Math 641-600 Current Assignment

Assignment 9 - Due Tuesday, November 27.

• Read sections 3.3 and 3.4.
• Do the following problems.
  1. Prove the polarization identity for u,v in a hilbert space H:
     ||u+v||² + ||u-v||² = 2(||u||² + ||v||²).
     

  2. Let M is a closed subspace of a Hilbert space H.
     1. Show that if h ∈ H and if
        α := inf_v∈M|| h- v||,
        then there is a sequnce of vectors {p_j} in M for which
        lim_j→∞|| h- p_j|| = α
        
     2. In class, we used the polarization identity to show that
        ||p_j − p_k||² = 2||h − p_j||² + 2||h − p_k||² − 4||h − ½(p_j+p_k)||².
        Use this identity to prove that {p_j} is a Cauchy sequence. From there, finish the proof of the Projection theorem.
     3. Show that M is closed if and only if M = (M^&perp)^&perp.

  3. Weak solutions. In this problem, let
     < f,g >_H = ∫₀¹(f ′(x) g ′(x) +P(x)f(x)g(x))dx,
     where f,g are continuous functions vanishing at x = 0 and x = 1, and having L² derivatives. In addition, we assume that P is continuous and strictly positive on [0,1]. Show that the boundary value problem (BVP) below has a unique weak solution in H, given that H is known to be a Hilbert space. (Hint: use the Riesz Representation Theorem.)
     -u''+P(x)u = f(x), u(0) = u(1) = 0, f ∈ L²[0,1].

  4. Section 3.2 problem 3(d), page 128. (Assume the appropriate operators are closed and that λ is real.)

  5. Section 3.3 problem 2, page 129. (Assume the appropriate operators are closed and that λ is real.)

  6. A sequence {f_n} in H is said to be weakly convergent to f∈H if and only if lim _{n → ∞} < f_n,g> = < f,g> for every g∈H. When this happens, we write f = w-lim f_n. For example, in class we showed that if {φ_n} is any orthonormal sequence, then φ_n converges weakly to 0. One can show that every weakly convergent sequence is a bounded sequence; that is, there is a constant C such that ||f_n|| ≤ C for all n. Prove the following:

     Let K be a compact linear operator on a Hilbert space H. Show that if f_n weakly converges to f, then Kf_n converges strongly to Kf – that is,
     lim _{n → ∞} || Kf_n - Kf || =0.

     Hint: Suppose this doesn't happen, then there will be a subsequence of {f_n}, say {g_k}, such that
     || Kg_k - Kf || ≥ ε
     for all k. Use this and the compactness of K to arrive at a contradiction.

Updated 11/15/07 (fjn).