Introductory Course: Fourier Analysis and its many uses

Solutions - Exercises 16.1-16.5, 16.9, 16.12, 16.13, 16.15, 16.16, 16.17, 16.20, 16.21, 16.23 from “A First Look at Fourier Analysis” by T.W. Körner

(prepared by Mihai Stoiciu)

Exercise 16.1. i) Take P (t) = \frac{f^{(n)}(0)}{n!} t^{n} + \frac{f^{(n−1)}(0)}{(n−1) !} t^{n−1} + ⋯ + f(0).

ii) Note first that g(0) = g′(0) = ⋯ = g^{(n)}(0) = 0. Since g(0) = g(t) = 0 it follows from Rolle’s theorem that there exists a c_{1} ∈ (0, t) such that g′(c_{1}) = 0. Observe now that g′(0) = g′(c_{1}) = 0 so Rolle’s theorem implies that there exists a c_{2} ∈ (0, c_{1}) such that g′′(c_{2}) = 0. Repeating this procedure (n + 1) times we get that there exists c = c_{n+1} ∈ (0, t) such that g^{(n+1)}(c) = 0. This immediately implies f(t) = P (t) + \frac{f^{(n+1)}(c)}{(n+1)!} t^{n+1}.

Exercise 16.2. A polynomial P of degree at most (2n + 1) with prescribed values for P (0), P′(0), ... P^{(n)}(0), P (1), P′(1), ... P^{(n)}(1) can be obtained by taking a linear combination of the polynomials \frac{1}{k!}x^{k}(x−1)^{n+1}, 0 ≤ k ≤ n and \frac{1}{k!}x^{n+1}(x−1)^{k}, 0 ≤ k ≤ n.

We prove now that P is unique. Let Q be a polynomial of degree at most (2n + 1) such that P (0) = Q(0), P′(0) = Q′(0), ...P^{(n)}(0) = Q^{(n)}(0), P (1) = Q(1), P′(1) = Q′(1), ... P^{(n)}(1) = Q^{(n)}(1). R = P − Q is a polynomial of degree at most 2n + 1 and R(0) = R′(0) = ⋯ = R^{(n)}(0) = R(1) = R′(1) = ⋯ = R^{(n)}(1) = 0. Therefore the polynomials X^{n+1} and (X − 1)^{n+1} divide R and since deg R ≤ 2n + 1 we obtain R = 0.

Let P be the unique polynomial of degree at most (2n + 1) such that P (0) = f(0), P′(0) = f′(0), ... P^{(n)}(0) = f^{(n)}(0), P (1) = f(1), P′(1) = f′(1), ... P^{(n)}(1) = f^{(n)}(1). For a fixed y ∈ (0, 1) let

g(x) = f(x) − P (x) − \frac{f(y) − P (y)}{y^{n+1}(y − 1)^{n+1}} x^{n+1}(x − 1)^{n+1}

Let’s observe that g(0) = g′(0) = ⋯ = g^{(n)}(0) = 0 and g(1) = g′(1) = ⋯ = g^{(n)}(1) = 0. Furthermore g(0) = g(y) = g(1) = 0. Therefore, from Rolle’s theorem it follows that there exists two points c_{1}^{1} ∈ (0, y) and c_{2}^{1} ∈ (y, 1) such that g′(c_{1}^{1}) = g′(c_{2}^{1}) = 0. Since g′(0) = g′(1) = 0 we can apply Rolle’s theorem again and we get three points c_{1}^{2} ∈ (0, c_{1}^{1}), c_{2}^{2} ∈ (c_{1}^{1}, c_{2}^{1}) and c_{3}^{2} ∈ (c_{2}^{1}, 1) such that g′′(c_{1}^{2}) = g′′(c_{2}^{2}) = g′′(c_{3}^{2}) = 0. Repeating this procedure (n + 1) times we get (n + 2) points c_{1}^{n+1}, c_{2}^{n+1}, ... c_{n+2}^{n+1} ∈ (0, 1) such that g^{(n+1)}(c_{1}^{n+1}) = g^{(n+1)}(c_{2}^{n+1}) = ⋯ = g^{(n+1)}(c_{n+2}^{n+1}) = 0.

Now we can apply Rolle’s theorem again and we get (n + 1) points

c_{1}^{n+2} ∈ (c_{1}^{n+1}, c_{2}^{n+1}), c_{2}^{n+2} ∈ (c_{2}^{n+1}, c_{3}^{n+1}), ... c_{n+1}^{n+2} ∈ (c_{n+1}^{n+1}, c_{n+2}^{n+1})

such that g^{(n+2)}(c_{1}^{n+2}) = g^{(n+2)}(c_{2}^{n+2}) = ⋯ = g^{(n+2)}(c_{n+1}^{n+2}) = 0. Repeating this procedure (n + 1) times we get that there exists a point c = c_{1}^{2n+2} ∈ (0, 1) such that g^{(2n+2)}(c) = 0.

Since g^{(2n+2)}(x) = f^{(2n+2)}(x) − \frac{f(y)−P (y)}{y^{n+1}(y−1)^{n+1}} (2n + 2) ! we immediately get

E(y) = f(y) − P (y) = \frac{f^{(2n+2)}(c)}{(2n + 2) !} y^{n+1}(y − 1)^{n+1}

Exercise 16.3. Taking imaginary parts in the de Moivre formula it is easy to see that

U_{n}(t) = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} (\begin{matrix}n + 1 \\ 2k + 1 \end{matrix}) (t^{2} − 1)^{k} t^{n−2k}

Taking \frac{∂}{∂θ} in T_{n}(cos θ) = cos nθ we get T′_{n}(cos θ) sin θ = n(sin nθ) which implies that T′_{n}(x) = nU_{n−1}(x).

