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Haoran Sun (haoransun@link.cuhk.edu.cn) Problem 1 (P117 Q4).
(a) Note that the formula for cosine coefficients is
1 l nπx
An = Z ϕ(x)cos dx
l −l l
Since ϕ(x) is odd and cos nπxl is even, we have ϕ(x)cos nπxl an odd function. Hence An = 0.
(b) Note that the formula for sine coefficients is
1 l nπx
Bn = Z ϕ(x)sin dx
l −l l
Since ϕ(x) even and sin nπxl odd, we have ϕ(x)sin nπxl odd. Hence Bn = 0.
Problem 2 (P117 Q8).
(a) Suppose f is even and differentiable on (−l,l). Then
f′(x0) = lim f(x0 + h) − f(x0) = lim −f(−x0) − f(−x0 − h) = −f′(−x0) h→0 h h→0 h
Then f′ is odd.
Following similar steps, we can also prove that f′ is even given that f is odd.
(b) Suppose f even and integrable on (−l,l). Then
F(x) =f(t)dt = Z −f(−t)d(−t) = −F(−x) x −x
0 0
Then F is odd.
Following similar steps, we can also prove that F is even given that f is odd.
Problem 3 (P117 Q10).
(a) limx→0+ ϕ(x) = 0
(b) limx→0+ ϕ(x) = 0 and limx→0+ ϕ′(x) exists.
(c) limx→0+ ϕ(x) exists.
(d) limx→0+ ϕ(x) exists and limx→0+ ϕ′(x) = 0.
Problem 4 (P123 Q10).
(a) Proof. Prove by induction. Easy to prove that (Z1,Z2) = 0. Suppose Z1,…,Zk are orthogonal to each other. Then
X
k
Yk+1 = Xk+1 −(Zi,XK+1)Zi
i=1 k
(Zj,Yk+1) = (Zj,Yk+1) − X(Zi,Xk+1)δij = (Zj,Yk+1) − (Zj,yk+1) = 0
k=1
Then Yk+1 is orthogonal to Z1,…,Zk. Suppose ∥Yk+1∥ ̸= 0, then Zk+1 exists and Z1,…,Zk+1 orthogonal to each other.
HW05 Haoran Sun

(b) Let f1(x) = cosx + cos2x and f2(x) = 3cosx − 4cos2x, then
π xdx = π ⇒ z
1 π
y Z
⇒ z
Problem 5 (P134 Q1).
Remark. ∀x ∈ (−1,1)
lim Sn = lim
n→∞ n→∞x2
(a) Note that
|Sn − S| = −1 +(−xx22)n = 1 +x2nx2 2N
Easy to show that for eachn xn∈converges pointwisely.(−1,1), ∀ϵ > 0, we can choose N ∈ N where x < ϵ s.t.
|S − S| < ϵ ∀n > N. Then S
(b) always findSn does not converges uniformly since there alwaysx0 = (1/2)1/2n where ∃ϵ = 1/8 > 0, s.t. ∀n > N ∈ N, we can
x2n 1 2n 1 1
|Sn(x0) − S(x0)| = 1 + x2 > 2x = 4 > ϵ = 8
(c) Sn converges in the L2 sense since ∀ϵ > 0, we can choose N ∈ N where N > 1/ϵ s.t.

Z 1 x2n 2! 1 1 2nϵ
x dx =< < ϵ
−1 1 + x42 + ϵ
Problem 6 (P134 Q3).
(a) ∀x /∈1[1/N </2 − 1|x/n,−112/|2)s.t.∪ (1∀n > N/2,1/2 + 1, we have/n], we havefn(x) = 0fn. Then(x) = 0f. Forall othern → 0 pointwisely.x, ∀ϵ > 0, ∃N ∈ N with
(b) ∃s.t.ϵ = 1|fn,(x∃)N|1=∈|γNn|s.t.> ϵ|= 1γn|. Then> ϵ = 1fn∀n > N→ 0 not uniformly.1 since γn → ∞. Then ∀n > N1, ∃xn = x + 1/2n
(c) Easy to show that
→ 0 as n → ∞
fn → 0 in L2 sense.
HW05 Haoran Sun

(d) Easy to show that
2 2 2
∥fn∥ = n = 2n → ∞
n
fn → 0 not in L2 sense.
Problem 7 (P134 Q7).
(a) We can verify that
1 1
c0 =Z ϕ(x)dx = 0
2 −1 cn = 1 Z 1 e−inπxϕ(x) = 1 Z 0 e−inπx(−1 − x)dx + Z 1 e−inπx(1 − x)dx = 1
2 −1 2 −1 0 inπ
(b) First non-zero term (traditional sine series) is
sinπx, sin2πx, sin3πx
(c) Note that
1
∥ϕ∥2 =ϕ(x)2 dx =
−1
N 2 1 4 N 1 2
c
n=X−N | n| Z−1 1dx = π2 nX=1 n2 → 3
Then ∥Sn∥ not converge to ∥ϕ∥, then it converges in the L2 sense.
(d) It converges pointwisely.
(e) It does not converges uniformly.