Norm of linear operator 3

Question

Define operator $A:L^2(0,1)\to \ell_2$, $$ Af=(\int_{\displaystyle(0,1)}f(x)\mbox{ d}x,\int_{\displaystyle(0,\frac{1}{4})}f(x)\mbox{ d}x,\int_{\displaystyle(0,\frac{1}{9})}f(x)\mbox{ d}x,\ldots).$$ What is the norm of operator A? Is the norm attained ? My try is following $$ \|Af\|^2=\sum_{n\in\mathbb{N}}\left(\int_{\displaystyle(0,\frac{1}{n^2})}f(x)\mbox{ d}x\right)^2\leq \sum_{n\in\mathbb{N}}\frac{1}{n^2}\int_{\displaystyle(0,\frac{1}{n^2})}f^2(x)\mbox{ d}x\leq \|f\|^2\sum_{n\in\mathbb{N}}\frac{1}{n^2}.$$ But I am not able to find sequence $f_n$ such that $\frac{\|Af_n\|}{\|f_n\|}\to \sqrt{\sum_{n\in\mathbb{N}}\frac{1}{n^2}}$. Isn't this estimate too big ? Thanks in advance for your help.

Answer 1

Hint: For the start, use the Cauchy-Schwartz inequality to bound $$\int_0^{\frac{1}{i^2}}f(x)\,dx=\int_0^1\chi_{[0,{\frac{1}{i^2}}]}\cdot f(x)\,dx\leq \left(\int_0^1\chi_{[0,{\frac{1}{i^2}}]}^2\,dx\right)^{1/2}\cdot \left(\int_0^1f^2(x)\,dx\right)^{1/2}=\frac{1}{i}\|f\|_{L^2}$$ so that $$\|A\|\leq \frac{\pi}{\sqrt{6}}.$$ It is easy to find a sequence of functions whose $l_2$ norms converge to that limit (take functions $f_i$ whose $L^2$ norm is $1$ and whose support is, say, on $[0,1/i^2]$ for each $i$), so that indeed that is the norm.

To show that it is not achieved work by contradiction.