Exercise 16.4. For any t ∈ ℝ with |t| < 1 and for any θ ∈ ℝ we have |te^{iθ}| < 1 so

∑_{n=0}^{∞} t^{n}e^{inθ} = \frac{1}{1 − te^{iθ}}

Since Re [t^{n}e^{inθ}] = t^{n} cos(nθ) and Re [\frac{1}{1−te^{iθ}} ] = \frac{1−t cos θ}{1−2t cos θ+t^{2}} we get that \frac{1−t cos θ}{1−2t cos θ+t^{2}} = ∑_{n=0}^{∞} T_{n}(cos θ) t^{n} for any |t| < 1 and any θ ∈ ℝ, which implies

\frac{1 − tx}{1 − 2tx + t^{2}} = ∑_{n=0}^{∞} T_{n}(x) t^{n}

for any |t| < 1 and |x| ≤ 1.

Exercise 16.5. Let’s observe first that for any r ∈ (0, 1) we have P_{r}(f, θ) = (f ∗P_{r})(θ) where P_{r} : [−π, π] → ℝ, P_{r}(t) = ∑_{n=−∞}^{∞} r^{|n|}e^{int} = \frac{1−r^{2}}{1+r^{2}−2r cos t}. Furthermore P_{r}(t) > 0 for any t ∈ [−π, π] and:

\frac{1}{2π} ∫_{−π}^{π} P_{r}(t) dt = \frac{1}{2π} ∑_{n=−∞}^{∞} ∫_{−π}^{π} r^{|n|}e^{int}dt = 1

Let ε > 0. For any r ∈ (0, 1) we have:

|P_{r}(f, θ) − f(θ)| = |(f ∗ P_{r})(θ) − f(θ)\frac{1}{2π} ∫_{−π}^{π} P_{r}(t) dt| = |\frac{1}{2π} ∫_{−π}^{π} [f(θ − t) − f(θ)] P_{r}(t) dt|

≤ \frac{1}{2π} ∫_{|t|<δ} |f(θ − t) − f(θ)| P_{r}(t) dt + \frac{1}{2π} ∫_{|t|≥δ} |f(θ − t) − f(θ)| P_{r}(t) dt

Since f ∈ C(𝕋) it follows that f is bounded and uniformly continuous. Let M > 0 such that |f(s)| < M for any s ∈ [−π, π] and let δ_{ε} > 0 such that |t − t′| < δ_{ε} implies |f(t) − f(t′)| < \frac{ε}{2}.

For any t with |t| > δ_{ε} we have \frac{1−r^{2}}{1+r^{2}−2r cos t} < \frac{1−r^{2}}{1+r^{2}−2r cos δ_{ε}} < \frac{1−r^{2}}{1−(cos δ_{ε})^{2}} and therefore

∫_{|t|≥δ_{ε}} P_{r}(t) dt < \frac{1 − r^{2}}{1 − (cos δ_{ε})^{2}}

Let r_{ε} > 0 such that r ∈ (r_{ε}, 1) implies \frac{1−r^{2}}{1−(cos δ_{ε})^{2}} < \frac{2πε}{4M}. For any r ∈ (r_{ε}, 1) we have:

\frac{1}{2π} ∫_{|t|<δ_{ε}} |f(θ − t) − f(θ)| P_{r}(t) dt < \frac{ε}{2}

and

\frac{1}{2π} ∫_{|t|≥δ_{ε}} |f(θ − t) − f(θ)| P_{r}(t) dt < \frac{1}{2π} 2M \frac{2πε}{4M} = \frac{ε}{2}

We can conclude now that |P_{r}(f, θ)−f(θ)| < ε for any r ∈ (r_{ε}, 1) and any θ ∈ [−π, π]. Therefore P_{r}(f, ·) → f uniformly as r → 1, r < 1.

Exercise 16.9. i) Let ε > 0. Since f has real values it follows from Theorem 8.1 ii) that there exists a real trigonometric polynomial

Q(t) = ∑_{n=0}^{N} (a_{n} sin nt + b_{n} cos nt)

such that ∥f − Q∥_{∞} < ε. Let Q_{1}(t) = Q(−t). Since f(t) = f(−t) it follows that ∥f − Q_{1}∥_{∞} < ε. For R(t) = \frac{1}{2}(Q(t) + Q_{1}(t)) we have ∥f − R∥_{∞} < ε. Furthermore a straightforward computation shows that:

R(t) = ∑_{n=0}^{N} b_{n} cos nt

ii) Let g : [−π, π] → ℝ defined by g(t) = F (| cos t|). Let ε > 0. Since g ∈ C(𝕋) and is real valued it follows from i) that there exists a real trigonometric polynomial R(t) = ∑_{n=0}^{N} c_{n} cos nt such that ∥f − R∥_{∞} < ε. Let’s observe now that

R(t) = ∑_{n=0}^{N} c_{n} cos nt = ∑_{n=0}^{N} c_{n} T_{n}(cos t) = Q(cos t)

where T_{n} is the n-th Chebyshev polynomial of the first kind and Q ∈ ℝ[X].

