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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?

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  • $\begingroup$ No. $A$ could be a singleton set, for example. $\endgroup$ – David Mitra Jan 19 at 9:07
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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 $L\ne 0$.

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