# Condition for an operator to be compact

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.

• 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? Commented Jun 14, 2020 at 10:55
• 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 Commented Jun 14, 2020 at 10:58
• @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 $Se_1 = 2 e_2 \not \in T(B(0,1))$. Commented Jun 14, 2020 at 11:08
• @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 $\|y_n - y_m\| \leq \|z_n - z_m\|$. Commented Jun 14, 2020 at 11:16
• @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. Commented Jun 14, 2020 at 11:23

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.
• @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 $T^*T - S^*S$). Commented Jun 14, 2020 at 11:29
• 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 $S^*$? Commented Jun 14, 2020 at 11:32