# uniform boundness principle (Banach- Steinhaus)

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.

• 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/… – ktoi Jan 10 at 17:52
• 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 $x$. – Rellek Jan 10 at 17:55

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 $$\|T_i\| for all $$i$$.
• to prove that if $$\{T_nx\}$$ converges for all $$x\in E$$, then $$Tx=\lim T_nx$$ defines a bounded operator.
• to prove the spectral radius formula: for $$T\in B(E)$$, $$\operatorname{spr}(T)=\lim_{n\to\infty}\|T^n\|^{1/n}.$$