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I am currently in the phase of my class on the analysis of metric spaces where we have transitioned to studying the normed vector space $B(V,W)$ — the space of all continuous linear maps $T : V \to W$ $($where $(V,||\cdot||_{V})$ and $(W, \|\cdot\|_{W})$$)$.

With this new transition, we have also considered the operator norm on $B(V,W)$, $$\|T\|_{B(V,W)} = \sup\limits_{\|x\|_{V} \neq 0} \frac{\|Tx\|_{W}}{\|x\|_{V}}.\tag1$$ While this is clearly a norm on $B(V,W)$, I have seen other authors write this norm as $$\|T\|_{B(V,W)} = \sup_{\|x\|_{V} = 1}\|Tx\|_{W}$$ or $$\|T\|_{B(V,W)} = \sup\{\|Tx\|_{W} : |\|x\|_{V} =1\}.$$ How is this the same as $(1)$? I understand that the only requirement here is that the $\|x\|_{V} \neq 0$, but I also understand that we can write $\|T\|_{B(V,W)}$ as $$\inf\{C : \|Tx\|_{W} \leq C\|x\|_{V}\quad\forall x \in V\}.$$ In other words,

$\|T\|_{B(V,W)}$ satisfies $\|Tx\|_{W} \leq \|T\|_{B(V,W)}\|x\|_{V}$, and $\|T\|_{B(V,W)}$ is the smallest constant satisfying this inequality

according to my textbook $($N.L. Corothers' Real Analysis$)$.

Note: I know that $B(V,W)$ may be written as $L(V,W)$, but I am just referring to the notation in my book.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Taylor Rendon
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1 Answers1

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The function $$ x \mapsto \frac{\|Tx\|}{\|x\|} $$ is invariant on the span of $x$ (sans zero) because $T$ is linear. (Exercise: check this.)

So taking the supremum of this function over all $x\in V$ is the same as taking the supremum over the unit circle.

More formally,

$$\sup_{x\in V} \frac{\|Tx\|}{\|x\|} = \sup_{x\in V} \bigg\|T\bigg(\frac{x}{\|x\|}\bigg)\bigg\| = \sup_{\|x\|=1}\|Tx\|$$

Neal
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