4
$\begingroup$

Let $\;H\;$ be a Hilbert space and $\;φ_1,φ_2, \dots ,φ_m\in H\;$ where $\;m \lt \infty\;$. Prove that the linear span of $\;φ_1,φ_2, \dots ,φ_m\; \equiv\langle φ_1,φ_2, \dots ,φ_m \rangle \;$ is a closed subset of $\;H\;$

My attempt

Consider $\;y_n \in \langle φ_1,φ_2, \dots ,φ_m \rangle \;$ such that $\;y_n \rightarrow y \in H\;$. It is sufficient to show $\;y \in \langle φ_1,φ_2, \dots ,φ_m \rangle \;$. Since $\;y_n \in \langle φ_1,φ_2, \dots ,φ_m \rangle \;$ , there are $\;a_i \in \mathbb C \; \forall 1\le i \le n\;$ such that $\;y_n=\sum_{i=1}^m a_iφ_i \;$ (*). But $\;y_n \rightarrow y \;$ and so $\;y=\sum_{i=1}^m a_iφ_i \;$. This means $\;y\in \langle φ_1,φ_2, \dots ,φ_m \rangle \;$

I'm a bit unsure if the above is right. I know it's something quite elementary but I've been stuck. If the dimension of $\;\langle φ_1,φ_2, \dots ,φ_m \rangle \;$ wasn't finite then my proof wouldn't be valid because $\;m\;$ in (*) would be $\; \infty\;$?

Any help would be valuable! Thanks in advance!

$\endgroup$
2
$\begingroup$

In your proof, you make $\{y_n\}$ a constant sequence. What you actually have is that for each $n\in\mathbb N$, there are scalars $a_{i,n}\in\mathbb C$ ($1\leq i\leq m$) such that $y_n=\sum_{i=1}^ma_{i,n}\varphi_i$.

It may help to assume (without loss of generality) that the $\varphi_i$ are linearly independent, and furthermore orthonormal.

$\endgroup$
4
  • 1
    $\begingroup$ Oh, I missed that. You're right! So if I want $\;y \in \;\langle φ_1,φ_2, \dots ,φ_m \rangle \;$ , I should show $\;a_{i,n} \rightarrow a_i\;$. Is this possible to prove? $\endgroup$ Jun 1 '17 at 19:23
  • $\begingroup$ if $\;φ_i\;$ are orthonormal, how does this help me? It's not very clear to me... I guess I 'm super silly! $\endgroup$ Jun 1 '17 at 19:26
  • 1
    $\begingroup$ if $\varphi_i$ is a orthonormal basis of the subspace then $a_{i,n} = \langle y_n, \varphi_i\rangle$. $\endgroup$ Jun 1 '17 at 19:30
  • $\begingroup$ @NathanaelSkrepek What a silly I am! Of course! Thanks a lot for your help $\endgroup$ Jun 1 '17 at 19:32
2
$\begingroup$

It is no harder to prove a more general result:

Let $\emptyset\neq Y=span\left \{ \phi_1,\cdots, \phi_n \right \}.$ Then, the map $\psi$ given by $c_1\phi_1+\cdots +c_n\phi_n\mapsto (c_1,\cdots, c_n)$ is an isomorphism of $Y$ onto $\mathbb R^n.$

Now, since all norms on spaces of dimension $n$ are equivalent, there are $C_1,C_2>0$ such that

$C_1\left \| \sum_{k=1}^{n}c_k\alpha_k \right \|\le \sum_{k=1}^{n}|c_k|\le C_2\left \| \sum_{k=1}^{n}c_k\alpha_k \right \|.$

Thus, a sequence $(\vec x_i)$ in $Y$ converges $\Leftrightarrow $ the sequence $\psi((\vec x_i))=(\vec c_i)$ of coefficients in $\mathbb R^n$ converges. Since $\mathbb R^n$ is complete, the result follows.

$\endgroup$
0
$\begingroup$

Let us prove a more general statement:

Proposition. Let $E$ be a normed vector space and let $F$ be a subvector space of $E$. If $F$ is finite-dimensional, then $F$ is complete and in particular closed in $E$.

Proof. Let $(f_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $F$, then there exists $N\in\mathbb{N}$ such that: $$\|f_n-f_0\|\leqslant 1.$$ Therefore, $(f_n-f_0)_{n\geqslant N}$ is Cauchy sequence of the unit ball of $F$, which is compact using Riesz theorem. The sequence $(f_n-f_0)_{n\in\mathbb{N}}$ converges toward $f\in B_F(0,1)$ and $(f_n)_{n\in\mathbb{N}}$ converges toward $f+f_0\in F$. Whence the result. $\Box$

Please note that I used that a compact space is complete, indeed:

Claim. Let $(X,d)$ be a metric space and $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence, then $(x_n)_{n\in\mathbb{N}}$ is convergent if and only if it has an accumulation point.

$\endgroup$
2
  • 1
    $\begingroup$ This is a little overkill^^ $\endgroup$ Jun 1 '17 at 19:28
  • 2
    $\begingroup$ @C.Falcon Thanks a lot for your time and your reply. I'll check it later and I 'll return with questions (if there are any). $\endgroup$ Jun 1 '17 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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