What's the difference between the operator norm and the sup norm over $C[0,1]$. a.k.a $\left\lVert x\right\rVert_\infty$ vs $\left\lVert x\right\rVert_{op}$
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1$\begingroup$ What do you mean by 'operator norm' in this context? $\endgroup$– BerciMay 21, 2019 at 14:59
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1 Answer
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These are two different norms for entirely different purposes.
The supremum norm over $C[0,1]$ is the norm of this particular Banach space. We have that $\|\cdot\|_{\infty}:C[0,1]\to[0,\infty)$ and this norm measures the size of a continuous function.
The operator norm is used for bounded linear operators that map the elements of one normed space to another normed space. For example, if $L$ is a bounded linear operator from $X$ to $Y$, where $X$ and $Y$ are two normed spaces, then the operator norm is defined as the smallest $M$ such that $\|Lv\|_Y\le M\|v\|_X$ (see here for more details). The operator norm measures the size of an operator.
I hope this is helpful.
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$\begingroup$ So whats the difference between $\left\lVert f-g\right\rVert_\infty$ and $\left\lVert f-g\right\rVert_{op}$ for $f,g\in C[0,1]$? $\endgroup$– EliTMay 21, 2019 at 15:10
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$\begingroup$ @ElkanaTovey Writing $\|f-g\|_{op}$ does not make sense. The input of $\|\cdot\|_{op}$ is an operator and $f-g$ is a continuous function. $\endgroup$– Cm7F7BbMay 21, 2019 at 15:11
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$\begingroup$ What if I considered $f-g$ as a bounded operator? $\endgroup$– EliTMay 21, 2019 at 15:13
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