Bounds for the spectral abscissa of an element in a Banach algebra

For an arbitrary element $x$ with spectrum $\operatorname{sp}(x)$ in a Banach algebra with identity e ne 0 we define the upper (lower) spectral abscissa $\sigma_{\substack{+\\(-)}}(x)=\max\limits_{\displaystyle(\min)}\operatorname{Re}\lambda$, $\lambda\in\operatorname{sp}(x)$. With the aid of the spectral radius $\rho(x)=\max\limits_{\lambda\in\operatorname{sp}(x)}|\lambda|=\lim\limits_{n\to+\infty}\|x^n\|^{1/n}$ we prove the following bounds: $\gamma_-(x)\le\sigma_-(x)\le\Gamma_-(x)\le\Gamma_+(x)\le\sigma_+(x)\le\gamma_+(x)$, где $\Gamma_{(\pm)}(x)=(2\delta_{(\pm)})^{-1}(\rho_{\delta_{(\pm)}}^2-\delta_{(\pm)}^2-\rho_0^2)$ $(\delta_{(\pm)}\ne0)$, $\gamma_{(\pm)}(x)=(\pm)\rho_{\delta_{(\pm)}}-\delta_{(\pm)}$, $\delta_+\ge0$, $\delta_-\le0$ и $\rho_{\delta_{(\pm)}}=\rho(x+e\delta_{(\pm)}$. We mention a case where equality is achieved, some corollaries,and discuss the sharpness of the bounds: for every $\varepsilon>0$ there is a delta: $\delta:|\delta|\ge\rho_0^2/2\varepsilon$, such that $\Delta:=|\gamma_{(\pm)}(x)-\Gamma_{(\pm)}(x)|<\varepsilon$ and conversely, if the bounds are computed for some $\delta\ne0$, then $\Delta\le\rho_0^2/2|\delta|$. An example is considered.