Integral operator doesn't attend norm of kernel in $L^2[0,1]$

Question

$H=L^2[0,1]$, $K \in L^2([0,1]\times[0,1]) $, $f \in L^2[0,1]$ . If I have $J: H \to H$ given by $$ (Jf)(s) \overset{a.e}{=} \int_0^1 K(s,t) f(t)dt $$ In what case $\| J \|_{\text{op}}< \| K \|_2$ ? Can you give an example?

Thanks

Answer 1

It's easy to see that in general $\|J\| \leq \| K \|_2$.

Here's an example in which the norm of $K$ is not attained. Define $$ K(s,t):= \chi_{[0,s]}(t) $$ Then, $$ \| K \|_2^2=\int_0^1 \int_{0}^1 |\chi_{[0,s]}(t)|^2 dt ds = \int_0^1 s ds = \frac{1}{2} $$ Thus, $\| K \|_2=\frac{1}{\sqrt{2}}$. Now, notice that in this case $$ (Jf)(s)=\int_0^s f(t) dt $$ You can check (with some work though) that $\| J \|=\frac{2}{\pi} < \frac{1}{\sqrt{2}}=\| K \|_2$.