Any linear transformation between two finite dimensional topological vector spaces is continuous.

Question

Theorem

Any linear transformation $T$ from one finite dimensional topological vector space $V$ into another finite dimensional topological vector $W$ space is necessarily continuous.

Unfortunately I don't be able to prove the statement so could someone prove it? Then if the statement is generally false is it false if $V=\Bbb R^m$ and $W=\Bbb R^m$ too? So could someone help me, please?

See Treves, Topological Vector Spaces, Distributions and Kernels. Probably it is proved there — Danilo Gregorin Afonso, Jul 21 '20 at 14:52
Then you get silly examples like the identity map $(\mathbb{R},\text{indiscrete})$ to $(\mathbb{R},\text{usual})$. — user10354138, Jul 21 '20 at 14:57
@user10354138 Umm...unfortunately $\Bbb R$ equipped with indiscrete topology is not a topological vector space! Is this incorrect? — Antonio Maria Di Mauro, Jul 21 '20 at 15:00
No! Since you didn't require a TVS to be $T_1$ any vector space with the indiscrete topology is a TVS (since every map to an indiscrete space is automatically continuous, in particular $+\colon X\times X\to X$ and $\cdot\colon\mathbb{R}\times X\to X$ are continuous) — user10354138, Jul 21 '20 at 15:48
@user10354138 Okay, so if I suppose that the topological vector spaces must be $T_1$ then is the theorem true? and so how to prove it? — Antonio Maria Di Mauro, Jul 21 '20 at 15:59
Any finite-dimensional $T_1$ TVS (therefore Hausdorff) must be the standard $\mathbb{F}^n$ for some $n$, and you know linear maps $\mathbb{F}^n\to\mathbb{F}^m$ are continuous ($\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$). — user10354138, Jul 21 '20 at 16:25
@BrianM.Scott I read it. However I don't understand why a finite dimensional topological vector space can be equipped with a norm which gives the same topology. Indeed sometimes ago I post an answer (look here) where I ask this but unfortunately nobody answered. So what can you say about? — Antonio Maria Di Mauro, Jul 21 '20 at 21:01
@AntonioMariaDiMauro: Okay, take a look at Section $3.4$ of this PDF. In Proposition $3.17$ take $\mathscr{X}$ itself to be $n$-dimensional and let $\mathscr{Y}=\mathscr{X}$. Any basis for $\mathscr{X}$ gives you an isomorphism from $\mathscr{F}^n$ to $\mathscr{X}$, and the theorem says that it’s a homeomorphism. Here $\mathscr{F}$ is the scalar field, and it’s pretty clear from what he says earlier in the PDF that he has $\Bbb R$ and $\Bbb C$ in mind. Thus, in effect you’re just looking at the normed space $\mathscr{F}^n$. — Brian M. Scott, Jul 21 '20 at 21:43

score -1 · Answer 1 · answered Jul 21 '20 at 21:09

-1

Let $T:V\longmapsto W$ a linear transformation, and $\mathcal{B}=\{e_1,\cdots,e_n\}$ a basis of V. In finite dimension all norms are equivalents, let $||x||=\sum_{i=1}^n\vert x_i\vert$, then $$||f(x)||=||f(x_1e_1+\cdots+x_ne_n)|| \leqslant |x_1|||f(e_1)||+...+|x_n||f(e_n)||$$ Let $M=\underset{i\in\{1,...,n\}}{sup}||f(e_i)||$, then $$ |x_1|||f(e_1)||+...+|x_n||f(e_n)|| \leqslant M(|x_1|+...+|x_n|)=M||x|| $$ $\implies $ $$||f(x)|| \leqslant M||x||$$

answered Jul 21 '20 at 21:09

H_K

694

1

Okay, and if $V$ and $W$ are not normed spaces? – Antonio Maria Di Mauro Jul 21 '20 at 21:12
Perhaps do you know the answer to this question? – Antonio Maria Di Mauro Jul 21 '20 at 21:13
You can see https://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous – H_K Jul 21 '20 at 21:51
There is the same problem there too – Antonio Maria Di Mauro Jul 21 '20 at 22:19

Any linear transformation between two finite dimensional topological vector spaces is continuous.

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