Therefore |g(t) − R(t)| = |F (| cos t|) − Q(cos t)| < ε for any t ∈ [−π, π] which immediately implies ∥F − Q∥_{∞} < ε.

iii) Since ∫_{0}^{1} g(t) t^{n} dt = 0 for any n it follows that ∫_{0}^{1} g(t) P (t) dt = 0 for any polynomial P ∈ ℝ[X]. Let P_{n} be a sequence of polynomials such that P_{n} → \overline{g} uniformly. Since

lim_{n→∞} ∫_{0}^{1} g(t)P_{n}(t)dt = ∫_{0}^{1} g(t) \overline{g(x)} dt

it follows that ∫_{0}^{1} g(t) \overline{g(x)} dt = ∫_{0}^{1} |g(t)|^{2} dt = 0 so g = 0.

Exercise 16.12. i) Since f is continuous it follows immediately that ∫_{−π}^{π} |f(t)|^{2}dt < ∞ and Theorem 8.2. implies that ∑_{n=−∞}^{∞} |f̂(n)|^{2} < ∞. We get lim_{n→∞} f̂(n) = 0 and lim_{n→−∞} f̂(n) = 0.

ii) We can write sin t = \frac{1}{2i} (e^{it} − e^{−it}) and therefore f_{1}(t) = \frac{1}{2i} (g_{1}(t)e^{it} − g_{1}(t)e^{−it}). This implies f̂_{1}(j) = \frac{1}{2i} (ĝ_{1}(j − 1) − ĝ_{1}(j + 1)) so for any n ≥ 1

S_{n}(f_{1}, 0) = ∑_{j=−n}^{n} f̂_{1}(j) = f̂_{1}(n) + f̂_{1}(n + 1) − f̂_{1}(−n) − f̂_{1}(−n − 1)

Using part i) we get lim_{n→∞} S_{n}(f_{1}, 0) = 0.

iii) For any t ≠ nπ we have g_{2}(t) = \frac{1}{sin t}f_{2}(t). We must prove that \frac{1}{sin t}f_{2}(t), t ≠ nπ can be extended to a continuous function on 𝕋. For any t ∈ (−π, π), t ≠ 0

\frac{1}{sin t} f_{2}(t) = \frac{t}{sin t} \frac{f_{2}(t)}{t} = \frac{t}{sin t} \frac{f_{2}(t) − f_{2}(0)}{t − 0}

Therefore lim_{t→0} \frac{1}{sin t}f_{2}(t) = f′_{2}(0), so the function \frac{1}{sin t}f_{2}(t) can be extended by continuity at t = 0. Similarly

lim_{t→π} \frac{1}{sin t} f_{2}(t) = lim_{t→π} \frac{1}{sin(π − t)} f_{2}(t) = lim_{t→π} \frac{t − π}{sin(π − t)} \frac{f_{2}(t) − f_{2}(π)}{t − π} = −f′_{2}(π)

lim_{t→−π} \frac{1}{sin t} f_{2}(t) = lim_{t→−π} \frac{−1}{sin(π + t)} f_{2}(t) = lim_{t→−π} \frac{−(t + π)}{sin(π + t)} \frac{f_{2}(t) − f_{2}(−π)}{t + π} = −f′_{2}(π)

so the function \frac{1}{sin t}f_{2}(t) can be extended to a continuous function g_{2} : 𝕋 → ℂ. Using ii) we get lim_{n→∞} S_{n}(f_{2}, 0) = 0.

iv) It is easy to see that f_{4} is continuous and differentiable at 0. Furthermore for any j = 2n, n ∈ ℤ we have:

f̂_{4}(j) = f̂_{4}(2n) = \frac{1}{2π} ∫_{−π}^{π} f_{3}(2t) e^{−2int}dt = \frac{1}{4π} ∫_{−2π}^{2π} f_{3}(s) e^{−ins}ds = f̂_{3}(n)

¿From part iii) we get that there exists a function g ∈ C(𝕋) such that f_{4}(t) = g(t) sin t and from ii) we get that f̂_{4}(2n) = \frac{1}{2i}(ĝ(2n − 1) − ĝ(2n + 1)). We can now conclude that

S_{N}(f_{3}, 0) = ∑_{n=−N}^{N} f̂_{3}(n) = ∑_{n=−N}^{N} f̂_{4}(2n) = \frac{1}{2i} (ĝ(−2n − 1) − ĝ(2n + 1))

Therefore lim_{N→∞} S_{N}(f_{3}, 0) = 0.

v) Suppose f is continuous and differentiable at x_{0} ∈ [−π, π]. Let g : [−π, π] → ℂ, g(x) = f(x + x_{0}) − f(x_{0}). Clearly g(0) = 0 and g is differentiable at 0. Using part iv) we get lim_{N→∞} S_{N}(g, 0) = 0. But

S_{N}(g, 0) = ∑_{n=−N}^{n=N} ĝ(n) = ĝ(0) + ∑_{n=−N n≠0}^{n=N} ĝ(n) = −f(x_{0}) + ∑_{n=−N}^{N} \frac{1}{2π} ∫_{−π}^{π} f(x + x_{0})e^{−inx}dx

= −f(x_{0}) + ∑_{n=−N}^{N} e^{inx_{0}} \frac{1}{2π} ∫_{−π}^{π} f(x + x_{0})e^{−in(x+x_{0})}dx = −f(x_{0}) + ∑_{n=−N}^{N} e^{inx_{0}} f̂(n)

= −f(x_{0}) + S_{N}(f, x_{0})

which implies lim_{N→∞} S_{N}(f, x_{0}) = f(x_{0}).

