MA3422 (Functional Analysis 2) Tutorial sheet 4
[February 13, 2015]
Name: Solutions
1. For a self-adjoint operator T (i.e. with T = T ∗ ) on H = Kn (the n-dimensional euclidean
space) show that the operator norm of T is the absolute value of the largest eigenvalue of
T . [Hint: the spectral theorem says that there is an orthonormal basis for Kn consisting of
eigenvectors for T .]
Solution: As in the hint, choose an orhonormal basis φ1 , φ2 , . . . , φn for Kn conisting of
eigenvectors for T . Say the eigenvalue for φj is λj ∈ K so that T φj = λj φj .
If we use coordinates
with respect to the basis φ1 , φ2 , . . . , φn every x ∈ Kn can be exPn
pressed x = j=1 hx, φj iφj and then
Tx =
n
X
hx, φj iT φj =
j=1
n
X
λj hx, φj iφj
j=1
So in the coordinates with respect to this basis, T is a multiplication operator as in question
2 of the previous sheet.
v
uX
u n
|λj hx, φj iφj |2
kT xk = t
j=1
v
uX
u n
= t
|λj |2 |hx, φj iφj |2
j=1
v
uX
u n
≤ t ( max |λk |2 )|hx, φj iφj |2
j=1
=
1≤k≤n
v
uX
u n
max |λk |t
|hx, φj iφj |2
1≤k≤n
j=1
= ( max |λk |)kxk
1≤k≤n
So kT k ≤ max1≤k≤n |λk |. Since kT φk k/kφk k = kλk φk k/1 = |λk |, we also have kT k ≥
|λk | for all k and so kT k = max1≤k≤n |λk |.
2. If T ∈ B(H) is a bounded operator on a Hilbert space, show that all eigenvalues of T ∗ T
are real and nonnegative.
Solution: Suppose T ∗ T x = λx with x 6= 0 (so that x is an eigenvector of T ∗ T with
eigenvalue λ). Then
hT ∗ T x, xi = hλx, xi = λhx, xi = λkxk2
But also
hT ∗ T x, xi = hx, T ∗ T xi = hT x, T xi = kT xk2 = kT xk2
So
kT xk2
λ = kxk = kT xk ⇒ λ =
≥0
kxk2
2
2
3. Let E = `2 ∩ `1 with the norm k · k2 and F = `1 . Define T : E → F by T x = x. Show
that T is not bounded but has closed graph. [Hint: consider x = (1, 1, . . . , 1, 0, 0, . . .) with
n ones and the rest of the terms 0 to show that T is not bounded.]
Solution:
As in the hint consider
xn = (1, 1, . . . , 1, 0, 0, . . .) with n ones. Then kxn k2 =
√
√
12 + 12 + · · · 12 + 0 = n while
√ kT xn k = kxn k1 = 1 + 1 + · · · + 1 + 0 = n. There is
no finite constant C with n ≤ C n for all n and hence T is noyt bounded.
Suppose now (xn , yn ) is in the graph of T for all n (different xn from above) and (xn , yn ) →
(x, y) in E × F (as n → ∞). So yn = T xn = xn . Also xn → x in E and yn = xn → y in
F . So kxn − xk2 → 0 as n → ∞ and kxn − yk1 → 0 as n → ∞.
Then x ∈ `2 and y ∈ `1 . But also, the k th term of xn converges (in K) to the k th term of x
(for each k ∈ N) and to the k th term of y.
That is because if xn,k is the k th term of xn and x∞,k is the k th term of x, then |xn,k −
x∞,k | ≤ kxn − xk2 and a similar fact holds for the norm k · k1 .
So x = y ∈ `2 ∩ `1 . That means (x, y) = (x, x) is in the graph of T and thus the graph is
closed.
Richard M. Timoney
2