Uniform boundedness principle on dense subspace

Question

In my thesis I encountered the following problem: Let $X,Y$ Banach spaces and $Z\subset X$ a norm-dense subspace. Suppose we have operators $\left\{T_n:n\in\mathbb{N}\right\}$ such that for all $x\in Z$ it holds that $\left\{\left\lVert T_nx\right\rVert:n\in\mathbb{N}\right\}$ is bounded. Can I somehow use the Uniform Boundedness principle to conclude that $\left\{\lVert T_n\rVert:n\in\mathbb{N}\right\}$ is bounded?

Thanks in advance!

Not without further assumptions. If $Z$ is barrelled, you can. But incomplete normed spaces are rarely (if ever) barrelled. — Daniel Fischer, Oct 05 '16 at 11:08
@DanielFischer Do you have a ref. for this? I am mainly interested in case of RKHSs. — Mohammad Khosravi, May 05 '21 at 19:13

score 7 · Accepted Answer · answered Oct 05 '16 at 12:30

It may not holds, even if $X$ is a separable Hilbert space and $Y=\mathbb R$. Indeed, consider a Hilbert basis $\left(e_n\right)_{n\geqslant 1}$ and $Y$ the vector space of (finite) linear combinations of $e_n$'s. Define $T_n(x):=n\langle x,e_n\rangle$. Then for each $x\in Y$, write $x=\sum_{j=1}^N\langle x,e_j\rangle e_j$. Then $T_n(x)=0$ if $n\gt N$ hence $\sup_n \left|T_n(x)\right|$ is finite. But $\lVert T_n\rVert=n$.

Uniform boundedness principle on dense subspace

1 Answers1