Exercise 16.13. - Reformulated

i) Prove that for any ε > 0 and any K > 0 we can find a continuous function f ∈ C(𝕋) and an integer M > 0 such that ∥f∥_{∞} < ε and |S_{M}(f, 0)| > K.

ii) Prove that the function f from part i) can be chosen to be a trigonometric polynomial.

iii) Prove that for any ε > 0, any K > 0 and any positive integer m > 0 we can find a trigonometric polynomial P and a positive integer M such that ∥P ∥_{∞} < ε, P̂(r) = 0 for any integer |r| ≤ m and |S_{M}(P, 0)| > K.

iv) For any nonzero trigonometric polynomial P (t) = ∑_{n=−N}^{N} a_{n}e^{int} we denote by deg P = max {n, n nonnegative integer s.t. a_{n} ≠ 0 or a_{−n} ≠ 0}. Prove that we can find a sequence of trigonometric polynomials {P_{n}}_{n≥1} such that the sequence m(n) = deg(P_{n−1}) is increasing and:

a) ∥P_{n}∥_{∞} < 2^{−n}

b) P̂_{n}(r) = 0 if |r| ≤ m(n) or |r| > m(n + 1)

c) |S_{deg P_{n}}(P_{n}, 0)| ≥ 2^{n} + ∑_{k=1}^{n−1} |S_{deg P_{k}}(P_{k}, 0)|

v) Prove that ∑_{n=1}^{∞} P_{n} is uniformly convergent to some continuous function f and that for any n and for any integer r such that m(n) + 1 ≤ r ≤ m(n + 1) we have f̂(r) = P̂_{n}(r).

vi) Deduce that |S_{m(n+1)}(f, 0)| ≥ 2^{n} for any n and therefore {S_{N}(f, 0)} diverges.

Solution i) Let M large enough so that εB log M > K. From Lemma 6.5. there exists a function g ∈ C(𝕋) such that ∥g∥_{∞} < 1 and |S_{M}(g, 0)| ≥ B log M > \frac{K}{ε}. Let f = εg. We have ∥f∥_{∞} < ε and |S_{M}(f, 0)| = ε|S_{M}(g, 0)| > K.

ii) Using part i) we can find a function f ∈ C(𝕋) and a positive integer M such that ∥f∥_{∞} < \frac{ε}{2} and |S_{M}(f, 0)| > (K + 1). Since the trigonometric polynomials are dense in C(𝕋) we can find a trigonometric polynomial P such that ∥P − f∥_{∞} < min\{\frac{1}{2M+1}, \frac{ε}{2}\}. Clearly ∥P ∥_{∞} < ε. Furthermore:

|S_{M}(f, 0) − S_{M}(P, 0)| = |S_{M}(f − P, 0)| ≤ ∑_{n=−M}^{M} \frac{1}{2π} |∫_{−π}^{π}(f − P)(t)e^{−int}dt| < 1

which shows that |S_{M}(P, 0)| > K.

iii) For any nonzero trigonometric polynomial P (t) = ∑_{n=−N}^{N} a_{n}e^{int} we denote by deg P = max {n nonnegative integer s.t. a_{n} ≠ 0 or a_{−n} ≠ 0}. Let Q be a trigonometric polynomial such that ∥Q∥_{∞} < ε and |S_{deg Q}(Q, 0)| > K. Let P (t) = e^{(deg Q+m+1)it}Q(t). Clearly ∥P ∥_{∞} < ε and P̂(r) = 0 for any integer r ≤ m. Let M = 2 deg P + m + 1. Then S_{M}(P, 0) = S_{deg Q}(Q, 0) so |S_{M}(P, 0)| > K.

iv) We will repeatedly use iii) to get the sequence of trigonometric polynomials {P_{n}}. Let P_{1} be a trigonometric polynomial such that ∥P_{1}∥_{∞} ≤ 2^{−1} and |S_{deg P_{1}}(P_{1}, 0)| ≥ 2. Let m(2) = deg P_{2}. Let now P_{2} be a trigonometric polynomial such that ∥P_{2}∥_{∞} ≤ 2^{−2}, |S_{deg P_{2}}(P_{2}, 0)| ≥ 2^{2} + |S_{deg P_{1}}(P_{1}, 0)| and P̂_{2}(r) = 0 for all integers r, −m(2) ≤ r ≤ m(2). Let m(3) = deg P_{2}. Repeating this procedure we get a sequence of trigonometric polynomials {P_{n}} and an increasing sequence of integers {m(n)}, m(n) = deg P_{n−1} for any n ≥ 2, such that ∥P_{n}∥_{∞} ≤ 2^{−n}, |S_{deg P_{n}}(P_{n}, 0)| ≥ 2^{n} + ∑_{k=1}^{n−1} |S_{deg P_{k}}(P_{k}, 0)| and P̂_{n}(r) = 0 for any r, |r| ≤ m(n) or |r| > m(n + 1).

v) Since ∥P_{n}∥_{∞} < 2^{−n} and the series ∑_{n=1}^{∞} 2^{−n} is convergent it follows that ∑_{n=1}^{∞} P_{n} converges uniformly to a continuous function f. For any integer r, m(n) + 1 ≤ |r| ≤ m(n + 1)

f̂(r) = \frac{1}{2π} ∫_{−π}^{π} f(t)e^{irt}dt = \frac{1}{2π} ∫_{−π}^{π} (∑_{n=1}^{∞} P_{n}(t)) e^{irt}dt

= \frac{1}{2π} ∑_{n=1}^{∞} ∫_{−π}^{π} P_{n}(t)e^{irt}dt = ∑_{k=1}^{∞} P̂_{k}(r) = P̂_{n}(r)

