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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.

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How do you mean linear on a metric space? – Berci Feb 6 '13 at 20:24
Presumably you mean on a vector space with a metric? – copper.hat Feb 6 '13 at 20:26
I corrected the question. It now reads "...metric linear space" instead of "metric space" – mas Feb 6 '13 at 20:30
Is the metric compatible with the linear structure, that is do you assume that the addition and scalar multiplication are continuous functions? – Theo Feb 6 '13 at 20:45
@Berci: no, it's not the same. In your example $(\mathbb{R},d')$ the entire line is bounded with respect to the metric, but it is not absorbed by every neighborhood of zero. There is no scalar $\lambda$ such that $\mathbb{R} \subseteq \lambda B$, where $B$ is the ball of radius $1/2$. – Martin Feb 6 '13 at 22:54

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


  • 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 $S\subset \rho B$.

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Taking the last bullet as given the proof is very easy. So, how do you show the last bullet? – mas Feb 6 '13 at 23:04
@mas In the view of Martin and Berci comments, I think you should clarify what you mean by bounded. – Theo Feb 7 '13 at 0:49

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.

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Why is $B(r)=\frac{r}{\epsilon}B(\epsilon)$? You have to have some extra assumptions about the metric, I think. – Theo Feb 6 '13 at 20:55
Exactly. You cannot assume that. You only know that you are in a metric linear space, and you cannot assume d(ax,0)=ad(x,0). – mas Feb 6 '13 at 21:03
Aha.. a bit unusual to me, I am going to think it over.. – Berci Feb 6 '13 at 21:27
I started to prove, then got to the point to see that it is not true. Also added counterexample. – Berci Feb 6 '13 at 22:06
I think there is a confusion between two largely unrelated notions of "bounded": bounded with respect to the metric and bounded in the topological vector space sense. For bounded sets in the second sense the assertion is true, but, as you observe, not for bounded sets in the first sense. // Your metric $d'$ is translation invariant, but it isn't homogeneous. – Martin Feb 6 '13 at 22:40

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