Integrability of trigonometric series. The estimation of the integral modulus of continuity

Let $a_m$ tend to zero and let the quantities
$$\begin{align*} B_n&=\sum_{m=1}^n\biggl(\frac mn\biggr)^k|\Delta a_m|+\sum_{m=n+1}^\infty|\Delta a_m|+ \\ &\qquad+\sum_{m=2}^n\biggl(\frac mn\biggr)^k\biggl|\sum_{i=1}^{[m/2]}\frac{\Delta a_{m-i}-\Delta a_{m+i}}i\biggr|+\sum_{m=n+1}^\infty\biggl|\sum_{i=1}^{[m/2]}\frac{\Delta a_{m-i}-\Delta a_{m+i}}i\biggr|. \end{align*}$$
be finite. We put $f(x)=\frac{a_0}2+\sum_{m=1}^\infty a_m\cos mx$ and $g(x)=\sum_{m=1}^\infty a_m\sin mx$.
It is shown that the integral modulus of continuity of $k$th order for the function $f$ satisfies the estimate $\omega_k\bigl(f,\frac1n\bigr)_L=O(B_n)$, and that if the series $\sum\frac{|a_m|}m$, converges then
$$\omega_k\biggl(g,\frac1n\biggr)_L=\frac{2^k}\pi\sum_{m=n}^\infty\frac{|a_m|}m+O(B_n).$$

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