## Show a linear operator is not compact

For $f\in L^2(0,\infty),$ define $(Tf)(x)=x^{-1}\int_0^x f(s)ds,$ for $x\in(0,\infty),$ then from hardy's inequality, $T\in B(L^2),$ my question is how to show that $T$ is not compact?

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If you don't mind me asking: (a) why do you want to know? mere curiosity, or as a precursor to some other question? (b) what goes wrong with a direct attempt to use the definition of (sequential) compactness? – Yemon Choi Mar 14 2010 at 9:06

For a natural number $j$ let $f_j$ be the indicator function of the interval $[0,1/j]$ times the square root of $j$. Then the $L^2$-norm of $f_j$ is one. A simple calculation shows that one has $||Tf_i-Tf_j||^2\ge\int_0^{1/j}(\sqrt i-\sqrt j)^2dx= (1-\sqrt{i/j})^2$ for $i\le j$, which implies that no subsequence can be Cauchy.

It is natural to look for an example around $x=0$ since that is where the kernel of $T$ fails to be $L^2$.

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Anton already gave a very clean answer. Another way to see it is to work backwards: start from a sequence of functions $F_j$ in $L^2$ that does is non-compact, and define $f_j(x) = \frac{d}{dx} (x F_j(x) )$.

For example, let $\phi(x)$ be an arbitrary smooth bump function supported in $[-1/4,1/4]$, then the sequence of functions $F_j(x) = 2^j \phi( 4^j x - 1)$ all have disjoint support, but all have the same $L^2$ norm, so obviously does not have a converging subsequence in $L^2$.

Now set $f_j = (xF_j)' = F_j(x) + 8^j x \phi' (4^j x - 1)$. Since $\phi'$ has support only in $[-1/4,1/4]$, on the support of $f_j$ we can bound $4^j x$ absolutely by, say, 2. So we have that $f_j$ is a bounded sequence in $L^2$, whose corresponding $F_j = Tf_j$ cannot have a Cauchy subsequence.

Edit: I should also provide some motivation: observe that the scaling argument also works the other way (replace $j$ by $-j$, so that you can dilate). The Hardy-type inequality that you are using is a scaling invariant inequality: you estimate $f/x$ in $L^2$ by its derivative $f'$. If we treat $x$ as having units of distance, then the two objects have the same units regardless of what units $f$ has. This gives scaling invariance of the estimate. In other words, the estimate is invariant under the natural scaling action of $\mathbb{R}_{+}$ on $L^2(\mathbb{R}_+)$, where the group operation for $\mathbb{R}_{+}$ is multiplication.

Observe that $(\mathbb{R}_+, \times)$ is a non-compact Lie group. Generally, if you have an inequality/operator that is invariant under the action of a non-compact Lie group, the inequality/operator cannot be compact. You just need to start with some test function and act on it by the Lie group action to generate a bounded sequence that runs off non-compactly in the "infinity dimension" direction. Terry summarised it in his Buzz http://www.google.com/buzz/114134834346472219368/9UseDXTJN74/There-are-three-ways-that-sequential-compactness a short while back.

This is, of course, closely related to the notion of concentration compactness.

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 Sorry about the display problem near the end of the first paragraph of "Edit". I can't figure out why it does not want to parse. – Willie Wong Mar 14 2010 at 12:25 I took the liberty of editing your answer to solve the latex problem. This is a well-known problem and is caused by underscores and asterisks and you can solve it by enclosing the latex expression (including the dollar signs) in backticks. – José Figueroa-O'Farrill Mar 14 2010 at 12:33 Thanks! I suspected it may have something to do with underscores, but the mathematician in me gets bewildered when things break sometimes, but not always. I'll keep that solution in mind for the future... oh duh! It is already summarized on the right hand side! Curse my tiny Netbook screen for not putting it in my viewport earlier. – Willie Wong Mar 14 2010 at 12:52

Let { $L_{n}$ } be the sequence of Laguerre polynomials, and let us define

$e_{n}(t)=\dfrac{L_{n}(\ln t)}{t}$ $\cdot\chi_{\left(1,\infty\right)}\left(t\right)$ $(n\in\mathbb{N\textrm{, t > 0}})$ . Then { $e_{n}$ } is an orthonormal system in $L^{2}\left(0,\infty\right)$, and it is not hard to see that $He_{n}=e_{n}-e_{n+1}$ $(n\in\mathbb{N})$, where $H$ stands for the Hardy averaging operator. Therefore, $H$ cannot be compact.

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 So your $H$ is the questions's $T$... :) – Mariano Suárez-Alvarez Mar 14 2010 at 20:59