2
$\begingroup$

I have to solve the following problem:

Let $X$ and $Y$ be Banach spaces. Consider a family of linear bounded operators $\{L_\alpha\}_{\alpha\in J}\subset B(X,Y)$, where $J\neq\emptyset$ is a subset of $[0,\infty)$. Prove that if there is an open non-empty set $A\subset X$ such that, for any $x\in A$, $\sup_{\alpha \in J}\|L_\alpha x\|_{Y}$ is bounded then there exists $M>0$ such that:

$$\sup_{\alpha \in J}\|L_\alpha\|_{B(X,Y)}\leq M$$

Hint: consider a ball and exploit the linearity of the operator

My solution:

Given the set $A$, there exists a ball of radius $r$ such that $\sup_{\alpha \in J}\|L_\alpha x\|_{Y}$ $\forall x \in B_r \subset A$. But every element of $X$ can be written as an element of the ball multiplied by a scalar: $\forall x \in X\ \exists x^* \in B_r : x=Kx^*$ for some $K\in \mathbb{R}$. Then for every $x \in X$:

\begin{align} \sup_{\alpha \in J}\|L_\alpha x\|_{Y} &= \sup_{\alpha \in J}\|L_\alpha (Kx^*)\|_{Y} \\ &= \sup_{\alpha \in J}\|K L_\alpha (x^*)\|_{Y} \\ &= |K|\sup_{\alpha \in J}\|L_\alpha (x^*)\|_{Y} < \infty \end{align}

So it is bounded $\forall x \in X$. Then by Banach-Steinhaus it follows that the norm of the operator is bounded.

Am I doing something wrong? Thank you.

$\endgroup$
  • 1
    $\begingroup$ This only works if $A$ contains the origin. You'll either have to prove that this assumption is okay or modify your approach. $\endgroup$ –  mheldman Jul 17 '18 at 21:13
0
$\begingroup$

It isn't true that every vector in $X$ can be written as a scalar multiple of an element in an arbitrary open ball.

Consider the ball $B((1,1), 1) \subseteq \mathbb{R}^2$ and the point $(1,-1)$. For $\lambda \in \mathbb{R}$ we have

$$\|\lambda(1, -1) - (1,1)\|^2 = |\lambda - 1|^2 + |-\lambda-1|^2 = 2\lambda^2 + 2$$

which is never $< 1$ so there isn't a multiple of $(1,-1)$ in the ball $B((1,1), 1)$.


Try this instead:

$A$ is a nonempty open set so there exists an open ball $B(x_0, r) \subseteq A$.

For a fixed $x \in X, x \ne 0$ we have

$$x = \frac{2\|x\|}{r}\left(\left(\frac{rx}{2\|x\|} + x_0\right) - x_0\right)$$

with $\frac{rx}{2\|x\|} + x_0 \in B(x_0, r) \subseteq A$.

Therefore for any $L_\beta$ we have

\begin{align} \|L_{\beta}x\| &= \left\|L_\beta\frac{2\|x\|}{r}\left(\left(\frac{rx}{2\|x\|} + x_0\right) - x_0\right)\right\|\\ &\le \frac{2\|x\|}{r}\left\|L_\beta\left(\frac{rx}{2\|x\|} + x_0\right)\right\| + \frac{2\|x\|}{r}\|L_\beta x_0\| \\ &\le \frac{2\|x\|}{r} \left[\sup_{\alpha \in J}\left\|L_\alpha\left(\frac{rx}{2\|x\|} + x_0\right)\right\| + \sup_{\alpha\in J}\|L_{\alpha} x_0\|\right]\\ &< +\infty \end{align}

so $\sup_{\alpha \in J}\|L_\alpha x\| < +\infty$.

Since $x$ was arbitrary, the uniform boundedness principle implies that $\sup_{\alpha \in J} \|L_\alpha\| < +\infty$.

$\endgroup$
0
$\begingroup$

Your argument can be used if you replace $A$ by $A-A\equiv \{x-y:x,y\in A\}$. For, then $A-A$ is open and contains $0$. By linearity the hypothesis holds for $A-A$ also.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.