Therefore for any integer n we have that

S_{m(n+1)}(f, 0) = ∑_{k=−m(n+1)}^{m(n+1)} f̂(k) = ∑_{k=−m(2)}^{m(2)} f̂(k) + ∑_{l=2}^{n} [  ∑_{m(l)+1≤|k|≤m(l+1)} f̂(k)]

= ∑_{k=−m(2)}^{m(2)} P̂_{1}(k) + ∑_{l=2}^{n} [ ∑_{m(l)+1≤|k|≤m(l+1)} P̂_{l}(k)] = S_{m(2)}(P_{1}, 0) + ∑_{l=2}^{n} S_{m(l+1)}(P_{l}, 0)

= S_{deg P_{n}}(P_{n}, 0) + ∑_{l=1}^{n−1} S_{deg P_{l}}(P_{l}, 0)

Since |S_{deg P_{n}}(P_{n}, 0)| ≥ 2^{n} + ∑_{k=1}^{n−1} |S_{deg P_{k}}(P_{k}, 0)| we get |S_{m(n+1)}(f, 0)| ≥ 2^{n} for any positive integer n which shows that \overline{lim}_{N→∞}|S_{N}(f, 0)| = ∞, so S_{N}(f, 0) ↛ f(0).

The function f constructed before has complex values. Since for any positive integer N we have S_{N}(f, 0) = S_{N}(Re(f), 0) + iS_{N}(Im(f), 0) it follows that at least one of \overline{lim}_{N→∞}|S_{N}(Re(f), 0)| and \overline{lim}_{N→∞}|S_{N}(Im(f), 0)| is ∞ so at least one of Re(f) and Im(f) is a real valued continuous function for which the Fourier series diverges at 0.

Exercise 16.15. Let’s observe that ⟨log_{10} n⟩ ∈ [0, 1/2] if and only if there exists a p ∈ ℕ such that 10^{p} ≤ n ≤ 10^{p+1/2}. Let

a_{N} = \frac{card\{1 ≤ n ≤ N, ⟨log_{10} n⟩ ∈ [0, 1/2]\}}{N}

Let S(k) = 10^{k} − 1 and T (k) = 3 · 10^{k}. Since 10^{1/2} < 4 we clearly have

lim sup_{k→∞} a_{S(k)} < 4/10

Also

lim inf_{k→∞} a_{T (k)} > lim_{k→∞} \frac{1}{3 · 10^{k}} (3 · 10^{k} − 10^{k}) = 2/3

which shows that {a_{N}} does not have a limit as N → ∞. In particular ⟨log_{10} n⟩ are not equidistributed in [0, 1].

For ε > 0, x ∈ [0, 1] and n ∈ ℕ n > 0 , |⟨log_{10} n⟩ − x| < ε if and only if there exists a p ∈ ℕ such that n ∈ [10^{p+x−ε}, 10^{p+x+ε}]. Since lim_{p→∞}(10^{p+x+ε} − 10^{p+x−ε}) = ∞ we can pick a p ∈ ℕ large enough such that the set M = [10^{p+x−ε}, 10^{p+x+ε}] ⋂ ℕ is not empty. Any n ∈ M will satisfy |⟨log_{10} n⟩ − x| < ε.

Exercise 16.16. i) This is a different proof for the ‘Riemann-Lebesgue’ lemma. Obviously if f is a trigonometric polynomial then f̂(n) = 0 if |n| is large enough.

Let ε > 0 and f ∈ C(𝕋). Let P (t) = ∑_{k=−N}^{N} a_{k}e^{ikt} be a trigonometric polynomial such that ∥f − P ∥_{∞} < ε. Then for any n ∈ ℤ such that |n| > N, we have

f̂(n) = \frac{1}{2π} ∫_{−π}^{π}(f(s) − P (s)) e^{ins}ds = \frac{1}{2π} ∫_{−π}^{π}(f(s) − P (s)) e^{ins}ds + \frac{1}{2π} ∫_{−π}^{π} P (s) e^{ins}ds

so

|f̂(n)| ≤ \frac{1}{2π} ∫_{−π}^{π} |(f(s) − P (s)) e^{ins}| ds < ε

which shows that lim_{n→∞} f̂(n) = 0 and lim_{n→−∞} f̂(n) = 0.

Let f(t) = 1. Then

∫_{0}^{2π} | sin nt| dt = \frac{1}{n} ∫_{0}^{2nπ} | sin u| du = \frac{1}{n} ∑_{k=0}^{n−1} ∫_{2kπ}^{2kπ+2π} | sin u| du = \frac{1}{n} 4n = 4

= \frac{4}{2π} ∫_{0}^{2π} 1 dt

Let now f(t) = e^{imt}, m ∈ ℤ, m ≠ 0 and n > m. Then, integrating by parts twice, we get:

I_{m,n} = ∫_{0}^{2π} e^{imt}| sin nt| dt = ∑_{k=0}^{2n−1}(−1)^{k} ∫ \frac{(k+1)π}{n} ; \frac{kπ}{n} e^{imx} sin nx dx

= ∑_{k=0}^{2n−1}(−1)^{k} ∫ \frac{(k+1)π}{n} ; \frac{kπ}{n} (\frac{e^{imx}}{im} )′ sin nx dx = \frac{n}{im} ∑_{k=0}^{2n−1}(−1)^{k+1} ∫ \frac{(k+1)π}{n} ; \frac{kπ}{n} e^{imx} cos nx dx

