Show a linear operator is not compact - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:39:12Z http://mathoverflow.net/feeds/question/18155 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18155/show-a-linear-operator-is-not-compact Show a linear operator is not compact gylns 2010-03-14T08:56:09Z 2010-03-14T20:31:12Z <p>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?</p> http://mathoverflow.net/questions/18155/show-a-linear-operator-is-not-compact/18159#18159 Answer by anton for Show a linear operator is not compact anton 2010-03-14T10:42:48Z 2010-03-14T10:42:48Z <p>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.</p> <p>It is natural to look for an example around $x=0$ since that is where the kernel of $T$ fails to be $L^2$.</p> http://mathoverflow.net/questions/18155/show-a-linear-operator-is-not-compact/18161#18161 Answer by Willie Wong for Show a linear operator is not compact Willie Wong 2010-03-14T11:22:24Z 2010-03-14T12:31:31Z <p>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 $f_j(x) = \frac{d}{dx} (x F_j(x) )$. </p> <p>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 $L^2$.</p> <p>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.</p> <p>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 <code>$\mathbb{R}_{+}$</code> on <code>$L^2(\mathbb{R}_+)$</code>, where the group operation for <code>$\mathbb{R}_{+}$</code> is multiplication. </p> <p>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 <a href="http://www.google.com/buzz/114134834346472219368/9UseDXTJN74/There-are-three-ways-that-sequential-compactness" rel="nofollow">http://www.google.com/buzz/114134834346472219368/9UseDXTJN74/There-are-three-ways-that-sequential-compactness</a> a short while back.</p> <p>This is, of course, closely related to the notion of concentration compactness. </p> http://mathoverflow.net/questions/18155/show-a-linear-operator-is-not-compact/18208#18208 Answer by Ady for Show a linear operator is not compact Ady 2010-03-14T20:31:12Z 2010-03-14T20:31:12Z <p>Let { $L_{n}$ } be the sequence of Laguerre polynomials, and let us define </p> <p>$e_{n}(t)=\dfrac{L_{n}(\ln t)}{t}$ $\cdot\chi_{\left(1,\infty\right)}\left(t\right)$ $(n\in\mathbb{N\textrm{, t > 0}})$ . 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. </p> <p>See also <a href="http://revistas.ucm.es/mat/11391138/articulos/REMA0606220467A.PDF" rel="nofollow">revistas.ucm.es/mat/11391138/articulos/REMA0606220467A.PDF</a>.</p>