Advice with applied math problem: Let M ⊂ H be closed linear supspace...

nikhil714

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

Prove that m = inf (f, u) is uniquely achieved.
u∈M
|u|=1

I have approached it with the cosine definition for inner product but that only works with Euclidean spaces. I want to know how to apply this using the
Cauchy-Schwartz inequality so it works for Banach Spaces. Can anyone help? I have placed my work below.



The Hilbert Projection Theorem says that there exists a unique object
in M that minimizes the distance to f. It also happens to be the projection.

So let f = p + q, where p is in M, and q is in M⊥. (We use the projection theorem on M and M⊥ s.t. p is the projection onto M, and q is the projection onto M⊥)

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

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

\(\displaystyle (u,\, p)\, =\, \Vert{u}\Vert\, \Vert{p}\Vert \, \cos(\theta)\, =\, \Vert{p}\Vert\, \cos(\theta)\)

this is uniquely minimized when \(\displaystyle \, \cos(\theta)\, =\, -1,\,\) so u is on the opposite side in M, but with a norm of 1 (it is one away from the origin and thus M⊥)

\(\displaystyle u\, =\, \dfrac{-p}{\Vert p \Vert}\)

\(\displaystyle m\, =\, (u,\, p)\, =\, \left(\dfrac{-p}{\Vert p \Vert},\, p\right)\, =\, \dfrac{-(p,\, p) }{ \Vert p \Vert }\, =\, \dfrac{-\Vert p \Vert^2}{\Vert p \Vert}\, =\, -\Vert p \Vert\)
 
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