What is the operator norm of $Tf(x) = x^2f(x)$?

Question

Let $H = L^2([0,1],\mathbb{R})$ and $T : H \to H,\, Tf(x) = x^2f(x) $.

$T$ is linear.

$$\|Tf\|_{L^2([0,1],\mathbb{R})} = \sqrt{\int_0^1x^4f^2(x)dx} \leq\sqrt{\int_0^1f^2(x)dx} = \|f\|_{L^2([0,1],\mathbb{R})} $$

$T$ is linear and bounded therefore it's continuous. Also $\|T|| \leq 1$.

I tried finding solution to $\|Tf\|_{L^2([0,1],\mathbb{R})} = \|f\|_{L^2([0,1],\mathbb{R})} $

I found $$f(x) = \sqrt{\frac{2x-1}{x^4-1}}$$

but it's not in $L^2$ so it doesn't work.

anyone knows an $f$ to reach $1$, I'm not even sure it's $1$.

any help will be greatly appreciated !

Answer 1

Consider $f_n(x)=x^n$ for $n\in\Bbb N$.

Then $$\|f_n\|_2=\sqrt{\frac1{2n+1}}$$ and $$\|Tf_n\|_2=\sqrt{\frac1{2n+5}}$$ Since $$\lim_{n\to\infty}\frac{\|Tf_n\|_2}{\|f_n\|_2}=1$$ we have that $\|T\|\ge1$.

Answer 2

Multiplication by $x^2$ on $[0,1]$. $T$ is a self-adjoint operator. The essential range of the multiplier $x^2$ is $[0,1]$, so the spectrum of $T$ is $[0,1]$. The spectral radius of $T$ is $1$, and (since $T$ is self-adjoint), the norm of $T$ is $1$.