantity The advantage of the constant $C(q)$ is that the cases $C(1)$ and $C(2)$ are unimprovable (see [18], [21]). However, there are two numbers $\widehat{q}_0$ and $\widehat{q}_1$ such that $\kappa'(q)\leqslant C(q)$ for any $q\in [\widehat{q}_0,\widehat{q}_1]$. Indeed, for $q=3/2$ we have
$$
\begin{equation*}
\kappa'\biggl(\frac32\biggr)= \frac{1}{2}\sqrt{\frac{3\pi^2}{2}-14}\approx 0.448444< C\biggl(\frac32\biggr)=\frac{\sqrt{2}}{\pi}\approx 0.450158,
\end{equation*}
\notag
$$
and, moreover,
$$
\begin{equation*}
\begin{gathered} \, \kappa'(1.3)\approx 0.469203<C(1.3)\approx 0.469469, \\ \kappa'(1.4) \approx 0.458389<C(1.4)\approx 0.459712, \\ \kappa'(1.6) \approx 0.439246<C(1.6)\approx 0.440803, \\ \kappa'(1.7) \approx 0.430701<C(1.7)\approx 0.431641. \end{gathered}
\end{equation*}
\notag
$$
Remark 1. Since $\kappa'(q)$ is continuous with respect to $q$, there exist some neighborhoods $[\widehat{q}_0,\widehat{q}_1]$ of these points such that Numerical calculations show that $\kappa'(q)<C(q)$ for any $q\in[1.2823044502226741$, $1.7950834115169314]$. At the end-points of this interval, $\kappa'(q)-C(q)$ is close to zero. At first sight, it may seem that the above strengthening of the constant $C(q)$ is immaterial. A comparison of our inequality and the one-dimensional version of the Avkhadiev–Wirths inequality (1.4) for $\nu=1/q$ (see [15] for details) shows that, for any $ q\in [0,2]$, the constant $k'(q)$ cannot be smaller than For instance, $\lambda(3/2)\approx 0.360891$. It should be also pointed out the estimate $\lambda(q)\leqslant \kappa'(q)$ is rather “rough”.§ 3. Spatial inequalitiesIn this section, we will obtain spatial Hardy-type inequalities in arbitrary domains, in Davies-regular domains, in domains satisfying the cone condition, in domains that are $\lambda$-close to convex domains, and in convex domains. Let us first introduce the main notation used in this section. Suppose $\Omega$ is an open connected proper subset of Euclidean space $\mathbb{R}^n$, $n\geqslant 2$, $|\mathbb{S}^{n-1}|$ is the surface area of the $(n-1)$-dimensional unit sphere $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$, $d\mathbb{S}^{n-1}(\nu)$ is the element of surface measure on the unit sphere, and $d\omega(\nu)= d\mathbb{S}^{n-1}(\nu)/|\mathbb{S}^{n-1}|$ is the normalized measure on the unit sphere. For any point $x\in\Omega$, $\nu \in \mathbb{S}^{n-1}$, let be the distance between a point $x$ and the nearest point from the boundary $\partial\Omega$ in the direction of $\nu$, be the distance from a point $x$ to the boundary of the domain $\Omega$. We also set
$$
\begin{equation*}
\begin{gathered} \, \rho_\nu(x):= \min\{\tau_\nu(x),\tau_{-\nu}(x)\}, \qquad \mu_\nu(x):= \max\{\tau_\nu(x),\tau_{-\nu}(x)\}, \\ D_\nu(x):= \tau_\nu(x)+\tau_{-\nu}(x), \qquad D(\Omega)=\sup_{x\in \Omega,\, \nu\in \mathbb{S}^{n-1}} D_\nu(x), \end{gathered}
\end{equation*}
\notag
$$
and define the mean distance
$$
\begin{equation*}
\delta^{-2}_M=n\int_{\mathbb{S}^{n-1}}(\rho_\nu(x))^{-2}\, d\omega(\nu)= \frac{n}{|\mathbb{S}^{n-1}|}\int_{\mathbb{S}^{n-1}}(\rho_\nu(x))^{-2}d\mathbb{S}^{n-1}.
