Let $k\in C([0,1]^2)$, Define $T_k:C[0,1]\to C[0,1]$ such that for $x\in C[0,1]$, defined by $(T_kx)(t)=\int_{0}^1k(t,s)x(s)ds$. I want to show that $T_k$ is bounded and calculate its operator norm. Is my following proof correct
By definition, $
|T_k|:=\sup\{||T_kx||_{\infty}\,\big|\,||x||_{\infty}\leq 1\}.
$ Let $x\in C[0,1]$ with $||x||_\infty\leq 1$, so $\sup\limits_{s\in[0,1]}|x(s)|\leq 1$. Then $$||T_k x||_{\infty}=\sup\limits_{t\in[0,1]}\bigg|\int_{0}^1 k(t,s)x(s)ds\bigg|\leq\sup\limits_{t\in[0,1]}\int_{0}^{1}|k(t,s)|\,|x(s)|\,ds\leq||x||_\infty\sup\limits_{t\in[0,1]}\int_{0}^{1}|k(t,s)|\,ds\leq\sup\limits_{t\in[0,1]}\int_{0}^{1}|k(t,s)|\,ds.$$
It follows that $$|T_k|\leq\sup\limits_{t\in[0,1]}\int_{0}^1|k(t,s)|\,ds.$$
Let $\epsilon>0$ and define for all $t\in[0,1]$ $x_\epsilon:[0,1]\to\mathbb{K}$ by $$x_\epsilon(s)=\frac{\overline{k(t,s)}}{|k(t,s)|+\epsilon}.$$ Then we remark that $x_\epsilon$ is continuous and $||x_\epsilon||_{\infty}\leq 1$. We see that
\begin{align*}
||T_k x_e||_{\infty}&=\sup\limits_{t\in[0,1]}\bigg|\int_{0}^{1}\frac{|k(t,s)|^2}{|k(t,s)|+\epsilon}\,ds\bigg|=\sup\limits_{t\in[0,1]}\int_{0}^1\frac{|k(t,s)|^2}{|k(t,s)|+\epsilon}\,ds\geq\sup\limits_{t\in[0,1]}\int_{0}^1\frac{|k(t,s)|^2-\epsilon^2}{|k(t,s)|+\epsilon}
,ds=\\&\sup\limits_{t\in[0,1]}\int_{0}^1(|k(t,s)|-\epsilon)\,ds=\sup\limits_{t\in[0,1]}\int_{0}^{1}|k(t,s)|\,ds-\epsilon.
\end{align*}
We see that $$|T_k|\geq\sup\limits_{\epsilon>0}||T_kx_\epsilon||_\infty\geq\sup\limits_{t\in[0,1]}\int_{0}^1|k(t,s)|\,ds.$$
Combining this with our previous estimate of $|T_k|$, we conclude that $|T_k|=\sup\limits_{t\in[0,1]}\int_{0}^{1}|k(t,s)|\,ds,$ which concludes our proof.
Best Answer
When you choose $$ x_\epsilon(s) = \frac{\overline{k(t,s)}}{|k(t,s)|+\epsilon}, $$ the $t$ appearing in the expression should be predetermined and remain fixed in the following calculation. This is what you should fix in your argument. To make this point clear, let us write $$ x_\epsilon(s) = \frac{\overline{k(t_0,s)}}{|k(t_0,s)|+\epsilon}, $$ for some $t_0\in [0,1].$ Then, we have $$ \lVert T_k[x_\epsilon]\rVert_\infty \geq |T_k[x_\epsilon](t_0)| = ∫_0^1 \frac{|k(t_0,s)|^2}{|k(t_0,s)|+\epsilon}ds\geq ∫_0^1 |k(t_0,s)|ds-\epsilon. $$ Accordingly, we can bound $\lVert T_k\rVert$ from below: since $\lVert x_\epsilon \rVert_\infty \leq 1$, we have $$ \lVert T_k\rVert\geq \lVert T_k[x_\epsilon]\rVert_\infty\geq ∫_0^1 |k(t_0,s)|ds-\epsilon. $$Since our choice of $t_0$ was $\textbf{arbitrary}$, by taking supremum over $t_0\in[0,1]$, we get $$ \lVert T_k\rVert \geq \sup_{t_0\in [0,1]}∫_0^1 |k(t_0,s)|ds-\epsilon, $$ for all $\epsilon>0$. Finally by taking $\epsilon \to 0$, we have $$ \lVert T_k\rVert= \sup_{t\in [0,1]}∫_0^1 |k(t,s)|ds, $$as desired.