= \frac{n}{(im)^{2}} ∑_{k=0}^{2n−1}(−1)^{k+1} [(e^{imx} cos nx) |_{x= \frac{kπ}{n}}^{x= \frac{(k+1)π}{n}} + n ∫_{0}^{2π} e^{imx} sin nx dx]

= \frac{n}{(im)^{2}} ∑_{k=0}^{2n−1} (e^{im \frac{(k+1)π}{n}} + e^{im \frac{kπ}{n}} ) + \frac{n^{2}}{(im)^{2}} ∑_{k=0}^{2n−1}(−1)^{k+1} ∫ \frac{(k+1)π}{n} ; \frac{kπ}{n} e^{imx} sin nx dx

= \frac{n}{(im)^{2}} (e^{im \frac{π}{n}} + 1) ∑_{k=0}^{2n−1} (e^{\frac{imπ}{n}} )^{k} + \frac{n^{2}}{m^{2}} I_{m,n} = \frac{n^{2}}{m^{2}} I_{m,n}

Therefore (1 − \frac{n^{2}}{m^{2}} )I_{m,n} = 0 so I_{m,n} = 0. Since ∫_{0}^{2π} e^{imx}dx = 0 for any m ≠ 0 we can conclude now that

lim_{n→∞} ∫_{0}^{2π} P (t)| sin nt| dt = \frac{4}{2π} ∫_{0}^{2π} P (t) dt

for any trigonometric polynomial P (t) = ∑_{n=−N}^{N} a_{n}e^{int}.

We can now finish the proof. Let f : 𝕋 → ℝ continuous and let ε > 0. Let P (t) = ∑_{n=−N}^{N} a_{n}e^{int} be a trigonometric polynomial such that ∥f − P ∥_{∞} < \frac{ε}{6π}. Let n_{0} large enough such that for any n ≥ n_{0} we have

|∫_{0}^{2π} P (t)| sin nt| dt − \frac{4}{2π} ∫_{0}^{2π} P (t) dt| < \frac{ε}{3}

Then for any n ≥ n_{0} we have

|∫_{0}^{2π} f(t) | sin nt| dt − \frac{4}{2π} ∫_{0}^{2π} f(t) dt| < ∫_{0}^{2π} |f(t) − P (t)| | sin nt| dt

+ |∫_{0}^{2π} P (t) | sin nt| dt − \frac{4}{2π} ∫_{0}^{2π} P (t) dt| + \frac{4}{2π} ∫_{0}^{2π} |f(t) − P (t)| dt < ε

and therefore

lim_{n→∞} ∫_{0}^{2π} f(t) | sin nt| dt = \frac{4}{2π} ∫_{0}^{2π} f(t) dt

Exercise 16.17. A simple computation shows that ∫_{α}^{β} e^{2πit}dt = 0 if and only if (α − β) ∈ ℤ. Without loss of generality we can assume that R = [0, a] × [0, b]. For any integer j, 1 ≤ j ≤ k we have R(j) = [x_{j}, x_{j} + a_{j}] × [y_{j}, y_{j} + b_{j}]. Furthermore:

∫ ∫_{R_{j}} e^{2πi(x+y)} dx dy = (∫_{x_{j}}^{x_{j}+a_{j}} e^{2πix} dx) (∫_{y_{j}}^{y_{j}+b_{j}} e^{2πiy} dy) = 0

since at least one of a_{j} and b_{j} is an integer. Therefore

(∫_{0}^{a} e^{2πix} dx) (∫_{0}^{b} e^{2πiy} dy) = ∫ ∫_{R} e^{2πi(x+y)} dx dy = ∑_{j=1}^{k} ∫ ∫_{R_{j}} e^{2πi(x+y)} dx dy = 0

which shows that at least one of a and b is an integer.

Exercise 16.20. i) This is the Cauchy Schwarz inequality for a = (a_{−N}, a_{−N+1}, ...a_{N−1}, a_{N}) and b = (b_{−N}, b_{−N+1}, ...b_{N−1}, b_{N}) in ℂ^{2N+1}.

ii) For any N ≥ 0

∑_{j=−N}^{N} |a_{j}b_{j}| ≤ ( ∑_{j=−N}^{N} |a_{j}|^{2})^{1/2} ( ∑_{j=−N}^{N} |b_{j}|^{2})^{1/2} ≤ ( ∑_{j=−∞}^{∞} |a_{j}|^{2})^{1/2} ( ∑_{j=−∞}^{∞} |b_{j}|^{2})^{1/2}

which implies that {∑_{j=−N}^{N} |a_{j}b_{j}|}_{N∈ℕ} converges and

∑_{j=−∞}^{∞} |a_{j}b_{j}| ≤ ( ∑_{j=−∞}^{∞} |a_{j}|^{2})^{1/2} ( ∑_{j=−∞}^{∞} |b_{j}|^{2})^{1/2}

iii) For any f ∈ C(𝕋) and any j ∈ ℤ we get, using integration by parts:

f̂′(j) = \frac{1}{2π} ∫_{−π}^{π} f′(t) e^{−ijt} dt = ij \frac{1}{2π} ∫_{−π}^{π} f(t) e^{−ijt} dt = ij f̂(j)

Since by Plancherel theorem (Theorem 8.2. (i)) we have:

