# Given a linear operator $\in B(X,Y)$ that is bounded on $A\subset X$, is the operator norm bounded?

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.

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

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 $B((1,1), 1)$.

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

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.