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

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$S^{*}S< T^{*}T$ 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.

[$S^{*}S<T^{*}T$ 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 $x$].

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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 $S$.

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