∑_{j=−∞}^{∞} |f̂′(j)|^{2} = \frac{1}{2π} ∫_{−π}^{π} f′(t)dt

we immediately get

∑_{j=−∞}^{∞} j^{2}|f̂(j)|^{2} = \frac{1}{2π} ∫_{𝕋} |f′(t)|^{2}dt

iv) Using ii) we get

∑_{j=−∞}^{∞} |f̂(j)| = |f̂(0)| + ∑_{|j|≥1} |f̂(j)| j \frac{1}{j} ≤ |f̂(0)| + (∑_{|j|≥1} |f̂(j)|^{2} j^{2})^{1/2} (∑_{|j|≥1} \frac{1}{j^{2}} )^{1/2}

≤ |f̂(0)| + (\frac{1}{2π} ∫_{−π}^{π} |f′(t)|^{2} dt)^{1/2} (\frac{π^{2}}{3} )^{1/2}

which shows that ∑_{j=−∞}^{∞} |f̂(j)| converges. Therefore ∑_{j=−∞}^{∞} f̂(j) e^{ijt} converges uniformly to a continuous function g. It remains to show that g = f. But ĝ(j) = f̂(j) for any j ∈ ℤ. Therefore, from Theorem 7.4 we get g = f.

Exercise 16.21. (Wirtinger’s inequality) i) Since by hypothesis \frac{1}{2π} ∫_{𝕋} u(t) dt = 0, we have û(0) = 0. From Plancherel’s theorem and part iii) of the previous problem we have:

\frac{1}{2π} ∫_{−π}^{π}(u(t))^{2} dt = ∑_{j=−∞}^{∞} |û(j)|^{2} = ∑_{|j|≥1} |û(j)|^{2} ≤ ∑_{|j|≥1} j^{2}|û(j)|^{2} = \frac{1}{2π} ∫_{−π}^{π}(u′(t))^{2}dt

Note that in the previous relation we have equality if and only if û(j) = 0 for any |j| ≥ 2. Therefore u(t) = û(1)e^{it} + û(−1)e^{−it}. Since u is real valued we must have u(t) = a cos(t) + b sin(t), a, b ∈ ℝ. If C^{2} = a^{2} + b^{2} ≠ 0 we have

u(t) = \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} cos t + \frac{b}{\sqrt{a^{2} + b^{2}}} sin t) = C cos(t + φ)

where φ ∈ [0, 2π) is chosen such that sin φ = −\frac{b}{\sqrt{a^{2}+b^{2}}} and cos φ = \frac{a}{\sqrt{a^{2}+b^{2}}} .

ii) Let v ∈ C^{1}([0, \frac{π}{2}]) be a real valued function with v(0) = 0 and v′(\frac{π}{2}) = 0. Let f : [−π, π] → ℝ defined by:

f(t) = \begin{cases}v(t) & if t ∈ [0, \frac{π}{2}) \\ v(π − t) & if t ∈ [\frac{π}{2}, π) \\ −v(−t) & if t ∈ [−\frac{π}{2}, 0) \\ −v(π + t) & if t ∈ [−π, −\frac{π}{2}) \end{cases}

It is easy to see that f ∈ C(𝕋). For any x_{0} ∈ 𝕋 let’s denote by f′_{−}(x_{0}) the lefthand derivative of f at x_{0} and by f′_{+}(x_{0}) the right-hand derivative of f at x_{0}. Simple computations show that f′_{+}(0) = f′_{−}(0) = v′(0), f′_{+}(\frac{π}{2}) = f′_{−}(\frac{π}{2}) = 0, f′_{−}(π) = f′_{+}(−π) = −v′(0) and f′_{+}(−\frac{π}{2}) = f′_{−}(−\frac{π}{2}) = 0. Therefore f ∈ C^{1}(𝕋). From part i) we get

∫_{−π}^{π}(f(t))^{2} dt ≤ ∫_{−π}^{π}(f′(t))^{2} dt

Since ∫_{−π}^{π}(f(t))^{2}dt = 4 ∫0^{\frac{π}{2}} (v(t))^{2}dt and ∫_{−π}^{π}(f′(t))^{2}dt = 4 ∫0^{\frac{π}{2}} (v′(t))^{2}dt we get

∫0^{\frac{π}{2}} (v(t))^{2} dt ≤ ∫0^{\frac{π}{2}} (v′(t))^{2} dt

for any v ∈ C^{1}(𝕋).

The previous inequality becomes equality if and only if there exist C ∈ ℝ and φ ∈ [0, 2π) such that f(t) = C cos(t + φ). When C ≠ 0, since f(0) = f′(\frac{π}{2}) = 0, it follows that cos φ = 0 and sin(\frac{π}{2} + φ) = 0. Therefore φ = \frac{π}{2} or φ = \frac{3π}{2}. In both cases we can conclude that f(t) = C_{1} sin t , so v(t) = C_{1} sin t for a constant C_{1} ∈ ℝ.

iii) Let w ∈ C^{1}([0, \frac{π}{2}]) with w(0) = 0. Let M > 0 such that |w(t)| < M and |w′(t)| < M for any t ∈ [0, \frac{π}{2}].

