Suppose we have an Hilbert space $X$ a compact operator $T$ and an operator $S$ such that $TT^*-SS^*\geq 0$, then $S$ will be a compact operator. Using the condition I can see that $||Tx||\geq ||Sx|| ,\forall x\in X$, and we know that $T(B_X(0,1))$ is relative compact so if I can show that $S(B_X(0,1))\subset T(B_X(0,1)).$ we will get that $S$ is compact, now I can't seem to see why we would get that inclusion of the sets, so any hint is aprecciated , thanks in advance.
$\begingroup$
$\endgroup$
12
-
$\begingroup$ If $(y_n)$ is a sequence in $S(B_X(0,1)$ then, for every $n\in \mathbb{N}$ there is a $z_n\in T(B_X(0,1))$ s.t. $\| y_n\| \leq \| z_n\|$, by the inequality you proved. Now use that $T(B_X(0,1))$ is compact. Also, in your attempt, wouldn't you also need $S(B_X(0,1)$ to be -not just subset- but closed subset? $\endgroup$– alphaomegaCommented Jun 14, 2020 at 10:55
-
$\begingroup$ Hm ok that does prove me that the operator is compact, but will it be true that the image of $S$ is contained in the image of $T$? @alphaomega $\endgroup$– SomeoneCommented Jun 14, 2020 at 10:58
-
2$\begingroup$ @something To see that the inequality $\|Tx\| \geq \|Sx\|$ for all $x$ need not imply the inclusion of those images, consider the case $X = \mathbb{R}^2$ with orthonormal basis $\{e_1, e_2\}$. Let $T e_1 = 2 e_1, Te_2 = e_2$ and $S e _1 = 2 e_2, S e_2 = e_1$. Then $S$ is a rotation of $T$ so that $\|Tx \| = \|Sx \|$ for all $x$ but $Se_1 = 2 e_2 \not \in T(B(0,1))$. $\endgroup$– Rhys SteeleCommented Jun 14, 2020 at 11:08
-
2$\begingroup$ @Something That's wrong. If it were right it would show that every norm $1$ sequence converges since if $\|y_n\| = 1$ then $\|y_n\| \leq \|y_1\|$ and $(y_1)_{n \geq 1}$ is a constant sequence and thus converges. The problem is you don't know that $\|y_n - y_m\| \leq \|z_n - z_m\|$. $\endgroup$– Rhys SteeleCommented Jun 14, 2020 at 11:16
-
1$\begingroup$ @alphaomega No but you do need something stronger than $\|y_n\| \leq \|z_n\|$. My next comment gives an example where $z_n$ is convergent and $\|y_n\| \leq \|z_n\|$ but you can even take $y_n$ to be a sequence with no convergent subsequence. $\endgroup$– Rhys SteeleCommented Jun 14, 2020 at 11:23
|
Show 7 more comments
1 Answer
$\begingroup$
$\endgroup$
3
Recall that a bounded linear operator is compact if and only if its adjoint is compact (Schauder's Theorem). I will prove that $S^*$ is compact.
Let $x_n$ be a bounded sequence. We want to show that $S^*$ has a convergent subsequence. To do this, note that $TT^* - SS^* \geq 0$ implies that $\|T^*(x_n - x_m)\| \geq \|S^*(x_n - x_m)\|$ for every $n, m > 0$. Now $T^*$ is compact so that $T^* x_n$ has a convergent subsequence, $T^* x_{n_k}$.
In particular, $T^* x_{n_k}$ is Cauchy. This implies that $S^*x_{n_k}$ is Cauchy since $$\|S^* x_{n_k} - S^* x_{n_j}\| \leq \|T^* x_{n_k} - T^* x_{n_j}\|$$ and hence $S^*x_{n_k}$ converges. Hence $S^*$ is compact.
answered Jun 14, 2020 at 11:25
-
$\begingroup$ @something I use Schauder's Theorem only because $TT^* - SS^* \geq 0$ means that $\langle T^*x, T^*x \rangle - \langle S^*x, S^*x \rangle \geq 0$. In the question you say you have proved that $\|Sx\| \leq \|Tx\|$ for all $x$. If so, you can replace $S^*$ with $S$ and similarly for $T$ in this answer and remove the use of Schauders Theorem. If I don't miss something, this is not the inequality you get from the assumption though (Maybe I do miss something or maybe you accidentally switched the order to $T^*T - S^*S$). $\endgroup$ Commented Jun 14, 2020 at 11:29
-
1$\begingroup$ Yeah maybe , just one thing in your proof you wrote that you want to show that $SS^*$ has a convergent subsequence , shouldn't it be $S^*$? $\endgroup$– SomeoneCommented Jun 14, 2020 at 11:32
-
1$\begingroup$ @Something Yes! Typo now fixed. $\endgroup$ Commented Jun 14, 2020 at 11:33