\end{equation*}
\notag
$$
The mean distance is also referred to as the Davis distance. By $|\Omega|$, we will denote the volume of $\Omega$, and by $\Omega_x$ we will denote elements of the set $\Omega$ which are “visible” from the point $x$, that is,
$$
\begin{equation*}
\Omega_x=\{y\in \Omega\colon x+t(y-x)\in\Omega \ \forall\, t\in [0,1]\}.
\end{equation*}
\notag
$$
Recall that
$$
\begin{equation*}
K(n)=n\biggl[\frac{|\mathbb{S}^{n-1}|}{n}\biggr]^{2/n}.
\end{equation*}
\notag
$$
Below we will use the notation and some arguments from [9] (see also [1], [10], [20], [30]). 3.1. Inequalities in arbitrary domains in terms of the mean distance Theorem 5. Let $\Omega$ be an arbitrary domain of Euclidean space $\mathbb{R}^n$ and $\nu \in (0,1]$. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, Proof. Since both inequalities have the same form and differ only by the constants, we will the constants will not be specified in our analysis.
We denote by $\partial_\nu$ the partial derivative in the direction $\nu$. Using one-dimensional inequalities from Theorem 3, we obtain
$$
\begin{equation*}
\begin{aligned} \, \int_\Omega |\partial_\nu g (x)|^2\, dx &\geqslant C_1\int_\Omega\frac{|g(x)|^2}{(\rho_\nu(x))^2}\, dx+ C_2\int_\Omega\frac{|g(x)|^2}{\rho_\nu(x)\mu_\nu(x)}\, dx \\ &\qquad +C_3\int_\Omega\frac{|g(x)|^2}{(\mu_\nu(x))^2}\, dx+C_4\int_\Omega\frac{|g(x)|^2\rho_\nu(x)}{(\mu_\nu(x))^3}\, dx. \end{aligned}
\end{equation*}
\notag
$$
Integrating this inequality with respect to the normalized measure $d\omega(\nu)$ and using the definition of the directional derivative
$$
\begin{equation*}
|\partial_\nu g|=|\nu\cdot\nabla g|=|\nabla g|\, |{\cos(\nu,\nabla g)}|,
\end{equation*}
\notag
$$
we get
$$
\begin{equation*}
\begin{aligned} \, &\int_\Omega |\nabla g (x)|^2 \int_{\mathbb{S}^{n-1}}|{\cos(\nu,\nabla g)}|^2\, d\omega(\nu)\, dx \\ &\geqslant C_1\int_\Omega |g(x)|^2\int_{\mathbb{S}^{n-1}} \frac{1}{(\rho_\nu(x))^2}\, d\omega(\nu)\, dx+ C_2\int_\Omega |g(x)|^2\int_{\mathbb{S}^{n-1}}\frac{1}{\rho_\nu(x)\mu_\nu(x)}\, d\omega(\nu)\, dx \\ &\ \ +C_3\int_\Omega |g(x)|^2\int_{\mathbb{S}^{n-1}}\frac{1}{(\mu_\nu(x))^2}\, d\omega(\nu)\, dx+ C_4\int_\Omega |g(x)|^2\int_{\mathbb{S}^{n-1}}\frac{\rho_\nu(x)}{(D_\nu(x))^3}\, d\omega(\nu) \, dx. \end{aligned}
\end{equation*}
\notag
$$
Balinsky, Evans, and Lewis (see [1], § 3.6) proved that
$$
\begin{equation*}
\begin{aligned} \, \int_{\mathbb{S}^{n-1}}\frac{1}{\rho_\nu(x)\mu_\nu(x)}\, d\omega(\nu) &\geqslant \biggl[\frac{n}{|\mathbb{S}^{n-1}|}|\Omega_x|\biggr]^{-2/n}, \\ \int_{\mathbb{S}^{n-1}}\frac{1}{(\mu_\nu(x))^2}\, d\omega(\nu) &\geqslant \frac{1}{2}\biggl[\frac{n}{|\mathbb{S}^{n-1}|}|\Omega_x|\biggr]^{-2/n}, \end{aligned}
\end{equation*}
\notag
$$
and Tidblom [20] showed that
$$
\begin{equation*}
\int_{\mathbb{S}^{n-1}}\cos(\nu,\nabla g)|^2\, d\omega(\nu)=\frac{1}{n}, \qquad \int_{\mathbb{S}^{n-1}}\biggl(\frac{2}{D_\nu(x)}\biggr)^3\, d\omega(\nu) \geqslant \biggl[\frac{n}{|\mathbb{S}^{n-1}|}|\Omega_x|\biggr]^{-3/n}.
