Math 641-600 — Fall 2023

Current Assignment

Assignment 5 -- These are problems to be discussed in class.

Read sections 3.3-3.5, and my notes on Compact Operators, and on the Closed Range Theorem.
Look at the following problems.
1. Let $k(x,y)$ be defined by
  \[ k(x,y) = \left\{ \begin{array}{cl} y, & 0 \le y \le x\le 1, \\ x, & x \le y \le 1. \end{array} \right. \] Show that $k(x,y)$ is a Hilbert-Schmidt kernel and that the corresponing operator $Ku(x)=\int_0^1 k(x,y)u(y)dy$ satisfies $\|K\|_{L^2\to L^2} \le \sqrt{\frac{1}{6}}$.
2. Consider the Hilbert space $\mathcal H=\ell^2$ and let $S=\{x=(x_{1}\ x_{2}\ x_3\ \ldots)\in \ell^2: \sum_{n=1}^\infty (n^2+1)|x_n|^2 <1\}$. Show that $S$ is a precompact subset of $\ell^2$.
3. Consider the finite rank (degenerate) kernel
  k(x,y) = φ₁(x)ψ₁(y) + φ₂(x)ψ₂(y), where φ₁ = 6x-3, φ₂ = 3x², ψ₁ = 1, ψ₂ = 8x − 6.
  Let Ku= ∫₀¹ k(x,y)u(y)dy. Assume that L = I-λ K has closed range,
  1. For what values of λ does the integral equation
    u(x) - λ∫₀¹ k(x,y)u(y)dy =f(x)
    have a solution for all f ∈ L²[0,1]?
  2. For these values, find the solution u = (I − λK)⁻¹f — i.e., find the resolvent.
  3. For the values of λ for which the equation does not have a solution for all f, find a condition on f that guarantees a solution exists. Will the solution be unique?
4. In the following, H is a Hilbert space and B(H) is the set of bounded linear operators on H. Let L be in B(H) and let N:= sup {|< Lu, u>| : u ∈ H, ||u|| = 1}.
  1. Verify the identity < L(u+αv), u+αv> − < L(u-αv), u-αv> = 2α< Lu,v>+2α< Lv,u>, where |α| = 1.
  2. Show that N ≤ ||L||.
  3. Let L be a self-adjoint operator on H, which may be real or complex. Use (a) and (b) to show that N= ||L||. (Hint: In the complex case, choose α so that α< Lu,v> = |< Lu,v>|. For the real case, use $\alpha=\pm 1$, as required.)
  4. Suppose that H is a complex Hilbert space. If L ∈ B(H), then use (a) and (b) to show that
    N ≤ ||L|| ≤ 2N.
  5. For the real Hilbert space, H = R², let $L = \begin{pmatrix} 0& 1\\ -1 & 0 \end{pmatrix}. $ Show that $\|L\| = 1$, but $N=0$.

Updated 9/25/2023.