Math 641600 — Fall 2018
Current Assignment
Assignment 10  Due Wednesday, November 14, 2018.
 Read sections 3.33.5, and my notes on Compact
Operators, and on the
Closed Range Theorem.
 Do the following problems.
 Section 3.3: 1 (Assume the appropriate
operators are closed and that λ is real.)
 Section 3.4: 2(b)
 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 $\ell^2$.
 Let $S$ be a bounded subset (not a subspace!) of a Hilbert space
$\mathcal H$. Show that $S$ is precompact if and only if every
sequence in $S$ has a convergent subsequence. (Note: If $S$ is just
precompact, the limit point of the sequence may not be in $S$, because
$S$ may not be closed.)
 Show that every compact operator on a Hilbert space is bounded.
 Consider the finite rank (degenerate) kernel
k(x,y) =
φ_{1}(x)ψ_{1}(y) +
φ_{2}(x)ψ_{2}(y),
where φ_{1} = 6x3, φ_{2} = 3x^{2},
ψ_{1} = 1, ψ_{2} = 8x − 6.
Let Ku= ∫_{0}^{1} k(x,y)u(y)dy. Assume that L =
Iλ K has closed range,

For what values of λ does the integral equation
u(x)  λ∫_{0}^{1} k(x,y)u(y)dy =f(x)
have a solution for all f ∈ L^{2}[0,1]?
 For these values, find the solution u = (I −
λK)^{−1}f — i.e., find the resolvent.
 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?
 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}.
 Verify the identity < L(u+αv), u+αv> − <
L(uαv), uαv> = 2α<
Lu,v>+2α< Lv,u>, where α = 1.
 Show that N ≤ L.
 Let L be a selfadjoint 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.)
 Suppose that H is a complex Hilbert space. If L ∈
B(H), then use (a) and (b) to show that
N ≤ L ≤ 2N.
 For the real Hilbert space, H = R^{2}, let $L =
\begin{pmatrix}
0& 1\\
1 & 0 \end{pmatrix}.
$
Show that $L = 1$, but $N=0$.
Updated 11/7/2018.