\end{equation*}
\notag
$$
Thus
$$
\begin{equation*}
\begin{aligned} \, \int_\Omega |\nabla g (x)|^2 \, dx &\geqslant C_1\int_\Omega \frac{|g(x)|^2}{(\delta_M(x))^2}\, dx+\biggl(C_2+\frac{C_3}{2}\biggr)K(n)\int_\Omega \frac{|g(x)|^2}{ |\Omega_x|^{2/n}}\, dx \\ &\qquad+ \frac{C_4 K(n)n}{8|\mathbb{S}^{n-1}|}\int_\Omega |g(x)|^2\frac{\delta(x)}{|\Omega_x|^{3/n}} \, dx, \end{aligned}
\end{equation*}
\notag
$$
proving the theorem. Arguing as in the proof of Theorem 5 and using the obvious inequalities
$$
\begin{equation*}
\begin{gathered} \, \int_{\mathbb{S}^{n-1}}\frac{1}{\rho_\nu(x)\mu_\nu(x)}\, d\omega(\nu)\geqslant \int_{\mathbb{S}^{n-1}}\frac{4}{(\rho_\nu(x)+\mu_\nu(x))^2}\, d\omega(\nu)\geqslant \frac{4}{(D(\Omega))^2}, \\ \int_{\mathbb{S}^{n-1}}\frac{1}{(\mu_\nu(x))^2}\, d\omega(\nu)\geqslant \frac{1}{(D(\Omega))^2}, \qquad \int_{\mathbb{S}^{n-1}}\frac{1}{(\mu_\nu(x))^3}\, d\omega(\nu)\geqslant \frac{1}{(D(\Omega))^3}, \end{gathered}
\end{equation*}
\notag
$$
we find that
$$
\begin{equation*}
\begin{aligned} \, \int_\Omega |\nabla g (x)|^2 \, dx &\geqslant C_1\int_\Omega \frac{|g(x)|^2}{(\delta_M(x))^2}\, dx+\frac{n(4C_2+C_3)}{(D(\Omega))^2}\int_\Omega |g(x)|^2\, dx \\ &\qquad+ \frac{nC_4}{(D(\Omega))^3}\int_\Omega |g(x)|^2\delta(x) \, dx. \end{aligned}
\end{equation*}
\notag
$$
This proves the following result. Theorem 6. Let $\Omega$ be an arbitrary bounded domain of the Euclidean space $\mathbb{R}^n$ and let $\nu \in (0,1]$. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, 3.2. The case of Davies-regular domainsFollowing Davies [31] (see also [32], [33]), we say that $\Omega\subset\mathbb{R}^n$ is (Davies-)regular if there exists a finite constant $c$ such that
$$
\begin{equation*}
\delta(x) \leqslant m(x) \leqslant c\delta(x) \quad\, \forall{x\in\Omega},
\end{equation*}
\notag
$$
where $m(x)$ is defined by
$$
\begin{equation*}
(m(x))^{-2} =\frac{1}{|\mathbb{S}^{n-1}|}\int_{\mathbb{S}^{n-1}} \frac{d\mathbb{S}^{n-1}(\nu)}{(\tau_\nu(x))^2}.
