Continuous linear functionals

Question

Let L be a continuous linear functional on a metric linear space X. Prove: L(S) is a bounded set for any bounded subset S of X. The metric is translation invariant.

Answer 1

Consider the inverse image of any bounded open neighborhood $W$ of $0$ in the target space (let it be $\Bbb R$ or $\Bbb C$ or anything). Its preimage, $L^{-1}(W)$ is an open neighborhood of $0\in X$, so it contains a ball $B(\varepsilon)$ around $0$ with radius some $\ \varepsilon >0$. Let us fix this $\varepsilon$.

Now, if we have any bounded set $S\subseteq X$, then there exists a ball that contains it, say with radius $r_0>0$ and midpoint $Q$, then $\ S\subseteq B(r)\ $ where $\ r=r_0+d(0,Q)$.

Now imagine the continuous mappings $f_\lambda:X\to X,\ \ v\mapsto \lambda\cdot v$. If $\lambda\neq 0$, it is a homeomorphism, mapping any ball $B(\rho)$ (around $0$) to an open set, which also contains a ball around $0$, say with radius $f(\lambda,\rho)>0$. So that $B(\lambda\cdot\rho)=f_\lambda(B(\rho))\supseteq B(f(\lambda,\rho))$...

Then, $L(S)\subseteq L(B(r))= L(f_{\frac r\varepsilon}( B(\varepsilon))) \subseteq f(\displaystyle\frac r\varepsilon)\cdot W $, bounded.

---

Update: Hmm, in this generality it is not true. Let $X=\Bbb R$ with a bounded metric, for example $d'(x,y):=\displaystyle\frac{d(x,y)}{1+d(x,y)} <1$, and let $L$ be the identity $X\to\Bbb R$. Then the whole $X$ is bounded, but its image is not bounded.

Answer 2

Note that this is true in general for $L:X\rightarrow Y$ linear continuous between two metric linear spaces.

Hints:

• fix an open ball centered at $0$ (the one of radius $1$ for instance)

• use continuity at $0$.

• prove that $L$ is bounded on some open ball $B$ centered at $0$.

• use that for every bounded set $S$, there is $\rho>0$ such that $S\subset \rho B$.