# Real Analysis, Folland problem 5.3.37 Application of the Uniform Boundedness Principle

The Uniform Boundedness Principle - Suppose that $\mathscr{X}$ and $\mathscr{Y}$ are normed vector spaces and $\mathcal{A}$ is a subset of $L(\mathscr{X},\mathscr{Y})$.

a.) If $\sup_{T\in\mathcal{A}}\lVert Tx\rVert < \infty$ for all $x$ in some nonmeager subset of $x$ subset of $\mathscr{X}$, then $\sup_{T\in \mathcal{A}}\lVert T\rVert < \infty$

b.) If $\mathscr{X}$ is a Banach space and $\sup_{T\in \mathcal{A}}\lVert Tx\rVert < \infty$ for all $x\in \mathscr{X}$, then $\sup_{T\in\mathcal{A}}\lVert A\rVert < \infty$

Problem 3.3.37 - Let $\mathscr{X}$ and $\mathscr{Y}$ be Banach spaces. If $T:\mathscr{X}\rightarrow \mathscr{Y}$ is a linear map such that $f\circ T\in \mathscr{X}^{*}$ for every $f\in\mathscr{Y}^{*}$, then $T$ is bounded.

The Closed Graph Theorem - If $\mathscr{X}$ and $\mathscr{Y}$ are Banach spaces and $T:\mathscr{X}\rightarrow \mathscr{Y}$ is closed linear map, then $T$ is bounded.

Thoughts: If we show that $T:\mathscr{X}\rightarrow \mathscr{Y}$ is a closed linear map then we can conclude that $T$ is bounded. I am not sure if this is hard to show but I don't have any idea how I would show that. Perhaps we need to define a new function given the fact with our current linear map has that $f\circ T\in \mathscr{X}^{*}$ for every $f\in \mathscr{Y}^{*}$ and then use the Uniform Boundedness Principle to show that $T$ is bounded.

This section for me has been quite a headache I have been trying to solve various problems at ad nauseam all day. Any suggestions or references in getting a better understanding of this section is much appreciated as well as how to prove this problem.

Further thoughts/ideas: Since $\mathscr{X}$ and $\mathscr{Y}$ are Banach spaces and every normed vector space can be embedded in Banach space is a dense subspace we can mimic the construction of $\mathbb{R}$ from $\mathbb{Q}$ via Cauchy sequences. Would this help us at all?

• Edited sorry about that – Wolfy Mar 21 '16 at 0:32
• As you suggested in your question, I think the Closed Graph Theorem is the best approach. – carmichael561 Mar 21 '16 at 0:59
• Any idea how I should proceed? – Wolfy Mar 21 '16 at 1:02
• @MorganWeiss Define $A = \{\hat{x} \circ T^\intercal\mid x \in X, \lVert x \rVert = 1\}$, where $\hat{x}$ is the canonical image of $x$ in $X^{**}$ and $T^\intercal$ is the dual map of $T$. – Henricus V. Mar 21 '16 at 1:07

Suppose that $x_n\to x$ and $Tx_n\to y$. We want to show that $Tx=y$.
Let $f$ be an element of $Y^*$. Since $f\circ T$ is continuous, $x_n\to x$ implies $f(Tx_n)\to f(Tx)$. Also $f$ is continuous, so $Tx_n\to y$ implies $f(Tx_n)\to f(y)$.
Therefore $f(Tx)=f(y)$ for all $f\in Y^*$. Since the continuous linear functionals separate points, this implies that $Tx=y$. Therefore $T$ is bounded by the Closed Graph Theorem.
• Why is $f\circ T$ continuous? Also, why is $f$ continuous? I like your approach but I am not understanding it fully – Wolfy Mar 21 '16 at 1:13
• @MorganWeiss $f$ is an element of the dual space of $Y$, so it is automatically continuous. $f \circ T$ is an element of the dual space of $X$, so it is continuous. – Henricus V. Mar 21 '16 at 1:14
• Where are those facts coming from? From my understanding, If $X$ is a normed vector space, the space $L(X,K)$ where $K = \mathbb{R}$ or $\mathbb{C}$ of bounded linear functionals on $X$ is called the dual space of $X$ and is denoted $X^{*}$. – Wolfy Mar 21 '16 at 1:18