Proving that an operator $K$ is bounded and $||K|| = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x,y)|dy\bigg\}$

Question

Define $K:C([0,1])\rightarrow C([0,1])$ by $$Kf(x) = \int_0^1 k(x,y)f(y)dy,$$ where $k:[0,1]\times [0,1]\rightarrow \mathbb{R}$ is continuous. Prove that $K$ is bounded and $$||K|| = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x,y)|dy\bigg\}.$$

This is what I have so far: Let $A = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x,y)|dy\bigg\} = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x_0,y)|dy\bigg\}$.

$|Kf(x)|\leq ||f||\int_0^1 |k(x,y)|dy \leq ||f||A$.

$\implies$ $||Kf|| = \sup_{0\leq x\leq 1}|Kf(x)| \leq A||f||$

$\implies$ $||K||\leq A$.

Now I know that I want to show that $||K||\geq A$ to have equality but I am not sure how to do so.

I should also note this is not a homework question. I am attempting to do all the problems in my book for better understanding.

Any help and comments would be appreciated. Thank you.

Answer 1

Sketch: Let $x$ be a point such that $\int_0^1|k(x,y)|dy$ is maximal. Then set $f(y):={\rm sgn} (k(x,y))$.
Then approximate $f$ by continuous functions $f_n$, proving that $\ \|Kf\|_\max\ge A-1/n\ $ for all $n\in\Bbb N$.

Answer 2

So you need to approximate $f_{\infty}$ with a sequence of functions $f_{n}$ such that $f_{n}$ is continuous. $f_{\infty}$ can be written as $\frac{x}{|x|}$. Now, you just have to modify the formula for $f_{\infty}$ by adding something small and you'll get a continuous $f_{n}$ that'll converge to $f_{\infty}$. Then if you take the limit you get the >= case, and you'll have equality.