# How to determine the supremum of a set with respect of all vectors of norm 1?

I'm having some troubles regarding the definitions that use the supremum of a set for all vectors with a specific norm.

At the moment the case in question is the operator norm, defined as $$\|T\|_{X'} = \sup\{|T(x)|: \|x\|_X=1\}$$, for some continuous linear operator $$T$$ with values in $$\mathbb{R}$$ and defined on a normed space $$X$$. (when I write $$X'$$ i mean $$\mathcal{L}_c(X,\mathbb{R})$$, instead of $$X^*$$)

Let's say now that I'm asked to find the norm of the operator $$T$$. Here I have problems, I don't really know how to even begin to set up a proof that works with all the vectors of norm 1. Can I always take a vector $$v$$ and consider $$\frac{v}{||v||}$$ instead? (i've taken the norm 1 as example but this question applies to all other equivalent definitions).

I know that $$|T(v)| \le \|T\|_{X'}\|v\|_X$$, let's say I find that $$|T(v)|\le M \|v\|_X$$ for some $$M>0$$ and some generic vector $$v$$.

How do I prove/disprove that $$M$$ is the supremum? Going through all possible vectors of norm 1, seems cumbersome enough in finite dimension spaces.

Are there some general approaches when tackling this kind of definitions? (ae using limits and whatnot).

In short, my issue with inequalities is making sure that i'm close enough and there isn't nothing in between that I'm missing, and using the raw definition of supremum with $$\epsilon$$ most time is just infeasible.

I'm sorry for the rambling.

• Of course you can't go through all vectors of norm 1 - there are infinitely many of them. In general you won't be able to avoid using the definition of supremum. Of course it helps if all the inequalities you used to get $M$ were sharp. – MaoWao Feb 13 '19 at 11:09
• @MaoWao If i managed to get $M$ with sharp inequalities (which i suppose mean non-strict ones) would then that mean that i can't find a smaller bound? Say i found it using only generic vectors and their properties, then there must be at least a vector with those properties that satisfy the equality part of $\le$, is this correct? – WhiteEyeTree Feb 13 '19 at 11:22

I guess the best way to do this is by giving simple examples:

1) Assume $$X=\mathbb{R}^n$$, $$\lVert x\rVert = \sum_{i=1}^n\lvert x_i\rvert$$ ($$l_1$$-Norm), $$T:\mathbb{R}^n \rightarrow \mathbb{R}$$, $$T(x)=x_1$$.

Assuming $$\lVert x\rVert = 1$$, this means $$\sum_{i=1}^n\lvert x_i\rvert=1$$ and henceforth $$\lvert x_1\rvert \le \sum_{i=1}^n\lvert x_i\rvert=1$$. So you know that

$$\lvert T(x) \rvert \le 1 \text{, for }\lVert x\rVert = 1$$

You have found your $$M$$, in this case $$M=1$$. So you know that $$\lVert T \rVert_{X'} \le 1$$.

To check if equality holds, try to find an $$x\in X, \lVert x\rVert = 1$$ such that the equality holds in $$\lvert x_1\rvert \le \sum_{i=1}^n\lvert x_i\rvert$$. That would be $$x_e=(1,0,\ldots,0)$$.

So you have $$\lvert T(x_e)\rvert=\lvert 1\rvert=1$$. Since you have now found an $$x\in X, \lVert x\rVert = 1$$ (namely $$x_e=(1,0,\ldots,0)$$) with $$\lvert T(x) \rvert=1$$, you know that $$\lVert T \rVert_{X'} \ge 1$$.

So alltogether, you have $$\lVert T \rVert_{X'} = 1$$.

2) Assume $$X=\mathbb{R}^n$$, $$\lVert x\rVert = \max_{i=1}^n\lvert x_i\rvert$$ ($$l_\infty$$-Norm), $$T:\mathbb{R}^n \rightarrow \mathbb{R}$$, $$T(x)=\sum_{i=1}^nx_i$$.

Assuming $$\lVert x\rVert = 1$$, this means $$\max_{i=1}^n\lvert x_i\rvert=1$$, henceforth $$\lvert \sum_{i=1}^nx_i \rvert \le \sum_{i=1}^n\lvert x_i \rvert \le \sum_{i=1}^n1 = n$$. So you know that

$$\lvert T(x) \rvert \le n \text{, for }\lVert x\rVert = 1$$

You have found your $$M$$, in this case $$M=n$$. So you know that $$\lVert T \rVert_{X'} \le n$$.

To check if equality holds, try to find an $$x\in X, \lVert x\rVert = 1$$ such that the equality holds in $$\lvert \sum_{i=1}^nx_i \rvert \le \sum_{i=1}^n1$$. That would be $$x_e=(1,1,\ldots,1)$$. Just as in case 1), you now conclude that $$\lVert T \rVert_{X'} = n$$.

Trying to find this M and trying to find the right $$x_e$$ is not simple and cannot be 'automated'. Just like most creative parts of math, getting a feeling for what might work comes with experience.

• The problem is that in general, the supremum is not attained and thus "the right $x_e$" may not exist at all. – MaoWao Feb 13 '19 at 12:21
• Then you need to find a sequence of $x_e^{(k)}$ such that $\lvert T(x_e^{(k)}) \rvert \to M$. The basic idea is the same: Find where your inequalities turn into equalities, at least in the limit. – Ingix Feb 13 '19 at 12:27
• Thank you, albeit these things can't be automated having some proof of concept to better understand the application of the notions you learn really helps me a lot to generalise to other problems. If i understand what's going on in simpler spaces i can at least spot the differences in stranger ones. – WhiteEyeTree Feb 13 '19 at 12:51