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Since $k$ is continuous and defined on $[0,1]\times[0,1]=J$ compact domain so it is bounded. Therefore there exists $c\in \mathbb R$ such that $|k(s,t)|\le c,\quad \forall (s,t)\in [0,1]\times[0,1]=J$

So $$||Tx||=\sup_{t\in [0,1]}\left|\displaystyle\int_0^1 k(s,t)x(s)ds\right|$$$$\le \sup_{t\in [0,1]}\displaystyle\int_0^1 \left|k(s,t)\right|\left|x(s)\right|ds$$$$\le c||x||$$

My gut says that we need to specify this constant $c$ in a way and then for a special $x$(which depends on the k(s,t)) and then we need to say that $\displaystyle\int_0^1 k(s,t)x_k(s)ds=c_k$

So that we finally find that $||T||=c_k$ for this $T$ defined as above.

But how to do that? If $k$ is given as in

Norm of $T:(C[0,1],||.||_{\infty})\to \mathbb R,$ $T(f)=\int_0^1tf(t)$

I can do it like in the answers but how to do that for an arbitrary $k(s,t)$ function in $[0,1]\times[0,1]$

Jale'de jaled
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    It seems to me you have already shown $T$ is bounded. You know $c$ exists from where you first introduced it. – aschepler Jun 08 '20 at 23:36
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    $\sup_{t \in J}$ doesn't make sense - you mean $\sup_{t \in [0,1]}$. – aschepler Jun 08 '20 at 23:37
  • yeah it is bounded above but how about the below yeah sorry I will edit it now, @aschepler – Jale'de jaled Jun 08 '20 at 23:40
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    Well, I'm not sure if $c_k$ is in fact equal to $|T|$. To figure out what the $s$ dimension is doing here, I would look at sequences of functions $x_n(s)$ which are zero or small outside of shrinking intervals in the $s$ dimension, like the Dirac delta idea or triangle functions. – aschepler Jun 08 '20 at 23:51

2 Answers2

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You have already shown that $\|T\|\leq \sup_\limits{t\in [0,1]}\displaystyle\int_0^1 \left|k(s,t)\right|ds$. Since $k(s,t)\in C([0,1]\times[0,1])$, then function $\displaystyle\int_0^1 \left|k(s,t)\right|ds$ is continuous by $t$ and, by Weierstrass, reaches its greatest value in the $[0,1]$. I.e. $\exists t_0\in[0,1]$ such that $\displaystyle\int_0^1 \left|k(s,t_0)\right|ds=\sup_{t\in [0,1]}\displaystyle\int_0^1 \left|k(s,t)\right|ds$. Denote $z(s)=\text{sign}k(s,t_0)$ and $x_n(s)=z(s)$ everywhere except points of some set $E_n$. Here $x_n$ -- be the sequence of continuous functions on $[0,1]$ such that $\|x_n\|=1$. Then, on $E_n$ $|x_n(s)-z(s)|\leq 2$, and $$\left|\displaystyle\int_0^1k(s,t)z(s)ds-\displaystyle\int_0^1k(s,t)x_n(s)ds\right|\le\sup_{t,s}|k(t,s)|\displaystyle\int_0^1|z(s)-x_n(s)|ds\le2\sup_{t,s}|k(t,s)|\mu(E_n).$$ Let $E_n$ be such that $\mu(E_n)\le\dfrac{1}{2n\sup\limits_{t,s}|k(t,s)|}$, then $\forall t\in[0,1]$ $$\left|\displaystyle\int_0^1k(s,t)z(s)ds-\displaystyle\int_0^1k(s,t)x_n(s)ds\right|\le\frac{1}{n},$$$$\displaystyle\int_0^1k(s,t)z(s)ds\le\frac{1}{n}+\displaystyle\int_0^1k(s,t)x_n(s)ds\le\frac{1}{n}+\|Ax_n\|\le\frac{1}{n}+\|A\|.$$ By $t=t_0$ we have $\displaystyle\int_0^1k(s,t_0)z(s)ds\le\frac{1}{n}+\|A\|$ or $\displaystyle\int_0^1|k(s,t_0)|ds\le\frac{1}{n}+\|A\|$. From here follows $\|A\|\ge\displaystyle\int_0^1|k(s,t_0)|ds-\frac{1}{n}$. It remains to go to the limit at $n\to\infty$.

thing
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Since $|Tx(t)| \le \int_0^1 |k(s,t)| ds \|x\|$ we see that $\|T\| \le \sup_t \int_0^1 |k(s,t)| ds$.

Let $t^*$ be a maximiser of $t \mapsto \int_0^1 |k(s,t)| ds$ and let $\sigma_n(x) = \max(-1, \min(1, nx))$. Note that $\sigma_n(x) \to \operatorname{sgn} x$ for all $x$ and $\sigma_n$ is continuous.

Then with $x_n(s)= \sigma_n(k(s,t^*))$ we have $\|x_n\| \le 1$ and $\lim_n |Tx_n(t^*)| = \lim_n | \int_0^1 k(s,t^*)x_n(s)ds | = | \int_0^1 \lim_n k(s,t^*)x_n(s)ds | = \int_0^1 |k(s,t^*)|ds$, hence $\|T\| \ge \int_0^1 |k(s,t^*)|ds$.

copper.hat
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