Let ε > 0, ε < 7M^{2} and let t_{0} = \frac{π}{2} − \frac{ε}{7M^{2}} . Define v : [0, \frac{π}{2}] → ℝ by

v(t) = \begin{cases}w(t) & if t ∈ [0, t_{0}] \\ w(t_{0}) + \frac{1}{2} w′(t_{0}) (\frac{π}{2} − t_{0}) − \frac{1}{2} w′(t_{0}) (\frac{π}{2} − t)^{2} (\frac{π}{2} − t_{0})^{−1} & if t ∈ (t_{0}, \frac{π}{2}] \end{cases}

It is easy to see that v ∈ C^{1}([0, \frac{π}{2}]), v(0) = w(0) = 0 and v′(\frac{π}{2}) = 0. Therefore, from part ii) we get:

∫0^{\frac{π}{2}} (v(t))^{2} dt ≤ ∫0^{\frac{π}{2}} (v′(t))^{2} dt (1)

It is also easy to see that ∥v∥_{∞} < 2M and ∥v′∥_{∞} < M. Therefore

|∫0^{\frac{π}{2}} (v(t))^{2} dt − ∫0^{\frac{π}{2}} (w(t))^{2} dt| = |∫t_{0}^{\frac{π}{2}} (v(t))^{2} dt − ∫t_{0}^{\frac{π}{2}} (w(t))^{2} dt|

≤ ∫t_{0}^{\frac{π}{2}} (v(t))^{2} dt + ∫t_{0}^{\frac{π}{2}} (w(t))^{2} dt ≤ 5M^{2} \frac{ε}{7M^{2}} = \frac{5 ε}{7} (2)

and

|∫0^{\frac{π}{2}} (v′(t))^{2} dt − ∫0^{\frac{π}{2}} (w′(t))^{2} dt| = |∫t_{0}^{\frac{π}{2}} (v′(t))^{2} dt − ∫t_{0}^{\frac{π}{2}} (w′(t))^{2} dt|

≤ ∫t_{0}^{\frac{π}{2}} (v′(t))^{2} dt + ∫t_{0}^{\frac{π}{2}} (w′(t))^{2} dt < 2M^{2} \frac{ε}{7M^{2}} = \frac{2 ε}{7} (3)

Using (1), (2) and (3) we obtain

∫0^{\frac{π}{2}} (w(t))^{2} dt ≤ ∫0^{\frac{π}{2}} (w′(t))^{2} dt + ε

Since ε is arbitrarily small we can conclude that

∫0^{\frac{π}{2}} (w(t))^{2} dt ≤ ∫0^{\frac{π}{2}} (w′(t))^{2} dt

Exercise 16.23. - (The Gibbs Phenomenon) i) Let λ = \frac{1}{2π}(f_{+}(0) − f_{−}(0)) and g : [−π, π] → ℝ defined by

g(t) = \begin{cases}f(t) − λF (t) & if t ≠ 0 \\ \frac{1}{2} (f_{+}(0) + f_{−}(0)) & if t = 0 \end{cases}

It is easy to see that g_{−}(0) = g_{+}(0) = \frac{1}{2}(f_{+}(0) + f_{−}(0)) and therefore g ∈ C(𝕋). Furthermore, g is continuously differentiable on 𝕋\{0}. We clearly have f = g + λF.

ii) For r ∈ ℤ, r ≠ 0 we have

F̂(r) = \frac{1}{2π} ∫_{−π}^{π} F (t) e^{−irt} dt = \frac{1}{2π} ∫_{−π}^{0}(−π − t) e^{−irt} dt + \frac{1}{2π} ∫_{0}^{π}(π − t) e^{−irt} dt

= \frac{1}{2ir} ∫_{−π}^{0}(e^{−irt})′ dt − \frac{1}{2ir} ∫_{0}^{π}(e^{−irt})′ dt + \frac{1}{2πir} ∫_{−π}^{π} t(e^{−irt})′ dt

= \frac{1}{2ir} (−2(cos rπ) + 2) + \frac{1}{2πir}2π cos rπ = \frac{1}{ir}

Since obviously F̂(0) = 0 we have

S_{n}(F, t) = ∑_{r=−n}^{n} F̂(r) e^{irt} = ∑_{r=1}^{n} (−\frac{1}{ir} e^{−irt} + \frac{1}{ir} e^{irt}) = 2 ∑_{r=1}^{n} \frac{sin rt}{r}

iii) For any τ > 0 we have that P_{n} = \{r\frac{τ}{n}, 1 ≤ r ≤ n\} is a partition of [0, τ] in n intervals of length \frac{τ}{n}. Therefore

S_{n}(F, \frac{τ}{n}) = 2 \frac{τ}{n} ∑_{r=1}^{n} \frac{1}{\frac{rτ}{n}} sin \frac{rτ}{n}

is the Riemann sum associated to the function φ : [0, τ] → ℝ, φ(x) = 2 \frac{sin x}{x}, the partition P_{n} of [0, τ] and the set of intermediate points obtained by taking the right-hand endpoint from each interval [\frac{r−1}{n}, \frac{r}{n}]. Since the function φ is Riemann integrable on [0, τ] we get

lim_{n→∞} S_{n}(F, \frac{τ}{n}) = 2 ∫_{0}^{τ} \frac{sin x}{x}dx

iv) The graph of the function G(τ) = ∫_{0}^{τ} \frac{sin x}{x} dx is This suggest that lim_{τ→∞} ∫_{0}^{τ} \frac{sin x}{x}dx exists. (Actually lim_{τ→∞} ∫_{0}^{τ} \frac{sin x}{x}dx = \frac{π}{2})

v) When τ is small and n is large we have that t = τ/n is small. The graph of S_{50}(F, t) is

We can observe the two “bumps” near the origin which give a point of maximum to the right of 0 and a point of minimum to the left of 0.

Let’s also observe that the graph of the function h : (0, ∞) → ℝ defined by

h(t) = S_{n}(F, t) − F (t)

is (for n = 50) We can see that this graph is similar to the graph of the function G from part iv).