# continuouty of operators

I was given a task to understand, wheter operators $$A$$ and $$B$$ are compact, $$\displaystyle A:\ell_2 \rightarrow L_1(\mathbb{R}), (Ax)(t) = \sum\limits_{k=1}^{+\infty}\frac{x(k)}{\cosh^2(kt)},$$ $$B:c_0 \rightarrow L_1(\mathbb{R}), (Bx)(t) = \sum\limits_{k=1}^{+\infty}\frac{x(k)}{k^4+t^2}.$$

But I cannot even understand, why these operators are continuous. I tried to do the following:

$$\displaystyle\lVert Ax\rVert = \left|\int\limits_{\mathbb{R}}\sum\limits_{k=1}^{+\infty}\frac{x(k)}{\cosh^2(kt)}dt\right| \le \int\limits_{\mathbb{R}}\lVert x\rVert_2\sqrt{\sum\limits_{k=1}^{+\infty}\frac{1}{\cosh^4(kt)}}dt$$. Is $$\displaystyle\int\limits_{\mathbb{R}}\sqrt{\sum\limits_{k=1}^{+\infty}\frac{1}{\cosh^4(kt)}}dt < \infty$$? How can one understand it?

$$\displaystyle\lVert Bx\rVert = \left|\int\limits_{\mathbb{R}}\sum\limits_{k=1}^{+\infty}\frac{x(k)}{k^4+t^2}dt\right| \le \int\limits_{\mathbb{R}}\lVert x\rVert_\infty\sum\limits_{k=1}^{+\infty}\frac{1}{2k^2|t|}dt$$. The problem there is that $$\displaystyle\frac{1}{|t|}$$ is not integrable. How can be continuity be shown in this case?

Let $$(e_k)_{k=1}^\infty$$ be a Schauder basis for a Banach space $$X$$ and let $$A:X\to Y$$ be a continuous operator. Then $$A$$ is compact iff $$\lim_N \|A|_{\text{span}\{e_k:k\geqslant N\}}\|=0.$$ I think that will be the best method to use here, and it will more or less come out of the proof of continuity.
Note that for $$t>0$$, $$4\cosh^2(kt)=[e^{kt}+e^{-kt}]^2\geqslant e^{2kt}.$$ For $$t<0$$, $$4\cosh^2(kt)=[e^{kt}+e^{-kt}]^2 \geqslant e^{-2kt}.$$ So $$\int_\mathbb{R} \frac{1}{\cosh^2(kt)}dt =O\Bigl(\int_0^\infty e^{-2kt}dt\Bigr)=O(k^{-1}).$$ By the triangle inequality, we have $$\Bigl\|\sum_{k=1}^\infty \frac{x(k)}{\cosh^2(kt)}\Bigr\|_1\leqslant \sum_{k=1}^\infty |x(k)|Ck^{-1}$$ for some constant $$C$$. Use Cauchy Schwartz to get that this is finite. This also allows us to get estimates on $$\|A|_{\text{span}\{e_k:k\geqslant N\}}\|$$ in terms of the $$\ell_2$$ norms of tails of $$(k^{-1})_{k=1}^\infty$$.
For $$a>0$$, since $$\frac{d}{dt}\arctan(t/a)=\frac{1}{a^2+t^2}$$, we can directly calculate $$\int_\mathbb{R}\frac{1}{k^4+t^2}dt = \frac{\pi}{k^2}.$$ Therefore $$\Bigl\|\sum_{k=1}^\infty \frac{x(k)}{k^4+t^2}\Bigr\|_1\leqslant \max_k |x(k)|\sum_{k=1}^\infty \frac{\pi}{k^2}=\frac{\pi^3}{6}\max_k |x(k)|.$$