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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.

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  • $\begingroup$ Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – ktoi Jan 10 at 17:52
  • $\begingroup$ There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $\implies$ uniform boundedness for families of bounded linear operators; that is, for each $x \in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$. $\endgroup$ – Rellek Jan 10 at 17:55
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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 $T\in B(E)$, $$\operatorname{spr}(T)=\lim_{n\to\infty}\|T^n\|^{1/n}.$$

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