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I am doing an introductory functional analysis course alongside a real analysis course, and the following concept showed up. I am looking for a proof that combines what I have learned from these two courses to prove the statement in the title.

Recall, a normed space $(V,\lVert \cdot \rVert )$ is a vector space (for now over $\mathbb{R}$) equipped with a norm, satisfying all the familiar vector norm properties.

A bounded operator is a linear map

$$ T:V\to V $$

Such that there is some $C\in \mathbb{0}$ such that for any $\mathbf{x} \in V$,

$$ \lVert T\mathbf{x} \rVert \leq C\lVert \mathbf{x}\rVert $$

We want to prove that any operator which is continuous is necessarily bounded.

Firstly, note that it is sufficient to prove this for all those $\mathbf{x} \in V$ with unit norm, i.e. $\lVert \mathbf{x} \rVert = 1$, since both the norm and the linear operator preserve scalar multiplication, and every element of $V$ is some scalar multiple of an element in $V$ with unit norm. Thus it is sufficient to prove that there is some $C\in \mathbb{R} $

$$ \lVert \mathbf{x} \rVert = 1 \implies \lVert T\mathbf{x} \rVert \leq C$$

Now if I were doing real analysis and $V$ was just $\mathbb{R}^n$, I could use the fact that a unit hyper sphere is always a closed and bounded set in $\mathbb{R}^n$, and thus the image of this set under a continuous function would be a closed and bounded subset of $\mathbb{R}$, which has a supremum, and thus we can let $C$ be that supremum and we are done. However, I am not sure how to extend this to an arbitrary $V$ a possible "counterexample" I could think of involves $\mathbb{Q}^n$ since then we could have a continuous function from the unit sphere into $\mathbb{R}$ which is unbounded (I think?), But of course $\mathbb{Q}^n$ is not a vector space over $\mathbb{R}$. One might then think to check if every vector space over $\mathbb{R}$ is homeomorphic to some $\mathbb{R}^n$, which is the case for finite dimensional vector spaces over $\mathbb{R}^n$, but not in general.

My question is this: Is there a some general property of a vector space that allows us to conclude something similar to what we can conclude in the case of $\mathbb{R}^n$? I.e. a property that ensures that the continuous image of a bounded set is again bounded in $\mathbb{R}$.

Secondly, is there a way to adapt this line of thinking to prove the converse, that every bounded operator is continuous? I am aware that different proofs for this exist, but I am specifically looking for something that connects this to topology/real analysis, and is by some metric "elegant"

Carlyle
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  • Hint: continuity implies that there exists some $\delta > 0$ such that whenever $\lVert x \rVert < \delta$, then $\lVert Tx \rVert < 1$. Try to use this along with linearity to show that for any $x$, you have $\lVert Tx \rVert \le \frac{2}{\delta} \lVert x \rVert$. – Daniel Schepler Sep 28 '23 at 21:47
  • Does this answer your question? Prove that if a linear operator is continuous, then it is bounded. – Anne Bauval Sep 28 '23 at 21:49
  • Your title ("is this proof correct?") is misleading since your post contains no such proof. – Anne Bauval Sep 28 '23 at 21:59
  • @AnneBauval noted, the title has been adapted, I looked at that question, and it seems that it mostly pointed out that this line of reasoning doesn't work if you try to use compactness as "the property" and then proceeded to produce proofs using different methods. I am however specifically interested in something that does use this line of reasoning, clearly it is "possible" since the homeomorphic image of the unit ball must be a bounded set (this is the theorem we are proving) but I am not entirely convinced that one can't for example weaken or tweak compactness to find a more elegant proof – Carlyle Sep 28 '23 at 22:06
  • Which "line of reasoning"? The most elegant proof is the most parsimonious one. – Anne Bauval Sep 28 '23 at 22:18

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Firstly, unfortunately there is no such result about continuous images of unit spheres being bounded in infinite dimensions. In fact, a result of Bessaga shows that an infinite dimensional Hilbert space is Homeomorphic to its own unit sphere.. So there is no hope for an argument along those lines - one has to use linearity somehow for the infinite dimensional case. [Edit: Looking at the comments, I see you also have interpreted this as a question about an image of the unit ball being bounded, not necessarily just a sphere. See this example for a counterexample for balls.]

As far as the converse goes, in your setting the easiest way to show continuity is to directly observe that if $\|Tx\|\leq C\|x\|$, then one can let $\delta=\frac{\epsilon}{C}$ in the $\epsilon$-$\delta$ definition of continuity. However, one could always translate that into an argument about in terms of open sets very quickly, if one wanted to, mimicking the proof here (which is stated for reals but generalizes to any metric space easily).

If you are interested in more general topological arguments, it is worth noting that in an arbitrary topological vector space, a linear map that is continuous must also take bounded sets to bounded sets. Here $S\subseteq X$ is "bounded" if every neighborhood $U$ of $0$ has a scalar $\lambda$ so that $\lambda U\supseteq S$. (Note that this coincides with the usual definition in normed vector spaces.)

There is a partial converse, namely that if $f$ is linear and $f(V)$ is bounded for some open set $V\subseteq X$, then $f$ is continuous. In the special case that a bounded open set exists (e.g., in a normed vector space), this then implies that maps are continuous if and only if they take bounded sets to bounded sets.

See the answers here for proofs of the topological vector space facts (note that one answer is corrected in the comments.)

M W
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