# Are the norms on a vector space unique?

I was watching an online lecture on bounded linear transformations $$T: \mathcal{C}[a,b] \rightarrow \mathcal{C}[a,b]$$

So the condition for $T$ to be bounded was that for all $f \in \mathcal{C}[a,b]$, there exists some $k >0$ such that $$||Tf|| \leq k ||f||$$

In the video, a student asked

What norm are we using here?

To which the lecturer responded

When working with functions in $\mathcal{C}[a,b]$, we always use the infinity norm, that is implied. But sure I can add in this.

And then the lecturer modified it to

$$||Tf||_\infty \leq k ||f||_\infty$$

The way that the lecturer responded implied to me that the norm was, in some sense, unique. Or that this is the 'best' norm to use.

Why is it that this is the norm for continuous functions on an interval?

Also, I had a quick side-question somewhat related to this. I hear the expressions "infinity norm" and "supremum norm". Are they more or less the same thing?

• No, the norms are not all equivalent, but the infinity norm is the default norm just like the $2$-norm is the default for euclidean spaces. – 5xum Jun 8 '16 at 12:29

The norm $\|\cdot\|_\infty$ is quite special for the space $C([a,b])$, since this space is complete with respect to this norm. In fact, the convergence in norm $\|\cdot\|_\infty$ implies the uniform convergence of functions in $C([a,b])$ and the uniform limit of continuous function is a continuous function. If one consider other norms, for example the norm "2": $$\|f\|_2=\sqrt{\int_a^b |f(x)|^2 dx},$$ than it can be shown that $(C([a,b]), \|\cdot\|_2)$ is not complete.