# Banach-Steinhaus (Uniform-boundedness theorem) application

Let $$X$$ and $$Y$$ be Banach spaces. Consider a family of linear bounded operators $$\{L_{\alpha}\}_{\alpha \in J} \subset \mathcal{B}(X,Y)$$ where $$J \neq \emptyset$$ is a given subset of $$[0, \infty)$$. Prove that if there is an open non-empty set $$A \subset X$$ such that, for any $$x \in A$$, $$sup_{\alpha \in J} \lVert L_{\alpha} x \rVert _{Y}$$ is bounded then there exists $$M > 0$$ such that

$$sup_{\alpha \in J} \lVert L_{\alpha} \rVert _{\mathcal{B}(X,Y)} \leq M$$

Does this also hold for closed non-empty sets?

• No. $A$ could be a singleton set, for example. – David Mitra Jan 19 at 9:07

By the assumption, there exists an open ball $$B(x_0,\delta)=\{x\in X:\|x-x_0\|<\delta\}$$ such that $$A(x):=\sup_{\alpha\in J}\|L_\alpha(x)\|_Y<\infty, \quad\forall x\in B(x_0,\delta).$$ Let $$x \in X$$ be given. We can see that $$z_x:=\frac{\delta x}{2\|x\|}+x_0\in B(x_0,\delta)$$. Hence, we have $$\frac{\delta }{2\|x\|}\Big\|L_\alpha\left(x\right)\Big\|=\Big\|L_\alpha\left(\frac{\delta x}{2\|x\|}\right)\Big\|\le \Big\|L_\alpha\left(z_x\right)\Big\|+\Big\|L_\alpha\left(x_0\right)\Big\|\le A(z_x)+A(x_0)$$ for all $$\alpha\in J$$. This gives $$A(x)\le \frac{2\|x\|}{\delta}\left(A(z_x)+A(x_0)\right)<\infty,\quad\forall x\in X.$$ By Banach-Steinhaus theorem, it follows that $$\sup\limits_{\alpha\in J}\|L_\alpha\|<\infty$$.
If $$A$$ is closed, then the statement is false. For example, we can take $$A=\{0\}$$ since $$T(0)=0$$ for any bounded linear operator $$T$$, but $$\{kL\}_{k\in\Bbb N}$$ is not bounded for $$L\ne 0$$.