if $L$ is the operatior $Lx=t\cdot x(t)$ over $L_2(0,1)$, what is the $\|L\|_2$

Question

I think this is equivalent to the maximizing problem $\int_{0}^{1} t^2x^2(t)dt$,

subjecting to $\int_{0}^{1} x^2(t)dt=1$, the answer is $1$ when x=$\sqrt{\delta(t-1)}$, but can anyone proof why this is the maximum??

Any advise would be appreciated.

Answer 1

I think it's pretty easy. Since $t \in [0,1]$ , we have $t^2 \leq 1$ so: $$ \int_0^1 t^2 x^2(t) dt \leq \int_0^1 x^2(t) dt = 1 $$ So the norm is no greater then $1$. To prove that the norm is one, look at functions $f_n = \sqrt{n}\chi_{[\frac{n-1}{n}, 1]}$. We have $\int_0^1 f_n^2(t) dt = 1$, and : $$ \int_0^1 t^2 f_n^2(t) dt \geq \frac{n-1}{n} \int_0^1 f_n^2(t) dt = \frac{n-1}{n} $$ So $$ \sup_{||f||_2 \leq 1} \int_0^1 t^2 f^2(t) dt = 1 $$