The “supremum” norm

by ayushkhaitan3437

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.