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Let $T \in L(X,Y), S \in L(Y, Z)$, where $X,Y,Z$ are normed vector spaces. Show that $\|S \circ T \| \leq \|S\|\|T\|$.

I tried to show that $\|\frac{S\circ T}{S}\|\leq \|T\|$, but I do not know how to evaluate $\|\frac{S\circ T}{S}\|$.

Thank you all.

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    For any $x\in X$, $\|STx\|\le \|S\|\|Tx\|\le \|S\|\|T\|\|x\|$. – 2017-01-23
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    @CheeHan I did not quite get it. Can you tell me more? – 2017-01-23
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    Do you mean $\|S\circ T\mathbf{x}\|\leq \|S\|\|T\mathbf{x}\|$? – 2017-01-23
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    I am assuming $T\in L(X,Y)$ means $T$ is a bounded linear operator from $X$ to $Y$. If so, then the following inequality holds: $\|Tx\|\le \|T\|\|x\|$ for any $x\in X$. – 2017-01-23
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    Yes, $STx$ means $S(Tx)$ :) – 2017-01-23
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    @CheeHan Yes I have already proved the statement you stated. – 2017-01-23
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    Great. The result is then obtained by simply using the stated inequality. – 2017-01-23
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    Since $S$ is a bounded linear operator from $Y$ to $Z$, and $Tx$ is an element in $Y$, it satisfies $\|S(Tx)\|\le \|S\|\|Tx\|$, but since $x\in X$ and $\|Tx\|\le \|T\|\|x\|$, we have $\|S(Tx)\|\le \|S\|\|T\|\|x\|$. Taking the supremum over all $x\in X$ with norm 1 yields the desired inequality $\|ST\|\le \|S\|\|T\|$. – 2017-01-23

2 Answers 2

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Using a definition $$ \|S\|=\sup_{\|x\|=1}\|Sx\|=\sup_{x\ne 0}\frac{\|Sx\|}{\|x\|}, $$ and a property, derived by the definition $$ \|Sx\|\le \|S\|\|x\|, $$ one obtains that, for every $x\in X$, $$ \|STx\|=\|S(Tx)\|\le \|S\|\|Tx\|\le \|S\|\|T\|\|x\|. $$ Hence $$ \|ST\|=\sup_{\|x\|=1}\|STx\|\le \|S\|\|T\|. $$

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See any book on Functional Analysis. In particular, you will understand that $||\frac{S\circ T}{S}||,$ generally, has no sense.