Condition for an operator to be compact

Question

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.

Answer 1

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.