# Homework Help: Projection Theorem

1. Jan 17, 2008

### P3X-018

[SOLVED] Projection Theorem

1. The problem statement, all variables and given/known data
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

$$x-y\in M^{\perp}$$

3. 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 "<="?

Last edited: Jan 17, 2008
2. Jan 17, 2008

### 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.

Last edited: Jan 17, 2008
3. Jan 18, 2008

### 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?

4. Jan 18, 2008

### 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.

Last edited: Jan 18, 2008
5. Jan 20, 2008

### 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

$$(t \|z\|^2 + 2Re(x-y,z))t < 0$$
$$0< t < -\frac{2Re(x-y,z)}{\|z\|^2}$$

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 $-2Re(x-y,z)/\|z\|^2>0$ or $Re(x-y,z)<0$.
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?