Fourier Analysis Second Semester 2008/9 Solutions to Selected Homework Questions

These solutions are only sketches. Any gaps must be completed to produce the full solution. #2, 2. For δ ∈ (0, π), we have defined f on [−π, π] to be

f(θ) = \{ \begin{matrix}0, & if |θ| > δ, \\ 1 − |θ|/δ, & if |θ| ≤ δ. \end{matrix}

(a) Graphs of functions should be of the correct shape and have the correct position, and the axes should be labelled.

(b) In order to prove

f(θ) = \frac{δ}{2π} + 2 ∑_{n=1}^{∞} \frac{1 − cos(nδ)}{n^{2}πδ} cos(nθ).

we will first compute the Fourier series of f and observe it equals the right-hand side of the above, and then we will prove that f is equal to its Fourier series. Now,

f̂(n) = \frac{1}{2π} ∫_{δ}^{δ}(1 − |θ|/δ)e^{inθ} dθ,

so f̂(0) = δ/2π and for n ≠ 0,

\begin{aligned}f̂(n) & = \frac{1}{πδ} ∫_{0}^{δ}(1 − θ)e^{inθ} dθ = \frac{1}{inπδ} ∫_{0}^{δ} e^{inθ} dθ + [\frac{1}{inπ} (1 − θ/δ)e^{inθ}]_{0}^{δ} \\ & = [\frac{−1}{n^{2}πδ} e^{inθ} + \frac{1}{inπ} (1 − θ/δ)e^{inθ}]_{0}^{δ} = \frac{1 − e^{inδ}}{n^{2}πδ} + \frac{i}{nπ} . \end{aligned}

Therefore the Fourier series can formally be written as

\frac{δ}{2π} + 2 ∑_{n=1}^{∞} \frac{1 − cos(nδ)}{n^{2}πδ} cos(nθ),

since f is an even function. Observe that the Fourier series are not absolutely convergent, so we are not able to conclude the Fourier series converges uniformly to f. However, f is continuous and locally of bounded variation (for example, f = f_{1} + f_{2} where

f_{1}(θ) = \{ \begin{matrix}0, & if θ < −δ, \\ 1 + θ/δ, & if − δθ ≤ 0, \\ 1, & if 0 < θ \end{matrix}

is non-decreasing and f_{2} is non-increasing), so we can apply Jordan’s criterion to conclude that the Fourier series converges to f at every θ.

#2, 6. We compute

|f̂_{k}(n) − f̂(n)| = |\frac{1}{2π} ∫_{π}^{π}(f_{k}(x) − f(x))e^{inx} dx| ≤ \frac{1}{2π} ∫_{π}^{π} |f_{k}(x) − f(x)| dx → 0

as k → ∞. #3, 1. Suppose A_{n} = {a_{n}^{N}}_{n} is a Cauchy sequence in l^{2}(Z). Therefore, for each n ∈ Z,

|a_{n}^{N} − a_{n}^{M}| ≤ ( ∑_{m∈Z} |a_{m}^{N} − a_{m}^{M}|^{2})^{\frac{1}{2}} → 0

as M, N → ∞, so {a_{n}^{N}}_{N} is a Cauchy sequence in R. Therefore, by the completeness of R, we can define a component-wise limit of A_{N} to be A = {a_{n}}_{n}, where a_{n} = lim_{N→∞} a_{n}^{N}.

Now fix ε > 0. There exists an number N_{0} such that, for N, M > N_{0}, we have

( ∑_{m∈Z} |a_{m}^{N} − a_{m}^{M}|^{2})^{\frac{1}{2}} < ε.

Now, by Fatou’s lemma,

(∑_{n∈Z} |a_{n}^{N} − a_{n}|^{2})^{\frac{1}{2}} ≤ lim inf_{M→∞} ( ∑_{m∈Z} |a_{m}^{N} − a_{m}^{M}|^{2})^{\frac{1}{2}} < ε,

for N > N_{0}. This proves both that A ∈ l^{2}(Z) and that lim_{N→∞} A_{N} = A in l^{2}(Z). #4, 4. Clearly f̂(0) = 1. Also, for ξ ≠ 0,

f̂(ξ) = ∫_{−1}^{1} e^{2πixξ} dx = [\frac{e^{2πixξ}}{2πiξ} ]_{−1}^{1} = \frac{e^{2πiξ}}{2πiξ} − \frac{e^{−2πiξ}}{2πiξ} = \frac{sin(2πξ)}{πξ}.

Computing the Fourier transform of g is similar. #4, 6. We have

f̂(ξ) − f̂(ξ + h) = (1 − e^{2πihξ})f̂(ξ) → 0

as h → 0, so f̂ is continuous. Also, using the given formula (which is easy to prove using the periodicity of e^{2πixξ})

|f̂(ξ)| ≤ \frac{1}{2} ∫ |f(x) − f(x − 1/(2ξ)| dξ → 0

as |ξ| → ∞. #5, 5. (a) We have

\begin{aligned}(Af, g) & = ∫_{R} f′(x)\overline{g(x)} + xf(x)\overline{g(x)} dx \\ & = ∫_{R} −f(x)\overline{g′(x)} + f(x)\overline{xg(x)} dx = (f, A^{∗}g). \end{aligned}

(b) Using (a),

(A^{∗}A(f), f) = (Af, Af) = ∥Af∥^{2} ≥ 0.

(c) This one was harder than I thought, sorry. Define

A_{t}f(x) = f′(x) + txf(x) and A_{t}^{∗}f(x) = −f′(x) + txf(x).

and observe that we can prove (A_{t}^{∗}A_{t}f, f) ≥ 0 just as before. Now,

\begin{aligned}0 & ≤ (A_{t}^{∗}A_{t}f, f) = ∫_{R} −f′′(x)\overline{f(x)} + x^{2}|f(x)|^{2} − |f(x)|^{2} dx \\ & = ∫_{R} |f′(x)|^{2} + t^{2}x^{2}|f(x)|^{2} − t|f(x)|^{2} dx \end{aligned}

and so viewing the right-hand side as a polynomial in t, it can’t have more than one real root, so we must have

(∫_{R} |f(x)|^{2} dx)^{2} − 4 (∫_{R} |f′(x)|^{2} dx) (∫_{R} |f(x)|^{2} dx) ≤ 0.

Rearranging this, using the normalisation condition and Plancherel’s Theorem gives the uncertainty principle. The other direction is proved by reversing these steps.