I am a beginner of functional analysis and I can't understand at all the Banach Steinhaus theorem:
Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i \in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F.$ Assume that
$$ \sup_{i \in I} \lVert T_ix \rVert<\infty \quad \forall x \in E. $$
then
$$ \sup_{i \in I} \lVert T_i \rVert< ∞ .$$
(from Brezis page 32)
my question is : if the family consists of a finite number of linear bdd operators the hypothesis:
$$\sup_{i \in I} \lVert T_ix \rVert < \infty \quad \forall x \in E$$
is always verified, isn't it?
and more generally, what is this theorem telling me?
I apologize for the banality of my question but I can't fully understand this theorem.
Best Answer
Yes, if the family is finite, then $\sup_i\|T_ix\|=\max\{\|T_1x\|,\ldots,\|T_nx\|\}<\infty$, and also $\sup_i\|T_i\|=\max\{\|T_1\|,\ldots,\|T_n\|\}<\infty$. The theorem is relevant when the family is infinite.
What they theorem says it what it says, there's no much philosophy there: if your family $\{T_i\}$ is bounded "pointwise" (at every $x$), then it is bounded uniformly: there exists $c>0$ such that $\|T_i\|<c$ for all $i$.
The theorem has many applications. Typical ones are
to prove that if $\{T_nx\}$ converges for all $x\in E$, then $Tx=\lim T_nx$ defines a bounded operator.
a weakly bounded set in a Banach space is norm-bounded; in particular, a weakly convergent sequence is bounded.
to prove the spectral radius formula: for $T\in B(E)$, $$\operatorname{spr}(T)=\lim_{n\to\infty}\|T^n\|^{1/n}.$$