\end{equation*}
\notag
$$
In [32], sufficient conditions for Davies-regularity were weakened and generalized to the case of multivariate domains. Using the definition of the functions $\rho_\nu$ and $\tau_\nu$, we obtain (see [32] for more details)
$$
\begin{equation*}
\begin{aligned} \, (\delta_M(x))^{-2} &= \frac{n}{|\mathbb{S}^{n-1}|}\int_{\mathbb{S}^{n-1}}\frac{d\mathbb{S}^{n-1}(\nu)}{(\rho_\nu(x))^2} =\frac{n}{|\mathbb{S}^{n-1}|}\int_{\mathbb{S}^{n-1}}\frac{d\mathbb{S}^{n-1}(\nu)} {(\min\{\tau_\nu(x),\tau_{-\nu}(x)\})^2} \\ &=\frac{n}{|\mathbb{S}^{n-1}|}\biggl(\int_{\mathbb{S}^{n-1}}\frac{d\mathbb{S}^{n-1}(\nu)} {(\tau_\nu(x))^2}+\int_{\mathbb{S}^{n-1}}\frac{d\mathbb{S}^{n-1}(\nu)}{(\tau_{-\nu}(x))^2}\biggr) \\ &=\frac{2n}{|\mathbb{S}^{n-1}|}\int_{\mathbb{S}^{n-1}}\frac{d\mathbb{S}^{n-1}(\nu)} {(\tau_\nu(x))^2}=\frac{2n}{(m(x))^2}\geqslant \frac{2n}{c^2(\delta(x))^2}. \end{aligned}
\end{equation*}
\notag
$$
Now the following result follows from Theorem 5. Theorem 7. Let $\Omega\subset\mathbb{R}^n$ be a regular domain and $\nu \in (0,1]$. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, 3.3. Domains satisfying the cone conditionA domain $\Omega\subset\mathbb{R}^n$ is said to satisfy the $\theta$-cone condition if every $x\in\partial\Omega$ is the vertex of a circular cone $C_x$ of angle $2\theta$ lying entirely in $\mathbb{R}^n\setminus{\Omega}$. The following result holds. Theorem 8. Let $\nu \in (0,1]$ and let $\Omega\subset\mathbb{R}^n$ satisfy the $\theta$-cone condition. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, Proof. If $\Omega$ satisfies the $\theta$-cone condition, then (see [1], § 3.3, for more details)
$$
\begin{equation*}
\delta_M(x)\leqslant 2\delta(x)\biggl[\frac{1}{ns((\sin \theta)/2)}\biggr]^{1/2}.
\end{equation*}
\notag
$$
It is clear that $|\Omega_x|\leqslant |\Omega|$ for any $x\in\Omega$. Now the required result follows by applying Theorem 5 and the above estimate for the mean distance.
3.4. Domains $\lambda$-close to convex domainsFollowing [34], we say a domain $\Omega$ is $\lambda$-close to a convex domain if $\Omega\neq\mathbb{R}^n$ and if, for any boundary point $y\in (\partial\Omega)\setminus\{\infty\}$, there exists a point $a_y\in\mathbb{R}^n\setminus\overline{\Omega}$ such that $|y-a_y|=\lambda$ and
$$
\begin{equation*}
B_y=\{x\in \mathbb{R}^n\colon |x-a_y|< \lambda\}\subset \mathbb{R}^n\setminus\overline{\Omega},
\end{equation*}
\notag
$$
that is, the point $y$ can be touched by a ball lying outside the domain $\Omega$. The following result holds. Lemma 3. If $\Omega$ is $\lambda$-close to a convex domain, then, for every $x \in \Omega$, where $\delta_0(\Omega)=\sup_{x\in\Omega}\delta(x)$ is the inner radius of the domain, and Proof. Let $x\in\Omega$ and $ y\in\partial\Omega$ be such that $\delta(x)=|x-y|$. Then there is a ball $B_y$ of radius $\lambda$ with centre at $a_y$ which touches the domain $\Omega$ at $y$, and $B_y=\{x\in \mathbb{R}^n\colon |x-a_y|< \lambda\}\subset \mathbb{R}^n\setminus\overline{\Omega}$ (see Fig. 1).
Consider the solid angle $\Lambda_x=\Lambda_x(\varphi_1,\dots, \varphi_{n-1})$ formed by a ball of radius $\lambda$ centred at a distance $\delta(x)+\lambda$ from the point $x$, where $\varphi_1\in [0,2\pi)$, $\varphi_2, \dots, \varphi_{n-2}\in[0,\pi]$ and $\varphi_{n-1}=[0,\arcsin (\lambda/(\lambda+\delta(x)))]$. Let $\mathbf{e}$ be a unit vector such that $\rho_\mathbf{e}(x)=\delta(x)$. We have
$$
\begin{equation*}
\begin{aligned} \, \frac{1}{(\delta_M(x))^2} &=\frac{n}{|\mathbb{S}^{n-1}|} \int_{\mathbb{S}^{n-1}}\frac{d\mathbb{S}^{n-1}(\nu)}{(\rho_\nu(x))^2}\geqslant \frac{n}{|\mathbb{S}^{n-1}|}\int_{\Lambda_x}\frac{d\mathbb{S}^{n-1}(\nu)}{(\rho_\nu(x))^2} \\ &\geqslant \frac{n}{|\mathbb{S}^{n-1}|(\delta(x))^2} \int_{\Lambda_x}(\cos(\mathbf{e},\nu))^2\, d\mathbb{S}^{n-1}(\nu) \\ &=\frac{n}{|\mathbb{S}^{n-1}|(\delta(x))^2}\int_0^{2\pi}d\varphi_{1}\int_0^{\pi}d\varphi_{2} \dotsb\int_0^{\pi}d\varphi_{n-2}\int_0^{\alpha} (\sin\varphi_{n-1})^{n-2} \\ &\qquad\times(\sin\varphi_{n-2})^{n-3}\dotsb \sin\varphi_{2} (\cos\varphi_{n-1})^2\, d\varphi_{n-1} \\ &=\frac{1}{(\delta(x))^2}\biggl(\int_0^{\alpha}(\sin\theta)^{n-2}(\cos\theta)^2 \, d\theta\biggr)\biggl(\int_0^{\pi}(\sin\theta)^{n-2}(\cos\theta)^2\, d\theta\biggr)^{-1}, \end{aligned}
\end{equation*}
\notag
$$
where $\alpha=\arcsin(\lambda/(\lambda+\delta(x))$.
Above we have used the equality (see [20]) The following results follow from Theorems 5 and 6. Theorem 9. Let $\nu \in (0,1]$ and let $\Omega\subset\mathbb{R}^n$ be $\lambda$-close to a convex domain. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, Theorem 10. Let $\nu \in (0,1]$ and let $\Omega\subset\mathbb{R}^n$ be $\lambda$-close to a convex domain. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, Proof. As mentioned in the introduction, the first eigenvalue $\lambda_1(\Omega)$ of the Laplacian under the Dirichlet boundary conditions is a sharp constant in the following Poincaré inequality
$$
\begin{equation*}
\lambda_1(\Omega)\int_\Omega|g(x)|^2\, dx\leqslant \int_\Omega |\nabla g(x)|^2\, dx \quad \forall\, g\in C_{\mathrm{c}}^1(\Omega).
\end{equation*}
\notag
$$
Now the corollary follows from inequality (3.1) and since $7(C_0(1))^2/4 \approx 2.75972$. 3.5. The case of convex domainsIn the case of convex domains, the formulas can be simplified. It is known that if $\Omega$ is a convex domain, then
$$
\begin{equation*}
|\Omega_x|=|\Omega| \quad\text{and}\quad \delta_M(x)\geqslant \delta(x).
