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The definition I have for the operator norm is:
$L(\Bbb U,\Bbb V)$ is the vector space of all linear transformations $f: \Bbb U \to \Bbb V$. Let $\Bbb U\ \&\ \Bbb V$ be inner product spaces. Define the operator norm on $L(\Bbb U,\Bbb V)$: $$|f|_O=\max_{|u| \le1} |f(u)|$$

First, I need to show that this IS a norm. I can prove all of the axioms except: If $f \neq 0$ then $|f|_O \gt 0$.
In particular, why couldn't we have a linear function f with the property that $f(u)=0$ for some interval (like $0 \le u \le 1$) and then nonzero outside of that interval? I don't see how that is not possible from $f(au)=af(u)$ or $f(u+v)=f(u)+f(v)$.

Second, I need to show that $|f(u)| \le |f|_O|u|$.
To me this looks like $|f|_O|u|=(max_{|u| \le1} |f(u)|)|u|=max_{|u| \le1} (|f(u)||u|)$ which I would think would be LESS than $|f(u)|$.

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  • $\begingroup$ Regarding your first doubt: If $f(u) \ne 0$ then what about $f(u/|u|)$? $\endgroup$ – user50948 Jun 5 '14 at 22:15
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If $f(u)=0$ for all $u$ with $|u|=1$, it is easy to show that $f(v)=0$ for all $v$. Hint: take $u=\frac{v}{|v|}$.

Regarding the second question, you made a mistake when writing $$\max_{|u|\leq 1}(|f(u)|)|u| = \max_{|u|\leq 1}(|f(u)||u|)$$

What you should write is $$\max_{|u|\leq 1}(|f(u)|)|u| = \max_{|v|\leq 1}(|f(v)||u|)$$

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