Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 $M$? –  user16299 Feb 1 '12 at 1:06
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 1 '12 at 1:55
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 $N$... –  user16299 Feb 1 '12 at 2:32
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. –  user16299 Feb 1 '12 at 4:22
@DylanMoreland: thanks, but you should replace "closed" in the title by "finite-dimensional", otherwise it states something false. –  user16299 Feb 1 '12 at 7:22

1 Answer 1

up vote 8 down vote accepted

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 $\mathcal X = \mathcal M\oplus \mathcal N$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.