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Suppose $X$ is a Hilbert space, $M\subset X$ is a closed subspace and $y\notin M$. Let $d = \inf\{ \|x-y\|:x\in M\}$ show that if $\{x_n\}_{1}^{\infty}$ and $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ then $\{x_n\}$ is Cauchy.

Attempted proof - Let $\{x_n\}_{1}^{\infty}\subset M$ such that $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ with $d = \inf\{ \|x-y\|:x\in M\}$. By the parallelogram law $$\lVert x_n - x_m\rVert^2 = 2\lVert x_n\rVert^2 + 2\lVert x_m\rVert^2 - \lVert x_n + x_m\lVert^2$$ If $M$ is convex then $\frac{1}{2}\lVert x_n + x_m\rVert\in M$. Therefore, $\lVert \frac{1}{2}(x_n + x_m)\rVert \geq d$ hence $$\lVert x_n - x_m\rVert^2 = 2\lVert x_n\rVert^2 + 2\lVert x_m\rVert^2 - 4d^2$$ As $m,n\rightarrow \infty$, this quantity goes to zero, therefore $\{x_n\}$ is a Cauchy sequence.

I am not sure if this is correct any suggestions would be greatly appreciated.

Wolfy
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  • There are some little mistakes, maybe typos. First, for $d$ you need to take an $\inf$. Then in the parallelogram law, the lhs should be a square (in at least 2 places). The idea is fine. – Thomas Apr 28 '16 at 16:42
  • (oh, and the last 'equality' is only an inequality) – Thomas Apr 28 '16 at 16:43
  • @Thomas Is it correct now, I changed the typos I made – Wolfy Apr 28 '16 at 16:47
  • The one thing I was concerned about was making $M$ convex I feel like that is a too much of an assumption to make what do you think?? – Wolfy Apr 28 '16 at 16:48
  • The result is not true if $M$ is not convex, not even in finite dimensions. Presumably 'subspace' means linear subspace. Still you don't get equality in your last equation. – Thomas Apr 28 '16 at 17:32
  • @Thomas so is the proof absolutely wrong? – Wolfy Apr 28 '16 at 17:33
  • No, the proof is correct if $M$ is assumed to be convex and you fix the small issues. (It's your lucky day, the last line is an inequality and not an equality, but the direction is the one you need). – Thomas Apr 28 '16 at 19:05
  • @Thomas I see but is it a big assumption to let $M$ be convex when in the problem it just says that $M$ is closed? – Wolfy Apr 28 '16 at 19:33
  • Convex sets are common objects of study (one reason is the property you are considering here). And, as I mentioned, the claim is wrong if convexity is dropped from the assumptions. – Thomas Apr 29 '16 at 04:43
  • I see so is there a way of proving the statement without using the assumption that we have convexity for $M$? – Wolfy Apr 29 '16 at 14:55
  • If the claim is wrong in that case then you cannot proof it. – Thomas Apr 29 '16 at 15:37

1 Answers1

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@Wolfi , your proof is correct (except for a minor typo). Please note that in the context of Hilbert and Banach space, "subspace" means "linear subspace" and any linear subspace $M$ of $X$ is a convex subset of $X$. In fact, if $M \subset X$ is a subspace, then for all $x,y \in M$ and all $\lambda \in [0,1]$, we have $\lambda x +(1-\lambda) y \in M$.

So actually your proof proves a generalization of the original question.

Original Question: Suppose $X$ is a Hilbert space, $M\subset X$ is a closed subspace and $y\notin M$. Let $d = \inf\{ \|x-y\|:x\in M\}$ show that if $\{x_n\}_{1}^{\infty}$ and $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ then $\{x_n\}$ is Cauchy.

Proof - Let $\{x_n\}_{1}^{\infty}\subset M$ such that $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ with $d = \inf\{ \|x-y\|:x\in M\}$. By the parallelogram law $$\lVert x_n - x_m\rVert^2 = 2\lVert x_n\rVert^2 + 2\lVert x_m\rVert^2 - \lVert x_n + x_m\lVert^2$$ Since $M$ is subspace of $X$ we have that $\frac{1}{2}( x_n + x_m)\in M$. Therefore, $\lVert \frac{1}{2}(x_n + x_m)\rVert \geq d$ hence $$\lVert x_n - x_m\rVert^2 = 2\lVert x_n\rVert^2 + 2\lVert x_m\rVert^2 - 4d^2$$ As $m,n\rightarrow \infty$, this quantity goes to zero, therefore $\{x_n\}$ is a Cauchy sequence.

Broader Question: Suppose $X$ is a Hilbert space, $M$ is a closed convex subset of $X$ and $y\notin M$. Let $d = \inf\{ \|x-y\|:x\in M\}$ show that if $\{x_n\}_{1}^{\infty}$ and $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ then $\{x_n\}$ is Cauchy.

Proof - Let $\{x_n\}_{1}^{\infty}\subset M$ such that $\lim_{n\rightarrow \infty}\|x_n - y\| = d$ with $d = \inf\{ \|x-y\|:x\in M\}$. By the parallelogram law $$\lVert x_n - x_m\rVert^2 = 2\lVert x_n\rVert^2 + 2\lVert x_m\rVert^2 - \lVert x_n + x_m\lVert^2$$ If $M$ is convex then $\frac{1}{2}(x_n + x_m)\in M$. Therefore, $\lVert \frac{1}{2}(x_n + x_m)\rVert \geq d$ hence $$\lVert x_n - x_m\rVert^2 = 2\lVert x_n\rVert^2 + 2\lVert x_m\rVert^2 - 4d^2$$ As $m,n\rightarrow \infty$, this quantity goes to zero, therefore $\{x_n\}$ is a Cauchy sequence.

Remark 1: As I mentioned in the beginning of this answer, the original question is a corollary of the broader question.

Remark 2: Since we want to prove that $\{x_n\}$ is a Cauchy sequence, we needed not use (in both questions) the assumption that $M$ is closed. Such assumption is required to conclude that $\{x_n\}$ converges to some $x\in M$.

Ramiro
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