\end{equation*}
\notag
$$
The following result is a corollary to Theorem 5. Theorem 11. Let $\nu \in (0,1]$ and let $\Omega$ be an open convex subset of $\mathbb{R}^n$ of finite volume. Then, for any function $g\in C_{\mathrm{c}}^1(\Omega)$, As an immediate consequence of this theorem for $\nu=1$ we have. Corollary 3. If $\Omega$ is a convex domain of fixed volume, then where $j'_1\approx 1.84118$ is the first positive root of the derivative $J_1'$ of the Bessel function $J_1$. Remark 2. Each multivariate inequality implies the corresponding estimate for the first eigenvalue $\lambda_1(\Omega)$. § 4. Nehari–Pokornii type sufficient univalence conditionsIn this section, we will obtain sufficient Nehari–Pokornii type univalence conditions. Sufficient conditions for analytic functions on the unit disc, on the exterior of the unit disc, and on the right half-plane will be considered. In this section, we will assume that
$$
\begin{equation*}
\kappa'(q)=\frac{1}{2}\biggl(\int_0^1 \biggl(B_{t}\biggl(\frac12,q-1\biggr)\biggr)^2 \frac{1-\sqrt{t}}{\sqrt{t}}\, dt \biggr)^{1/2},
\end{equation*}
\notag
$$
where $B_t(x,y)=\int_0^t\tau^{x-1}(1-\tau)^{y-1}\, d\tau$ is the incomplete beta function, and $\widehat{q}_0$ and $\widehat{q}_1$ are constants such that $\kappa'(q)\leqslant 2^{3q-4}\pi^{2(1-q)}$ for any $q\in[\widehat{q}_0, \widehat{q}_1] $. 4.1. Sufficient conditions on the unit discSuppose that $f(z)$ is a meromorphic function on the unit disc $\mathbb{D}=\{z\in \mathbb{C}\colon |z|<1\}$, and $S_f(z)=f'''/f'-(3/2)(f''/f')^2$ is the Schwartz derivative (or the Schwarzian function) of $f$. F. G. Avkhadiev proved the following theorem (see [22], [23])). Theorem A. A meromorphic function $f(z)$ on $\mathbb{D}$ is univalent in $\mathbb{D}$ if, for some tuple of real nonnegative parameters $n$, $a_k$ and $q_k$, $k=1,\dots,n$, where $a_1+a_2+ \dots +a_n\leqslant 1$, $0\leqslant q_1\leqslant q_2\leqslant \dots \leqslant q_n\leqslant 2$, and the Pokornii constants have the form For $n=0$ and $q=0$ or $q=2$, this result was proved by Nehari [21]; the case $n=1$ is due to Pokornii [35]. Later in [26], Yamashita proved that if $n=1$ and $q\in [0,1]$, then $A(q)=2(1+q)$, that is, the sufficient univalent condition was weakened. In [36], the author of the present paper improved Yamashita’s result and strengthened Theorem A for $q\in [0,1]$. In this section, we will obtain new results for $q\in[\widehat{q}_0,\widehat{q}_1]$, because in this case Recall that $\widehat{q}_0$ and $\widehat{q}_1$ were defined in Remark 1 as constants such that $\kappa'(q)\leqslant 2^{3q-4 }\pi^{2(1-q)}$ for any $q\in[\widehat{q}_0, \widehat{q}_1]$.The proof of Theorem A is based on a relation between the univalence of the function $f(z)$ in the domain $\mathbb{D}$ and the non-oscillation of the solutions of the differential equation (see [21]–[23] for more details) Consider the case $n=1$. The case of an arbitrary $n$ is dealt with similarly. If $f(z_1)=f(z_2)$, $z_1\neq z_2$, and where $S(z) =A(q)/(1-|z|^2)^{2-q}$, then there exists a solution $w_0$ of this differential equation such that $w_0(-\rho)=w_0(\rho)=0$ and the inequality
$$
\begin{equation*}
\frac{A(q)}{2}\int_{-\rho}^{\rho}\frac{|w_0(t)|^2}{(1-t^2)^{2-q}}\, dt\geqslant \int_{-\rho}^{\rho}|w'_0(t)|^2\, dt,
\end{equation*}
\notag
$$
holds for all $\rho\in(0,1)$. But this inequality contradicts the Hardy-type inequality
