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Suppose that $\{e_1,e_2,\dotsc,e_n\}$ is a finite orthonormal set in a Hilbert space. I want to prove that $M= \operatorname{span}(\{e_1,e_2,\dotsc,e_n\})$ is closed. So what I was thinking is if I can show that $M$ is finite then I have that it is a finite subspace of my Hilbert space therefore it is complete and therefore closed.

But I'm wondering if it's true that if my set is finite then the span of my set is finite. And if thats true how do I go about proving it?

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  • $\begingroup$ $\mathbb{C}^n$ (or $\mathbb{R}^n$ depending on your field) is isometrically isomorphic to $M$. $\endgroup$ – DisintegratingByParts Feb 11 '18 at 23:05
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$\mathbb{R}$ is the span of $\{1\}$, so your attempt won't work. But write that $i:M\to H$ given by $i(e_j)=e_j$ is an isometry, and so send the complete space $M$ (because finite dimensional) to the complete and so closed set $M\subset H$, and the end of your proof works well.

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  • $\begingroup$ i just dont understand why it has to be finite dimensional to be complete and if M is the span how do you know it finite?? $\endgroup$ – Sasha Feb 11 '18 at 15:58
  • $\begingroup$ It has not to be finite dimensional to be complete, but it is sufficient. It is finite by your hypothesis: $m=\mathrm{span}(e_1,\dots,e_n)$. $\endgroup$ – Balloon Feb 11 '18 at 16:00
  • $\begingroup$ yes i just dont understand how the span of a finite set is finite $\endgroup$ – Sasha Feb 11 '18 at 16:00
  • $\begingroup$ i don't really understand your proof... you're saying that because M is a subset of H then it is closed? $\endgroup$ – Sasha Feb 11 '18 at 16:02
  • $\begingroup$ It is not finite, it is finite dimensional (over the fields $\mathbb{R}$ or $\mathbb{C}$, the only finite vector space is $\{0\}$). No, I am saying that $M$ is a complete subset of $H$ and so closed, as you say at the end of your first paragraph. $\endgroup$ – Balloon Feb 11 '18 at 16:04
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Let $X$ be a normed linear space over $S$ where $S=\Bbb R$ or $S=\Bbb C$.

THEOREM. If $Y$ is a closed linear sub-space of $X$ and $v\in X\setminus Y$ then $Z=\{y+sv:y\in Y\land s\in S\}$ is closed.

Proof: Suppose a sequence $(y_n+s_nv)_n$ converges in norm to $z,$ with $y_n\in Y$ and $s_n\in S.$

Case (i). If $(s_n)_n$ is not a bounded sequence then it has a sub-sequence $(s_{n_j})_j$ with $|s_{n_j}| \to \infty$ as $j\to \infty,$ and with each $s_{n_j}\ne 0.$ Then $$0=\lim_{j\to \infty} \|(s_{n_j})^{-1}y_{n_j}+v-(s_{n_j})^{-1}z\|$$ $$\text { and }\quad 0=\lim_{j\to \infty}\| (s_{n_j})^{-1}z\|.$$ $$ \text {But these imply that }\quad \lim_{j\to \infty}\|(s_{n_j})^{-1}y_{n_j}+v\|$$ which in turn implies that $(-v)\in \overline Y=Y,$ and hence $v\in Y$ (because $Y$ is a vector space), contradicting $v\not\in Y.$ So this case is impossible.

Case (ii). If $(s_n)_n$ is a bounded sequence then it has a sub-sequence $(s_{n_i})_i$ converging to some $s\in S.$ Then as $i\to \infty$ we have $y_{n_i}+(s_{n_i}-s)v \to z-sv\;$ and we have $(s_{n_i}-s)v\to 0. \;$ So $y_{n_i} \to z-sv.\;$ So with $y=z-sv$ we have $y\in \overline Y=Y,$ and $$z=(z-sv)+sv=y+sz\in Z.$$

COROLLARY: If $Z$ is a finite dimensional vector sub-space of $X$ then $Z$ is closed.

Proof: If not, then take a counter-example $Z$ with the least possible dimension, $m$ . Clearly $m\ne 0.$ (Note: $\dim Z=0$ iff $Z=\{0\}.$) So let $Z$ be the linear span of $B=(\{y_j: 1\leq j\leq m\},$ with $m>0.$ By the minimality of $m,$ the linear span $Y$ of $B\setminus \{y_m\}$ is closed. (Note: In case $m=1,$ the linear span of $\emptyset$ is defined to be $\{0\}.$) And $y_m\in X\setminus Y.$ By the theorem, $Z=\{y+sy_m:y\in Y\land s\in S\}$ is closed, contrary to $Z$ being a counter-example.

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