Krdytkyu

Let $kin C([0,1]^2)$, Define $T_k:C[0,1]to C[0,1]$ such that for $xin C[0,1]$, 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 $xin C[0,1]$ with $||x||_inftyleq 1$, so $suplimits_{sin[0,1]}|x(s)|leq 1$. Then $$||T_k x||_{infty}=suplimits_{tin[0,1]}bigg|int_{0}^1 k(t,s)x(s)dsbigg|leqsuplimits_{tin[0,1]}int_{0}^{1}|k(t,s)|,|x(s)|,dsleq||x||_inftysuplimits_{tin[0,1]}int_{0}^{1}|k(t,s)|,dsleqsuplimits_{tin[0,1]}int_{0}^{1}|k(t,s)|,ds.$$
It follows that $$|T_k|leqsuplimits_{tin[0,1]}int_{0}^1|k(t,s)|,ds.$$
Let $epsilon>0$ and define for all $tin[0,1]$ $x_epsilon:[0,1]tomathbb{K}$ 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}&=suplimits_{tin[0,1]}bigg|int_{0}^{1}frac{|k(t,s)|^2}{|k(t,s)|+epsilon},dsbigg|=suplimits_{tin[0,1]}int_{0}^1frac{|k(t,s)|^2}{|k(t,s)|+epsilon},dsgeqsuplimits_{tin[0,1]}int_{0}^1frac{|k(t,s)|^2-epsilon^2}{|k(t,s)|+epsilon}
,ds=\&suplimits_{tin[0,1]}int_{0}^1(|k(t,s)|-epsilon),ds=suplimits_{tin[0,1]}int_{0}^{1}|k(t,s)|,ds-epsilon.
end{align*}
We see that $$|T_k|geqsuplimits_{epsilon>0}||T_kx_epsilon||_inftygeqsuplimits_{tin[0,1]}int_{0}^1|k(t,s)|,ds.$$
Combining this with our previous estimate of $|T_k|$, we conclude that $|T_k|=suplimits_{tin[0,1]}int_{0}^{1}|k(t,s)|,ds,$ which concludes our proof.

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 $t_0in [0,1].$ Then, we have
$$
lVert T_k[x_epsilon]rVert_infty geq |T_k[x_epsilon](t_0)| = ∫_0^1 frac{|k(t_0,s)|^2}{|k(t_0,s)|+epsilon}dsgeq ∫_0^1 |k(t_0,s)|ds-epsilon.
$$ Accordingly, we can bound $lVert T_krVert$ from below: since $lVert x_epsilon rVert_infty leq 1$, we have
$$
lVert T_krVertgeq lVert T_k[x_epsilon]rVert_inftygeq ∫_0^1 |k(t_0,s)|ds-epsilon.
$$Since our choice of $t_0$ was $textbf{arbitrary}$, by taking supremum over $t_0in[0,1]$, we get
$$
lVert T_krVert geq sup_{t_0in [0,1]}∫_0^1 |k(t_0,s)|ds-epsilon,
$$ for all $epsilon>0$. Finally by taking $epsilon to 0$, we have
$$
lVert T_krVert= sup_{tin [0,1]}∫_0^1 |k(t,s)|ds,
$$as desired.