# If T and S are bounded linear operators and T is compact and S*S<T*T, show that S is compact.

Exercise 14 section 9.3 Erwin Kreyszig. I don't know how to prove it, please help.

## 2 Answers

$$S^{*}S< T^{*}T$$ implies $$\|Tx\|^{2} \geq \|Sx\|^{2}$$ for all $$x$$. Let $$\{x_n\}$$ be a bounded sequnece. There exists $$n_k$$ increasing to $$\infty$$ such that $$Tx_{n_k}$$ converges It follows that $$\|Sx_{n_k}-Sx_{n_j}\| \leq \|Tx_{n_k}-Tx_{n_j}\| \to 0$$. Thus $$(Sx_{n_k})$$ is Cauchy, hence convergent. This proves that $$S$$ is compact.

[$$S^{*}S means $$\langle S^{*}Sx, x \rangle \leq \langle T^{*}Tx, x \rangle$$ which means $$\langle Sx, Sx \rangle \leq \langle Tx, Tx \rangle$$ or $$\|Sx||^{2} \leq \|Tx\|^{2}$$ for all $$x$$].

Kavi's answer is the straightforward one and it should be accepted. Here I want to show a less obvious way of proving it.

From this answer we see that there exists a contraction $$C$$ such that $$(S^*S)^{1/2}=C(B^*B)^{1/2}$$. Now write $$S=V(S^*S)^{1/2}$$ via the Polar Decomposition. Then $$S=VC(T^*T)^{1/2},$$ and as $$T$$ is compact, so is $$T^*T$$, so is $$(T^*T)^{1/2}$$, and thus so is $$S$$.