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Let $(X,\|\cdot \|_X)$ and $(Y,\|\cdot \|_Y)$ be real Banach spaces. Let $\{ T_\alpha\}_{\alpha\in \Delta}\subseteq B(X,Y)$, where $B(X,Y)$ is the space of all bounded linear maps. Define a new norm on $X$ by $$\|x\|'=\|x\|_X+\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y.$$ Prove that the uniform boundedness principle can be derived from the closed graph theorem.

My trial

By assumption,

  1. $\{ T_\alpha\}_{\alpha\in \Delta}\subseteq B(X,Y)$
  2. $\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y=\|x\|'-\|x\|_X<\infty.$

Let ${\alpha\in \Delta}$, define \begin{align} G(T_\alpha)=\big\{(x,T_\alpha x):x\in X \big\} \subseteq X\times Y .\end{align} Clearly, $G(T_\alpha)$ is a closed subset of $X\times Y$ which is Banach. Define \begin{align} \Pi_1:&\left(G(T_\alpha), \|\cdot \|_{X\times Y}\right)\to (X,\|\cdot \|_X)\\&(x,y)\mapsto \Pi_1(x,y)=x . \end{align} Clearly, $\Pi_1$ is bijective, linear and bounded. So, $\Pi_1^{-1}$ is continuous, and there exists $M\geq 0$ such that $$\|\Pi_1^{-1}(x) \|_{X\times Y}\leq M\|x\|_X.$$ This means that $$\|x\|_X+\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y=\|(x,y) \|_{X\times Y}\leq M\|x\|_X.$$ This results to $$\sup\limits_{\alpha\in \Delta}\|T_{\alpha}\|_Y\leq M.$$

Please, I'm I right? If no, then alternative proofs are highly welcome.

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Notice that the $\Pi_1$ you define also depends on $\alpha$ and hence so does the constant $M$. In particular, you cannot just take the $\sup$ to conclude that $$\|x\|_X+\sup\limits_{\alpha\in \Delta}\|T_{\alpha}x\|_Y=\|(x,y) \|_{X\times Y}\leq M\|x\|_X.$$ Instead, consider the space $Z = \{f: \Delta \to Y \mid \sup_{\alpha \in \Delta} \|f(\alpha)\|_Y < \infty\}$ of bounded functions from $\Delta$ to $Y$ with the $\sup$-norm $\|f\|_\infty = \sup_{\alpha \in \Delta} \|f(\alpha)\|_Y$. It is a standard exercise to check that $Z$ is a Banach space (similar to checking that the space of continuous real-valued functions is a Banach space for the $\sup$-norm).

Define $\Psi:X \to Z$ by $\Psi(x)(\alpha) = T_\alpha x$. By assumption, this map is well-defined and is clearly linear. If $(x_n, \Psi(x_n)) \to (x,f)$ in $X \times Z$, then for $\alpha \in \Delta$, $$f(\alpha) = \lim_{n \to \infty} \Psi(x_n)(\alpha) = \lim_{n \to \infty} T_\alpha x_n = T_\alpha x$$ since each $T_\alpha$ is bounded. Hence $f = \Psi(x)$ and so $\Psi$ has closed graph and hence is bounded by the closed graph theorem. Finally, for each $\alpha$, $$ \|T_\alpha x\|_Y = \|\Psi(x)(\alpha)\|_Y \leq \|\Psi(x)\|_\infty \leq \| \Psi \| \cdot \|x\|_X$$ and so $\sup_{\alpha \in \Delta} \|T_\alpha\| \leq \|\Psi\| < \infty$ (and, in fact, one can check easily that $\sup_{\alpha \in \Delta} \|T_\alpha\| = \|\Psi\|$).

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  • $\begingroup$ Sorry, you didn't use the new norm defined on $X$. $\endgroup$ – Omojola Micheal Feb 27 '19 at 15:44
  • $\begingroup$ @OmojolaMicheal This is a somewhat strange objection to a proof. If you really want to insist on using this norm for some reason then you should check that $X' = (X, \| \cdot \|')$ is also a Banach space, that $\operatorname{Id}: X' \to X: x \mapsto x$ is a continuous bijection and then apply the inverse mapping theorem (which is equivalent to the closed graph theorem) to conclude that $\operatorname{Id}$ has continuous inverse. This gives that $\|\cdot\|$ and $\|\cdot\|'$ are equivalent norms which in turn obviously implies the result. $\endgroup$ – Rhys Steele Feb 27 '19 at 17:02
  • $\begingroup$ @OmojolaMicheal Notice though that $X'$ is isometrically isomorphic to $G(\Psi)$ with the norm inherited from $\|(x, f)\|_{X \times Z} = \|x\|_X + \|f\|_Z$ which makes $X \times Z$ a Banach space. This means that checking that $X'$ is a Banach space is really nothing but a complicated way of checking that $G(\Psi)$ is closed and the rest of the argument is nothing but a complicated way of checking that $\Psi$ is bounded (without ever introducing $\Psi$). $\endgroup$ – Rhys Steele Feb 27 '19 at 17:04
  • $\begingroup$ I think the hint is fine. Let me work on it. If I have a challenge, I'll get back to you. $\endgroup$ – Omojola Micheal Feb 27 '19 at 17:16
  • $\begingroup$ Considering the space $Z = \{f: \Delta \to Y \mid \sup_{\alpha \in \Delta} \|f(\alpha)\|_Y < \infty\}$. We need to consider Banach spaces to Banach spaces but $\Delta$ is an arbitrary set, not necessarily normed space neither is it Banach. How can we rectify this? $\endgroup$ – Omojola Micheal Feb 28 '19 at 14:46

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