Solved: Projection Theorem in Hilbert Space

[P3X-018]

[SOLVED] Projection Theorem

Homework Statement

If M is a closed subspace of a Hilbert space H, let x be any element in H and y in M, then I have to show that

\|x-y\| =\inf_{m\in M}\|x-m\|

implies (equivalent to) that

x-y\in M^{\perp}

The Attempt at a Solution

I have shown the implication "<=", ie that x-y\in M^{\perp} implies the 1st statement. And I've been told that the implication (=>) I now want to show is basically in the that proof.

The proof for "<=" goes like: If x-y\in M^{\perp} then (x-y,z) = 0 for all z in M, so by Pythagoras thm

\|x-y+z\|^2 = \|x-y\|^2 +\|z\| \geq \|x-y\|^2

M subspace => m = y-z \in M. So \|x-y\| \leq \|x-m\| for all m in M. So this gives the implication the other way around.

But I don't see how I can go back, nor how the 'proof' for "=>" is basically contained in my this proof.
And how or where should I use that M is closed, I feel like it should have been used to conclude the implication "<=" from knowing \|x-y\| \leq \|x-m\|, or is it not needed for "<="?

 

[StatusX]

Here's a summary of what you've already shown (assume x, y are fixed vectors, with y in M, and z is an arbitrary vector in M, so that w=z-y is also an arbitrary vector in M):

x-y \in M^\perp \Rightarrow (x-y,z)=0 \Rightarrow ||x-y+z||^2 = ||x-y||^2 + ||z||^2 \Rightarrow ||x-y|| \leq ||x-w|| \Rightarrow ||x-y|| = \inf_{m\in M}\|x-m\|

Now to get the other direction, start by seeing how many of those \Rightarrow's you can turn into \Leftarrow's (or to get both directions at once, \Leftrightarrow's). The one that should give you the most difficulty is:

||x-y+z||^2 = ||x-y||^2 + ||z||^2 \Leftarrow||x-y|| \leq ||x-y+z||

The condition can be written as:

||x-y||^2 \leq ||x-y+z||^2 = ||x-y||^2 + 2 \Re (x-y,z) + ||z||^2

\Leftrightarrow 0 \leq ||z||^2 + 2\Re (x-y,z)

Think about what happens when z is really small.

 

[P3X-018]

So using Cauchy Schwartz, since (x-y,z) must be real, I get

0 \leq \|z\|^2 + 2(x-y,z) \leq \|z\|(\|z\| + 2\|x-y\|)

So that \|z\|^2 + 2(x-y,z) gets 'squeezed' to zero, and hence (x-y,z) too? But if z varies it'll affect (x-y,z) too, so that thing can get small too. So how does varying z helps me in this?

 

[StatusX]

Sorry, I'm not following your last post. One way to continue would be to write:

f(t) = ||tz||^2 + 2 \Re (x-y,tz)

Given the above inequality, we can write:

0 \leq f(t) = t^2||z||^2 + 2 \Re (x-y,z) t

But's it's easy to see that this will be negative at some points unless \Re(x-y,z)=0. There's a trick to get the same thing for the imaginary part. But I have a feeling there's a simpler way to do this.

 

[P3X-018]

That was my mistake I thought (x-y,z) should be real so I used Cauchy-Schwartz, but never mind that.

I see that there could be some points where that inequality could fail, but doesn't that need to be proved first? t would have to fulfill

(t \|z\|^2 + 2Re(x-y,z))t &lt; 0
0&lt; t &lt; -\frac{2Re(x-y,z)}{\|z\|^2}

So that f(t) would be negative. Don't we need to show that such a z in M would exist? Ie any z such that -2Re(x-y,z)/\|z\|^2&gt;0 or Re(x-y,z)&lt;0.
Otherwise it is clear how Re(x-y,z) has to be 0.

But I'm a little clueless on how to show Im(x-y,z) = 0, what is the trick you were talking about?