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My definition:

Let $(V,||\cdot||),(W,||\cdot||)$ be normed spaces over $\mathbb{F}$.

Let $T:V\rightarrow W$ be a continuous linear transformation.

Then $||T||_{op}\triangleq \sup\{||Tv||_W:v\in V, ||v||_V=1\}$.

Is this the standard definition of the operator norm?

If not, how far can i extend the domain of this definition? That is, what would be a concept that would replace "continuity" in this definition?

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    $\begingroup$ I believe I remember continuity being equivalent to possessing a finite operator norm. $\endgroup$ – Dustan Levenstein Mar 11 '14 at 4:31
  • $\begingroup$ This is the standard definition. Note that $T$ is continuous at each point of $X$, or at each point it is not. Moreover $T$ is continuous if and only if it is bounded, i.e. if there exists $M \in R^+$ such that $||T(x)||\leq M||x||$. $\endgroup$ – Bento Mar 11 '14 at 4:47
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    $\begingroup$ Note that you need $\|v\| \leq 1$ instead of $\|v\| = 1$ if you want the formula to continue to work in the case that $V = 0 = \{0\}$ is the zero space. $\endgroup$ – Ingo Blechschmidt Mar 11 '14 at 13:37
  • $\begingroup$ Infact $\sup$ is defined on sets $\neq \varnothing$, and in the case $V=\{0\}$, being $||0||=0$, $\{v : ||v||=1\}=\varnothing$. $\endgroup$ – Bento Mar 11 '14 at 15:27
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If you do not assume continuity, for what I have said in the comment, you can not put this definition, that is, $T$, not being bounded (on $X$) and discontinuous at each point in $||x|| \leq 1$, will not have in general the $\sup$.

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