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Prove: If $X$ is a locally convex space, $L \leq X$, $L$ has finite dimension, $M\leq X$ Then $L+M$ is closed.

What I know: If $L$ is a finite dimensional subspace, then $L$ is closed.

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By $L \le X$ and $M \le X$, I suppose you mean $L$ and $M$ are linear subspaces of $X$? –  Robert Israel Jun 5 '12 at 18:38
    
Is there an assumption that $M$ is closed as well? It's certainly not true for arbitrary linear subspaces. –  Robert Israel Jun 5 '12 at 18:40
    
Related to your question math.stackexchange.com/questions/42993/… –  userNaN Jun 5 '12 at 18:49
    
Hint: $L+M = \pi^{-1}[\pi(L)]$ if $\pi\colon X \to X/M$ is the quotient map. –  t.b. Jun 5 '12 at 19:02

1 Answer 1

up vote 4 down vote accepted

If $M$ is not closed then this is not true in general. Indeed, consider $$ X=\ell_1(\mathbb{N}),\quad L=\mathrm{span}\{e_1\},\quad M=\mathrm{span}\{e_n:n>1\} $$ Obviously, $$ \mathrm{dim}(L)=1,\quad \overline{M}=\{x\in X: x(1)=0\} $$ $$ L+M=\mathrm{span}\{e_n:n\in\mathbb{N}\} $$ $$ \overline{L+M}=\ell_1(\mathbb{N}) $$

If we additionally assume that $M$ is closed then this statement is true. This proof is taken from Rudin's Functional analysis theorem 1.42. Since $M$ is closed then $X/M$ is Hausdorff locally convex space. Consider quotient map $$ \pi:X\mapsto X/M:x\mapsto x+M $$ Since $L$ is finite dimensional, then does $\pi(L)$. Since $\pi(L)$ is finite dimensional then it is closed. Since $\pi$ is continuous then $\pi^{-1}(\pi(L))$ is closed. It is remains to note that $M+L=\pi^{-1}(\pi(L))$

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