Operator norm of X

Question

I've been trying to prove that the operator norm of $\hat{X}\phi(x) = x\phi(x)$ with $\phi(x) \in L^2[0,1]$ is given by: $$ ||\hat{X}|| = \sup_{||\phi(x)||=1} ||\hat{X}\phi(x)||=1 $$ However i havent been able to do it. I asked my professor for help and he suggested that i consider a series of functions $\phi_n(x)$ of unit norm with support on$[a_n,1]$ with $a_n \in (0,1)$ and $\lim_{n\to \infty} a_n=1$. Honestly that left me even more confused. Everywhere i searched seems to give this almost as a defnition.

Any help is greatly appreciated!

Answer 1

Let $0 \leq a < 1$ and suppose $\phi \in L^2[0,1]$ is supported on some subset of $[a,1]$. Then $$\int_a^1a^2 \phi(x)^2 \,dx \leq \int_a^1 x^2\phi(x)^2\,dx \leq \int_a^1 1^2\phi(x)^2\,dx$$ which, since $\phi(x) = 0$ for $0 \leq x < a$, is equivalent to $$a^2\int_0^1\phi(x)^2\,dx \leq \int_0^1(\hat{X}\phi)(x)^2\,dx \leq \int_0^1\phi(x)^2\,dx$$ and by taking square roots, $$a\|\phi\| \leq \|\hat{X}\phi\| \leq \|\phi\|.$$

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Now, any $\phi \in L^2[0,1]$ is supported on some subset of $[0,1]$, so by setting $a=0$, we have that $\|\hat{X}\phi\| \leq \|\phi\|$ for any $\phi \in L^2[0,1]$, and therefore $\|\hat{X}\| \leq 1$.

On the other hand, given any $0\leq a < 1$, there will be some nonzero $\phi \in L^2[0,1]$ which is supported on $[a,1]$, so $$\|\hat{X}\| \geq \frac{\|\hat{X}\phi\|}{\|\phi\|} \geq a$$ and by letting $a\to 1^-$, we have $\|\hat{X}\| \geq 1$.

Answer 2

Let $\phi_{n}(x)=\dfrac{1}{\sqrt{1-a_{n}}}\chi_{[a_{n},1]}$, then $\|\phi_{n}\|_{L^{2}[0,1]}=1$ and \begin{align*} \int_{0}^{1}(x\phi_{n}(x))^{2}dx=\dfrac{1}{1-a_{n}}\int_{a_{n}}^{1}x^{2}dx=\dfrac{1}{1-a_{n}}\dfrac{1}{3}(1-a_{n})(1+a_{n}+a_{n}^{2})=\dfrac{1+a_{n}+a_{n}^{2}}{3}, \end{align*} so \begin{align*} \|x\phi_{n}(x)\|_{L^{2}[0,1]}=\sqrt{\dfrac{1+a_{n}+a_{n}^{2}}{3}}\rightarrow 1. \end{align*}