The “supremum” norm

Today I shall speak on a topic that I feel is important.

All of us have encountered the “ $\sup$ ” norm in Functional Analysis. In $C[a,b]$ , $\|f_1-f_2\|=\sup|f_1x-f_2x|$ for $x\in X$ . In the dual space $B(X,Y)$ , $\|T_1x-T_2x\|=\sup\limits_{x\in X}\frac{\|(T_1-T_2)x\|}{\|x\|}$ . What is the utility of this “sup” norm? Why can’t our norm be based on “inf” or the infimum? Or even $\frac{\sup+\inf}{2}$ ?

First we’ll talk about $C[a,b]$ . Let us take a straight line on the $X-Y$ plane, and a sequence of continuous functions converging to it. How do we know they’re converging? Through the norm. Had the norm been $\inf$ , convergence would only have to be shown at one point. For example, according to this norm, the sequence $f_n=|x|+\frac{1}{n}$ converges to $y=0$ . This does not appeal to our aesthetic sense, as we’d want the shapes of the graphs to gradually start resembling $y=0$ .

Now let’s talk about $B(X,Y)$ . If we had the $\inf$ norm, then we might not have been able to take every point $x\in X$ and show that the sequence $(T_n x), n\in\Bbb{N}$ is a cauchy sequence. So what if we cannot take every $x\in X$ and say $(T_n x)$ is a cauchy sequence? This would crush the proof. How? Because then $\lim\limits_{n\to\infty} T_n$ would not resemble the terms of the cauchy sequence at all points in $X$ , and hence we wouldn’t be able to comment on whether $\lim\limits_{n\to\infty} T_n$ is linear for all $x\in X$ or not. Considering that $B(X,Y)$ contains only bounded linear operators, the limit of the cauchy sequence $(T_n)$ not being a part of $B(X,Y)$ would prove that $B(X,Y)$ is not complete. Hence, in order for us to be able to prove that $\lim\limits_{n\to\infty} T_n$ is linear and that $B(X,Y)$ is complete, we need to use the $\sup$ norm.