Hilbert Spaces Help

[nikhil714]

Let M ⊂ H be a closed linear subspace that is not reduced to {0}. Let
f ∈ H,f /∈ M⊥.

1. Prove that

m = inf(f, u)
u∈M
|u|=1


is uniquely achieved.

2. Let ϕ1, ϕ2, ϕ3 ∈ H be given and let E denote the linear space spanned by
{ϕ1, ϕ2, ϕ3}. Determine m in the following cases:
(i) M = E,
(ii) M = E⊥.

3. Examine the case in which H = L2(0, 1), ϕ1(t) = t,ϕ2(t) = t2, and ϕ3(t) = t3

[hypervalent_iodine]

!

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[nikhil714]

Let M ⊂ H be a closed linear subspace that is not reduced to {0}. Let

f ∈ H,f /∈ M⊥.

 

1. Prove that

 

m = inf(f, u)

u∈M

|u|=1

 

 

is uniquely achieved.

 

2. Let ϕ1, ϕ2, ϕ3 ∈ H be given and let E denote the linear space spanned by

{ϕ1, ϕ2, ϕ3}. Determine m in the following cases:

(i) M = E,

(ii) M = E⊥.

 

3. Examine the case in which H = L2(0, 1), ϕ1(t) = t,ϕ2(t) = t2, and ϕ3(t) = t3

So this is what I have been working with based off of the definitions and examples I have seen.

 

I can decompose objects in a Hilbert space into the part in M and the part in Mperp

 

The Hilbert Projection Theorem says that there exists a unique object in M that minimizes the distance to f, which is the projection.

 

So let f = p + q, where p is in M, and q is in Mperp. (I think I can use the projection theorem on M and Mperp such that p is the projection onto M, and q is the projection onto Mperp)

 

(u,f) = (u, p+q) = (u,p) + (u,q)

 

Now (u,q) = 0 because u in M, q is in Mperp so (u,f) = (u,p).

 

As you can see, inner producting with something from a subspace only takes into account the part in that subspace.

 

Now, I'm not too sure what to do with this:

 

(u,p) = ||u|| ||p|| cos(theta) = ||p|| cos(theta)

 

 

 

Still quite lost with parts 2 and 3 of the problem.

[hoponpop]

You may be overthinking this, or I may be under thinking it. Lol

 

I think you're right to decompose and go to the projection theorem, but from there isn't it just a matter of proving the projection theorem for each M and Mperp? : /