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Problem:

Let $H=L^{2}\left([0,1]\right)$ and $\varphi\in L^{\infty}\left([0,1]\right)$

Defined :

$$T_{\varphi}~~:~H~\to~H$$ $$T_{\varphi}f=\varphi f$$ Prove that $T_{\varphi}$ are bounded and find the $~~\|T_{\varphi}\|$

My attempts :

$\|T_{\varphi}f\|^{2}_{L^{2}}=\displaystyle\int\limits_{0}^{1}|T_{\varphi}f(x)|^{2}dx$

And we have :

$$|T_{\varphi}f(x)|^{2}=|\varphi(x)f(x)|^{2}$$ $$~~~~~~~~~~~~~~~~~\leq\|\varphi\|^{2}_{\infty}|f(x)|^{2}$$

So :

$\|T_{\varphi}f\|^{2}_{L^{2}}\leq\|\varphi\|^{2}_{\infty}\displaystyle\int\limits_{0}^{1}|f(x)|^{2}dx$

$~~~~~~~~~~~~~~\leq\|\varphi\|^{2}_{\infty}\|f\|^{2}$

$$\implies T_{\varphi}\in\mathcal{L}(H)$$ and $\|T_{\varphi}\|\leq~\|\varphi\|_{\infty}$

My problem now with how I find $\|T_{\varphi}\|$

i.e

How I find $f\in~H$ such that $\|f\|_{H}=1~\text{or}\leq 1$ Then give $\|T_{\varphi}f\|_{H}=\|\varphi\|_{\infty}$ ??

I always have problem to calculat norm of operator!!

Thanks

Ellen Ellen
  • 2,397

1 Answers1

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Assume $\varphi\neq 0.$ For $0<\delta <1$ let $$A_\delta=\{ x\,:\, |\varphi(x)|\ge (1-\delta) \|\varphi\|_\infty \}$$ Then $|A_\delta|>0,$ where $|\cdot|$ denotes the Lebesgue measure. Let $$f_\delta(x)=\begin{cases} 1 & x\in A_\delta \\ 0 & x\notin A_\delta \end{cases}$$ Then $$\displaylines{\|T_\phi(f_\delta)\|^2 =\int\limits_{A_\delta} |\varphi(x)|^2\,dx\\ \ge (1-\delta)^2\, \|\varphi\|_\infty^2\, |A_\delta|\\ = (1-\delta)^2 \|\varphi\|_\infty^2\,\|f_\delta\|^2 }$$ Therefore $$\|T_\varphi\|\ge (1-\delta)\, \|\varphi\|_\infty $$ As $\delta$ is arbitrary we get $$\|T_\varphi\|\ge \|\varphi\|_\infty $$

Remark If the set $$A_0=\{ x\,:\, |\varphi(x)|=\|\varphi\|_\infty \}$$ has the Lebesgue measure equal $0,$ there is no function $0\neq f\in L^2$ such that $$\|T_\varphi f\|_2=\|T_\varphi \|\,\|f\|_2$$

Ryszard Szwarc
  • 44,675