nikhil714

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 Mperp. (We use the projection theorem on M and Mperp s.t. 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). (u,p) = ||u|| ||p|| cos(theta) = ||p|| cos(theta) this is uniquely minimized when 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 Mperp) u = -p/(||p||) m = (u,p) = (-p/||p||,p) = -(p,p)/||p|| = -||p||^2 / ||p|| = -||p||

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.

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