What's the difference between the operator norm and the sup norm

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 $$\left\lVert x\right\rVert_{op}$$

• What do you mean by 'operator norm' in this context? – Berci May 21 at 14:59

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.

• So whats the difference between $\left\lVert f-g\right\rVert_\infty$ and $\left\lVert f-g\right\rVert_{op}$ for $f,g\in C[0,1]$? – EliT May 21 at 15:10
• @ElkanaTovey Writing $\|f-g\|_{op}$ does not make sense. The input of $\|\cdot\|_{op}$ is an operator and $f-g$ is a continuous function. – Cm7F7Bb May 21 at 15:11
• What if I considered $f-g$ as a bounded operator? – EliT May 21 at 15:13
• @ElkanaTovey How would you define it? – Cm7F7Bb May 21 at 15:13
• $f,g\in C[0,1]$ such that they are bounded – EliT May 21 at 15:14