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Let $E$ and $F$ be two locally convex TVS and $T:E\to F$ be a continuous linear function.

Let $C\subseteq E$ be a closed convex and bounded set. Is it true that $T(C)$ is closed convex and bounded too?

I am not sure on how to approach this proof, and I could not find anything else than this which seems to be specifically designed for $F=\mathbb R^n$.

P. Quinton
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    Take $E=c_0$ (the real sequences with limit $0$ endowed with the maximum norm), $T:c_0 \to \mathbb{R}$, $T(x_n)=\sum_{n=1}^\infty x_n/2^n$ and $C$ the closed unit ball. Then $T(C)$ is not closed, as $1 \in \overline{T(C)}$ but $1 \notin T(C)$. – Gerd Jul 26 '24 at 09:34
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    @BrunoB My apologies, it seems that MS stored the title of a previously aborted (because stupid) question of mine. I updated, thanks for the catch! – P. Quinton Jul 26 '24 at 18:05

2 Answers2

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As pointed out in the comments $T(C)$ does not have to be closed. But it is always bounded and convex.

For convexity:
Choose $x,y\in T(C)$ arbitrary. Then there are $a,b\in C$ such that $T(a)=x,T(b)=y$. We know by convexity of $C$, that: $$\forall \lambda \in [0,1]: \lambda a + (1-\lambda) b \in C.$$ This implies by linearity of $T$: $$\forall \lambda \in [0,1]: T(\lambda a + (1-\lambda) b )=\lambda T(a) + (1-\lambda) T(b)=\lambda x + (1-\lambda) y \in T(C).$$ And therefore $T(C)$ is convex.

For boundedness:
Take any zero neighborhood in $F$, say $U\in \mathcal{N}_0(F)$. Then since $T$ is continuous $T^{-1}(U)\in \mathcal{N}_0(E)$ is a zero neighborhood in $E$. So there exists $r>0$ with $C\subseteq rT^{-1}(U)$. This yields after applying $T$ and again using linearity: $$T(C)\subseteq rU.$$ Which shows that $T(C)$ is bounded. Also see this question.

Föölücks
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Here is the standard counter-example to show that $T(C)$ is not closed. This example is posed in $\mathbb R^3$.

Let $K$ be the so-called ice-cream cone, $$ K:=\left\{x\in \mathbb R^3 : \ x_3 \ge \sqrt{x_1^2 + x_2^2}\right\}. $$ Define $A = \begin{pmatrix}1&0&-1\\0&1&0\end{pmatrix}$. Then $$ A(K):=\{Ax: \ x\in K\} = \{y\in \mathbb R^2: \ y_1<0\} \cup \{0\} $$ is not closed.

daw
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  • 1/ Your set $K$ is not bounded.
    2/ There is an even simpler example. Let $E$ be the epigraph of $1/x$, $x>0$. Then $E$ is a nonempty closed convex subset of $\mathbb{R}^2$. Let $T$ be the projection onto the $x$-axis. Then $T(C)$ is not closed.     – max_zorn Sep 21 '24 at 06:12