Download Sample Midterm Problems and Solutions for Math 201B, Winter 2007 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!
Sample Midterm Problems
Brief Solutions
Math 201B, Winter 2007
Problem 1. Let E ⊂ R be a measurable subset of R. Define a linear
subspace M of L
2 (R) by
M =
f ∈ L
2 (R) | f (x) = 0 a.e. in E
Find M
⊥
. What is the orthogonal projection of f ∈ L
2 (R) onto M?
Solution.
c = R \ E be the complement of E. Then
M
⊥
f ∈ L
2 (R) | f (x) = 0 a.e. in E
c
- The direct-sum decomposition of f is
f = χEc f + χE f,
where χE is the characteristic function of E, and χEc f is the orthogonal
projection of f onto M.
Problem 2. Let
E =
(x, y) ∈ R
2 | 0 < x < ∞, 0 < y < 1
Prove that (^) ∫
E
y sin xe
−xy dxdy =
log 2.
Solution.
- To apply Fubini’s theorem, first check that
∫
E
∣y sin xe−xy
∣ (^) dxdy ≤
E
ye
−xy dxdy
0
0
ye
−xy dx
dy
- Then compute the iterated integral
0
0
y sin xe
−xy dx
dy.
Problem 3. Let T, S ∈ L
2 (T) be the triangular and square waves, defined
respectively by
T (x) = |x| if |x| < π, S(x) =
1 if 0 < x < π,
− 1 if −π < x ≤ 0,
Compute the Fourier coefficients of T , S. Show that T ∈ H
1 (T) and T
′ = S.
Show that S /∈ H
1 (T).
Solution.
- From the definition of the Fourier coefficient,
fˆ n =^
2 π
∫ (^) π
−π
f (x)e
−inx dx,
we compute that
Tˆ
n =
π
[
n − 1
n^2
]
n 6 = 0, Tˆ 0 =
π
2
√ 2 π
S^ ˆ
n =^ i
π
[
n − 1
n
]
n 6 = 0, Sˆ 0 = 0.
n∈Z
n
2
Tˆ
n
2
converges, since
1 /n
2 converges, so T ∈ H
1 (T). Since Sˆn = in Tˆn,
we have S = T
′ .
n∈Z
n
2
Sˆ
n
2
does not converge, since the terms do not approach zero as n → ∞, so
S /∈ H
1 (T).
Problem 5. An indexed set of vectors {uα | α ∈ A} in a Hilbert space H is
said to be stable if there exist constants m, M > 0 such that for all
{cα | cα ∈ C, α ∈ A} ∈ `
2 (A)
we have
m
α∈A
|cα|
2 ≤
α∈A
cαuα
2
≤ M
α∈A
|cα|
2 .
(a) Show that a stable set is linearly independent, and an orthonormal set is
stable.
(b) Suppose that {uα | α ∈ A} is a set of normalized vectors (‖uα‖ = 1) such
that (^) ∑
α 6 =β
|〈uα, uβ 〉|
2 < 1.
Show that {uα | α ∈ A} is stable.
(c) Let {en | n ∈ Z} be an orthonormal set, and define
un =
en + en+ √ 2
Show that {un | n ∈ Z} is not stable.
Solution.
- (a) If {uα | α ∈ A} is stable and
∑
α∈A
cαuα = 0,
then ∑
α∈A
|cα|
2 = 0.
Hence cα = 0 for all α ∈ A, so a stable set is linearly independent. (In
fact, what we have shown is stronger than linear independence, since
we did not consider only finite linear combinations of the vectors.)
- For an orthonormal set ∥ ∥ ∥ ∥ ∥
α∈A
cαuα
2
α∈A
|cα|
2
so the set is stable, with m = M = 1.
α∈A
cαuα
2
α,β∈A
cαcβ 〈uα, uβ 〉.
Using the Cauchy-Schwarz inequality, we get
∑
α,β∈A
cαcβ 〈uα, uβ 〉 =
α∈A
|cα|
2
α 6 =β
cαcβ 〈uα, uβ 〉
α∈A
|cα|
2
α 6 =β
|cαcβ |
2
α 6 =β
|〈uα, uβ 〉|
2
α∈A
|cα|
2
α∈A
|cα|
2
α 6 =β
|〈uα, uβ 〉|
2
α∈A
|cα|
2 ,
and
∑
α,β∈A
cαcβ 〈uα, uβ 〉 =
α∈A
|cα|
2
α 6 =β
cαcβ 〈uα, uβ 〉
α∈A
|cα|
2 −
α 6 =β
|cαcβ |
2
α 6 =β
|〈uα, uβ 〉|
2
α 6 =β
|〈uα, uβ 〉|
2
α∈A
|cα|
2 .
- (c) Consider, for example, cn = (−1)
n for 1 ≤ n ≤ N and cn = 0
otherwise. Then (^) ∑
n∈Z
|cn|
2 = N,
and (^) ∥ ∥ ∥ ∥ ∥
n∈Z
cnun
2
N eN +1 − e 1 √ 2
2
Since N ∈ N is arbitrary, there is no constant m > 0 with the required
property for stability.