$$
\begin{equation}
\frac{A(q)}{2}\int_{-\rho}^\rho\frac{(y(t))^2}{(1-t^2)^{2-q}}\, dt<\int_{-\rho}^\rho (y'(t))^2\, dt,
\end{equation}
\tag{4.2}
$$
which holds for any absolutely continuous on $(0,1)$ function $y(t)\,{\not\equiv}\, 0$ such that $y(-\rho)=y(\rho)=0$, $\rho\in(0,1)$, and $y'\in L^2(-\rho,\rho)$ (for more details, see [22], [23]). Using the above algorithm and an inequality obtained in § 2.3, we get the following theorem. Theorem 12. A meromorphic function $f(z)$ is univalent in $\mathbb{D}$ if $f'(z)\neq 0$ and if, for some tuple of real nonnegative parameters $n$, $a_k$ and $q_k$, $k=1,\dots,n$, Lemma 4 (see. [24]). Suppose that $f(z)$ is a meromorphic function on $D$, and $P(z)$, $Q(z)$, and the Schwarzian $S_f(z)$ $f(z)$ are all regular and satisfy Assume that $f(z)$ is not univalent in $D$, that is, $f(z_1)=f(z_2)$ for $z_1\neq z_2$, $z_1, z_2\in D$. Then there exists a nontrivial solution $w_0(z)$ of the equation vanishing at $z_1$ and $z_2$. We also will use the inequality
$$
\begin{equation*}
\frac{R(q)}{2}\int_{-\rho}^\rho \frac{(y(t))^2}{(1-t^2)^{2-q}}\, dt<\int_{-\rho}^\rho (y'(t))^2\, dt,
\end{equation*}
\notag
$$
which is obtained by combining (4.2) and the inequality of Theorem 4. Multiplying both sides of this inequality by positive numbers $a_k$, $k=1,\dots,n$, such that $a_1+\dots+a_n\leqslant 1$, we get
$$
\begin{equation*}
\frac{a_k R(q)}{2}\int_{-\rho}^\rho \frac{(y(t))^2}{(1-t^2)^{2-q}}\, dt< a_k\int_{-\rho}^\rho (y'(t))^2\, dt.
\end{equation*}
\notag
$$
Adding these inequalities and taking into account that $a_1+\dots+a_n\leqslant 1$, we obtain
$$
\begin{equation}
\sum_{k=1}^{n}\frac{a_k R(q)}{2}\int_{-\rho}^\rho \frac{(y(t))^2}{(1-t^2)^{2-q}}\, dt< \int_{-\rho}^\rho (y'(t))^2\, dt
\end{equation}
\tag{4.3}
$$
for any absolutely continuous function $y$ such that $y(-\rho)=y(\rho)=0$. Let us use the above arguments for the majorant
$$
\begin{equation*}
S(z)= \sum_{k=1}^{n}\frac{a_k R(q)}{2} (1-|z|^2)^{q-2}.
\end{equation*}
\notag
$$
If $f$ is nonunivalent in $|z|<1$, then, by local univalence, there exists a disc $|z|<\rho$ of maximum radius $\rho\in(0,1)$ such that $f$ is univalent in this disc, but $f$ is not univalent in any other disc $| z|< \rho'$, where $\rho'\in(\rho,1)$. We fix $\rho'\in(\rho,1)$. There exist two distinct points $z_1$ and $z_2$, $|z_1|=|z_2|=\rho'$ such that $f(z_1)=f(z_2)$. If there are no such points, then this would mean that the image of the circle $|z|=\rho'$ would be a simple curve, and therefore the function $f$ would be univalent in $|z|<\rho'$ by the argument principle (see, for example, [25]). However, this contradicts the maximality of $\rho$. Without loss of generality, we can assume that the points $0$, $z_1$ and $z_2$ lie on the same straight line or even on the real diameter of the circle. This can be achieved using the corresponding linear fractional transformation (see, for example, [23], § 6). By Lemma 4, for $P(z)=0$ and $Q(z)= S_f(z)/2$, there exists a function $w_0$ such that $w_0(x)\not\equiv 0$, $w_0(z_1)=w_0(z_2)=0$ and Multiplying both parts of the last equality by $\overline{w_0}(z)\, dz$ and integrating over the interval connecting the points $z_1$ and $z_2$, we get
$$
\begin{equation}
\int_{z_1}^{z_2} \overline{w_0}(z)\, dw'_0(z)+\int_{z_1}^{z_2} \frac{1}{2}|w_0(z)|^2 S_f(z)\, dz=0.
