So last week we talked about the Baire Category Theorem, and this important tool will give us three equivalent theorems about the qualitative properties of operators on Banach spaces. Today we will begin with the first theorem, uniform boundedness. The proof of this theorem is quite simple, and is just an application of Baire’s theorem.

Theorem 1 (Uniform Boundedness) Let {X} be a Banach space, {Y} a normed vector space, and {\{A_{\lambda}\}_{\lambda\in\Lambda}\subset L(X,Y)} for some index set {\Lambda}. If

\displaystyle \sup_{\lambda\in\Lambda}\norm{A_{\lambda}x}_Y<+\infty ~\forall x\in X,


\displaystyle \sup_{\lambda\in\Lambda}\norm{A_{\lambda}}_{L(X,Y)}<+\infty .

Proof: We first define the sets

\displaystyle E_n=\{x\in X: \norm{A_{\lambda}x}\leq n~ \forall \lambda\in\Lambda \}

and observe that they are closed. Indeed, take a sequence {\{x_k\}_{k\i\mathbb{N}}} with {x_k\rightarrow x\in X}. Then it follows that

\displaystyle \norm{Ax}_X=\lim_{k\rightarrow\infty}\norm{Ax_k}\leq n.

Furthermore, by the point wise boundedness condition

\displaystyle X=\bigcup_{n\in\mathbb{N}} E_n,

and so by Baire there exists an {n_0\in\mathbb{N}} and a ball {B_r(x_0)} such that

\displaystyle B_r(x_0)\subset E_{n_0}.

The claim is shown once we observe that

\displaystyle \sup_{\lambda\in\Lambda}\norm{A_{\lambda}}=\sup_{\lambda\in\Lambda}\sup_{\norm{x}\leq 1}\norm{A_{\lambda}x}\leq \norm{\frac1r A_{\lambda}rx-x_0}+\norm{\frac1r A_{\lambda}x_0}\leq \frac{2n_0}{r}.


Whenever we are faced with a theorem, it is natural to ask, what happens if we drop an assumption. In this case, what happens if we drop the completeness of {X}? It turns out the outcome is not pleasant. Consider the space {c_{00}=\{x\in\ell^{\infty}\vert~\exists N\in\mathbb{N}: x_n =0~ \forall n\geq N \}.} This space is not complete (find the example yourself!). Now consider {A_n x=nx_n} and note that they are point wise bounded but not uniformly bounded since {\norm{A_n}=n}.

An application of the uniform boundedness theorem is that if a sequence of continuous linear operators converge point wise, then the limit itself is a continuous linear operator; indeed,

\displaystyle \norm{A}\leq \liminf_{n\rightarrow\infty} \norm{A_n}\leq C.

However, this does not mean that the sequence converges uniformly. Take for example {A_n\in L(\ell^2)} where {A_n x = (0,0,\dots,0,x_n,0,\dots)}. We easily see that {A_n\rightarrow 0} point wise as {x_n\rightarrow 0} for an {x\in\ell^2}. However,

\displaystyle \norm{A_n-0}_{L(\ell^2)}=\norm{A_n}_{L(\ell^2)}=1.

That’s all I have time for today, we have this new tool in our toolbox, let’s see when we can use this!

%d bloggers like this: