Projection Theorem. If $M$ is a closed subspace of the Hilbert space $H$ and $x\in H$, then (i) there is a unique element $x'\in M$ such that $$ \lVert x-x'\rVert=\inf_{y\in M}\lVert x-y\rVert, $$ and (ii) $x'\in M$ and $\lVert x-x'\rVert=\inf_{y\in M}\lVert x-y\rVert$ if and only if $x'\in M$ and $(x-x')\in M^{\bot}$.

I have two questions, one concerning the proof of (i) and another concerning the proof of (ii).

(i) The proof of $(i)$ starts the following way:

If $d=\inf_{y\in M}\lVert x-y\rVert^2$, then there is a sequence $\left\{y_n\right\}$ of elements of $M$ such that $\lVert y_n-x\rVert^2\to d$. [...]

Why does such a sequence exist? Is it since we can associate a decreasing sequence $(\lVert y_n-x\rVert^2)$ which is bounded and therefore has to converge, say to $K$ and $K$ has to be identical to $d$? - Since if $K<d$, then $d$ isn't the infimum and if $K>d$, then for a sufficient small neighborhood of K we could have an $N$ such that there can be at least one element $\lVert y_n-x\rVert^2$ not in this neighborhood for $n\geq N$?


One direction of (ii) is proven as follows:

If $x'\in M$ and $(x-x')\notin M^{\bot}$ then $x'$ is not the element of $M$ closest to $x$ since $$ x''=x'+ay/\lVert y\rVert^2 $$ is closer, where $y$ is any element of $M$ such that $\langle x-x',y\rangle\neq 0, a=\langle x-x',y\rangle$. To see this, we write $$ \lVert x-x''\rVert^2=\langle x-x'+x'-x'',x-x'+x'-x''\rangle\\ =\lVert x-x'\rVert^2+\lvert a\rvert^2/\lVert y\rVert^2 +2\Re\langle x-x',x'-x''\rangle\\ =\lVert x-x'\rVert^2-\lvert a\rvert^2/\lVert y\rVert^2\\ <\lVert x-x'\rVert^2. $$

I do not understand the third equality, i.e. $$ \lVert x-x'\rVert^2+\lvert a\rvert^2/\lVert y\rVert^2 +2\Re\langle x-x',x'-x''\rangle=\lVert x-x'\rVert^2-\lvert a\rvert^2/\lVert y\rVert^2\ $$

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    $\begingroup$ Considering (i): Basically you got it. This follows from the definition of the infimum as seen here. Had exactly the same question when seeing this proof for the first time. :) $\endgroup$ – Piwi Oct 22 '15 at 16:17
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    $\begingroup$ Considering (ii): To see that this is true, note that this equality is equivalent to $\frac{|a|^2}{\lVert y\rVert ^2}+\mathfrak{R}\langle x-x',x'-x''\rangle=0$. Replace $x''$ in the product by its definition $x''=x'+ay/\lVert y\rVert^2$, pull $-\frac{a}{\lVert y\rVert^2}$ out of the second argument of the product, note that the remaining product reads $\langle x-x',y\rangle$ which is by definition the same as $a$. So in the end you have $\frac{|a|^2}{\lVert y\rVert^2}-\frac{|a|^2}{\lVert y\rVert^2}$ on the left side which vanishes. A bit sketchy but I'm confident this should put you on track. $\endgroup$ – Piwi Oct 22 '15 at 16:30
  • $\begingroup$ I get $2\Re\langle x-x',x'-x''\rangle=2\Re(\overline{\left(-\frac{a}{\lVert y\rVert^2}\right)}\cdot a)=2\Re (\overline{-\frac{1}{\lVert y\rVert^2}}\cdot\overline{a}\cdot a)=2\Re (\overline{-\frac{1}{\lVert y\rVert^2}}\cdot\lvert a\rvert^2)$ $\endgroup$ – M. Meyer Oct 22 '15 at 17:23
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    $\begingroup$ $-\frac{1}{\lVert y\rVert^2}$ is real, isn't it? Same for $|a|^2$. So $2\mathfrak{R} (\overline{-\frac{1}{\lVert y\rVert^2}}\cdot\lvert a\rvert^2)=-2\frac{|a|^2}{\lVert y\rVert^2}$. $\endgroup$ – Piwi Oct 22 '15 at 17:48
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    $\begingroup$ Just a small remark/reminder: If you know that $\lVert\cdot \rVert$ is a norm, you know that it maps to $\mathbb{R}$ (even $\mathbb{R}_{\geq 0}$). This should be clearly stated in any definition of "norm" (see Wikipedia for example). So technically you don't need to check again that $\lVert \cdot \rVert$ is real. You already did this by verifying that it indeed is a norm. $\endgroup$ – Piwi Oct 23 '15 at 0:17

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