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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!

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3 Answers 3

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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.

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    $\begingroup$ 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? $\endgroup$ Commented Jun 1, 2017 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! $\endgroup$ Commented Jun 1, 2017 at 19:26
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    $\begingroup$ if $\varphi_i$ is a orthonormal basis of the subspace then $a_{i,n} = \langle y_n, \varphi_i\rangle$. $\endgroup$ Commented Jun 1, 2017 at 19:30
  • $\begingroup$ @NathanaelSkrepek What a silly I am! Of course! Thanks a lot for your help $\endgroup$ Commented Jun 1, 2017 at 19:32
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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.

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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.

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    $\begingroup$ This is a little overkill^^ $\endgroup$ Commented Jun 1, 2017 at 19:28
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    $\begingroup$ @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). $\endgroup$ Commented Jun 1, 2017 at 19:34

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