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Let $X$ and $Y$ be Banach spaces, $T_j \in L(X,Y)$ for each $j$ and let $E_n = \left\lbrace x \in X: \sup_{j \geq 1} \|T_jx\| \leq n\right\rbrace$.

  1. Show $E_n$ is closed for each $n$
  2. If there exists $x_0 \in X$ such that $\sup\limits_{j \geq 1} \|T_jx_0\| = \infty$, then each $E_n$ is nowhere dense in $X$.

Here $L(X,Y)$ is the set of bounded linear operators from $X$ to $Y$.

First fix $n$ and a sequence $\left\lbrace x_k\right\rbrace_{k=1}^\infty \in E_n$. I need to show that $x = \lim\limits_{k \rightarrow \infty} x_k$ is in $E_n$. I know that $\|T_j(x_k)\| \leq n$ for each $k$. Since $T_j$ is continuous and $Y$ is Banach, then it's complete so $\lim_k Tj(x_k) = T_j(x) \in Y$

I feel like I'm running around in circles on this problem and don't see an obvious reason why $x \in E_n$. It should have something to do with the continuity of $T_j$, but I'm unable to see why.

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  • $\begingroup$ I think we need to look at the function space and show that the sup || T || is a norm in this space and that it meets definition of a normed vector space. Then prove this function space is also complete. Incidentally taking a class just like this so very interested in the result here. $\endgroup$ – Phillip Hamilton Feb 20 '16 at 0:14
  • $\begingroup$ Note that the converse is also true: If there isn't $x_0$ for which $\sup\limits_{j \ge 1}{\|T_j x_0\|}=\infty$, then $\cup E_n=X$ and as $X$ is complete, by Baire Theorem follows that there is some $n_0$ such that $E_{n_0}$ has nonempty interior (and the uniform boundedness principle follows) $\endgroup$ – Svetoslav Feb 20 '16 at 1:14
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  1. $T$ and $\| \|$ are continuous so $\|\|\circ T$ is continuous and $E^j_n=(\|\|\circ T_j)^{-1}([0,n])$ is closed. This implies that $E_n =\cap_jE^j_n$ is closed.

Suppose there exists $x$ such that $sup_{j\geq 1}\|T_j(x)\|=+\infty$ and the interior of the adherence $\bar E_n$ of $E_n$ is not empty. Let $U$ an open subset contained in $\bar E_n$. Consider $B(y,r)\subset U$. We deduce for every $z\in B(y,r)$, $\|T_j(z)\|\leq n$. Let $u\in B(0,r)$, $\mid \|T_j(y)\|-\|T_j(u)\|\mid\leq \|T_j(y+u)\|$. This implies that $\| T_j(u)\|\leq \|T_j(y)\|+\|T_j(y+u)\|\leq 2n$. Thus $\| T_j\| \leq {{2n}\over r}$.

This implies that $\| T_j(x)\|\leq \| T_j\|\|x\|\leq {{2n}\over r}\|x\|$. This is a contradiction with the fact that $sup_{j\geq 1}\|T_j(x)\|=+\infty$.

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  • $\begingroup$ Sorry what does the the interior of the adherence $E_n $ of $E_n $ mean? $\endgroup$ – Phillip Hamilton Feb 20 '16 at 1:01
  • $\begingroup$ The maximal open subset contained in the adherence $\endgroup$ – Tsemo Aristide Feb 20 '16 at 1:02
  • $\begingroup$ Hm, I can not figure it out: if $u:\|u\|\leq 1$ then $\|ur\|\leq r\Rightarrow \|T_jur\|\leq 2n\Rightarrow \|T_ju\|\leq 2n/r$. But on the other hand, if you take $u:\|u\|\leq r$ then $\|\frac{u}{r}\|\leq \frac{r}{r}=1\Rightarrow \|T_j\frac{u}{r}\|\leq \frac{2n}{r}$. What is going on here.. $\endgroup$ – Svetoslav Feb 20 '16 at 1:03
  • $\begingroup$ Obviously nothing, I should go to sleep. .. +1 $\endgroup$ – Svetoslav Feb 20 '16 at 1:06
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Because each $T_j$ is continuous and $x_k\to x$, you have $T_jx_k\to T_jx,\,\forall j\ge 1\Rightarrow \|T_jx_k\|\to \|T_jx\|$. But $\|T_jx_k\|\leq n,\,\forall k\in\mathbb N\,\forall j\ge 1\Rightarrow \|T_jx\|\leq n,\,\forall j\ge 1\Rightarrow x\in E_n$

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