\end{equation}
\tag{4.4}
$$
Since $|z_1|=|z_2|=\rho<1$, we have $z_1=e^{i\alpha}(x_0+iy_0)$ and $z_2= e^{i\alpha}(x_0-iy_0)$, where $\alpha$, $x_0$, $y_0$ are real numbers, and $x_0^2+y_0^2=\rho^2<1$. The parametric equation of the integration interval in (4.4) can be written in the form where $t\in[-\rho,\rho]$, $\rho=|y_0|/\sqrt{1-x_0^2}$. Now, using equality (4.4) and the estimate $|S_f(z)|\leqslant S(z)$, and transforming the function $w_0$ satisfying the conditions $w_0(-\rho)=w_0(\rho)=0$, we have
$$
\begin{equation*}
\sum_{k=1}^{n}\frac{a_k R(q)}{2}\int_{-\rho}^\rho \frac{|w_0(t)|^2}{(1-t^2)^{2-q}}\, dt\geqslant \int_{-\rho}^\rho |w'_0(t)|^2\, dt,
\end{equation*}
\notag
$$
which contradicts (4.3) (see also [21]–[24])). This proves the theorem. 4.2. Sufficient conditions in non-disc simply connected domainsBased on the previous theorems, one can obtain sufficient conditions for univalence also for simply connected domains distinct from the disc. Suppose that $F(\zeta)$ is a meromorphic function on a simply-connected domain $\mathfrak{D}$, and $\varphi(\zeta)$ is a function mapping univalently $\mathfrak{D}$ onto the unit disc $\mathbb{D}$. Then $F$ and $f(z)=F^{-1}(\varphi(\zeta))$ are either both univalent or both fail to be univalent. The following equality is known
$$
\begin{equation*}
S_f(z)=(\varphi'(\zeta))^{-2}(S_F(\zeta)-S_\varphi(\zeta)), \qquad\zeta\in \mathfrak{D},
\end{equation*}
\notag
$$
which takes a simpler form in the case $\varphi$ is a linear fractional transformation, that is, $S_\varphi(\zeta)\equiv 0$. Consequently, the sufficient condition of Theorem 12 can be rewritten as
$$
\begin{equation*}
|S_F(\zeta)-S_\varphi(\zeta)|\leqslant \sum_{k=1}^n a_k R(q_k) \frac{|\varphi'(\zeta)|^2}{(1-|\varphi(\zeta)|^2)^{2-q_k}} \quad \forall\, \zeta\in \mathfrak{D}.
\end{equation*}
\notag
$$
If, for example, $\varphi(\zeta) =1/\zeta$ and $\varphi(\zeta)= (\zeta-1)/(\zeta+1)$, then we obtain, respectively, the following results. Theorem 13. A function $F(\zeta)$, meromorphic on the exterior of the unit disc $\mathbb{D}^{-}=\{\zeta\in\mathbb{C}: |\zeta|> 1\}$, is univalent in $\mathbb{D}^{-}$ if $F'(\zeta)\neq 0$ in $\mathbb{D}^-$ and if, for some tuple of real nonnegative parameters $n$, $a_k$ and $q_k$, $k=1,\dots,n$, Theorem 14. A function $F(\zeta)$, meromorphic on the right half-plane $H_{+}=\{\zeta\in\mathbb{C}\colon \operatorname{Re}\zeta=\xi>0\}$, is univalent in $H_{+}$ if $F'(\zeta)\neq 0$ in $H_+$ and if, for some tuple of real nonnegative parameters $n$, $a_k$ and $q_k$, $k=1,\dots,n$, The author is grateful to Professor Farit Gabidinovich Avkhadiev for valuable advice on improving this article. The author also is grateful to the referee for a number of valuable comments which have greatly improved the exposition. 
Citation in format AMSBIB
\Bibitem{Nas23} |
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