0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I'm puzzled by the definition of the map $A$. Do you assign the $i$th component of $Af$ by integrating $f(x)$ over $(0,i^{-2})$? The upper limit of integration is hard to read in its present form, but an edit would make it legible. $\endgroup$
    – hardmath
    Nov 28, 2016 at 18:29

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This example of sequence does not work. I think that my estimate can be optimized. When we look at my estimate, when we can write equalities instead of inequalities ? First inequality is equality when $|f| =1$ but in second one $f=0$ on $(\frac{1}{n^2},1)$. $\endgroup$
    – elliptic
    Nov 29, 2016 at 20:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .