Let $H = L^2([-1,1],\lambda)$ with $\lambda$ the Lebesgue measure. Consider $$ T: H \to H: f \mapsto Tf $$ such that $(Tf)(x) = x^2f(x)$. Compute the norm of $T$.
Seems easy enough, but I am only able to show the norm is $\leq 1$. I know that the answer is that the norm is equal to 1. (In a later problem, we showed that the spectrum is $[0,1]$ meaning the norm is also at least 1). I showed that for any function $f$, we have $$||Tf||_2^2 = \int x^4 |f(x)|^2 dx \leq \int |f(x)|^2 dx = ||f||_2^2.$$
I also know that $T$ is self-adjoint, meaning we have $||T|| = \sup_{||f||\leq 1} \langle Tf,f \rangle$ and that $\langle Tf, f \rangle = \int x^2|f(x)|^2 dx \leq ||f||_2^2$. Do we find a function $f$ such that $||f|| = 1$ and $||Tf|| = 1$? Or is there another way of showing the norm equals one?
Let $f_n=\sqrt n1_{(1-\frac 1 n,1)}$. Show that $\|f_n\|=1$ and compute $\|Tf_n\|$. Show that $\|Tf_n||\to 1$ as $n \to \infty$. This is enough to show that $\|T\|\ge 1$ because $\|T\|\ge \|Tf_n\|$ for every $n$.