14

Here is a problem in functional analysis from Folland's book:

If $\mathcal{M}$ is a finite-dimensional subspace of a normed vector space $\mathcal{X}$, then there is a closed subspace $\mathcal{N}$ such that $\mathcal{M}\cap \mathcal{N} = 0$ and $\mathcal{M}+\mathcal{N} = \mathcal{X}$.

I tried the following approach: I am trying to define a projection map $\pi_{\mathcal{M}}$ from $\mathcal{X}$ to $\mathcal{M}$, which would be continuous and hence taking the inverse of any closed set would give a closed set in $\mathcal{X}$. I am confused about what the projection map would be. Please suggest some approach.

Dylan Moreland
  • 19,819
user24367
  • 1,286
  • Hint: if $M$ and $N$ were two different vector spaces, how would you go about defining a projection from $M\oplus N$ onto the embedded copy of $M$? –  Feb 01 '12 at 01:06
  • how will you define $N$ in terms of $\mathcal{M}+\mathcal{N}$ when $\mathcal{X}$ is infinite dimensional and $\mathcal{M}$ is finite, without using dot product. Thanks – user24367 Feb 01 '12 at 01:55
  • I didn't mention inner products: I was referring to the direct sum of vector spaces. As for how you define $N$, well the first sentence of your question seems to specify some $N$... –  Feb 01 '12 at 02:32
  • Oh, now I understand: is what you state in your first sentence what you are trying to prove? If so, perhaps you could reword the question to make that clearer. Not everyone has a copy of Folland on their desk or in their head, so it wasn't clear to me which part of your post was the question and which part the idea you were suggesting. –  Feb 01 '12 at 04:22
  • @DylanMoreland: thanks, but you should replace "closed" in the title by "finite-dimensional", otherwise it states something false. –  Feb 01 '12 at 07:22
  • @Yemon Ah, right. I was clarifying a few titles (I don't like seeing a front page full of "linear algebra question") and mixed up some assumptions and implications. Should be fixed now. Thanks for the ping. – Dylan Moreland Feb 01 '12 at 07:28

1 Answers1

16

Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in \mathcal M$ has then a unique representation $x = \alpha_1(x)e_1 +...+ \alpha_n(x)e_n$. Each $\alpha_i$; is a continuous linear functional on $\mathcal M$ (a linear map from finite dimensional space is always continuous) which extends to a member of $\mathcal X^*$, by the Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let $\mathcal N$ be the intersection of the null spaces of these exten­ sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.

azarel
  • 13,120
  • Hey, but how does this prove that $N$ is closed? – asdf Nov 29 '17 at 22:04
  • 1
    @asdf, because it's an intersection of closed sets. Each null space is a closed set, because it's a preimage of a single point under a continuous map. – Hasek Oct 03 '19 at 22:16