Math 597. Homework 12 Bryden Cais

(1) Suppose that f_{j} → f in L^{p} and g_{j} → g in L^{q} with 1/p + 1/q = 1 and p ≥ 1. Then we have

|∫ f_{j}g_{j} − fg dμ| ≤ ∫ |f_{j}g_{j} − f_{j}g + f_{j}g − fg| dμ
≤ ∫ |f_{j}||g_{j} − g| dμ + ∫ |g||f_{j} − f| dμ
≤ ∥f_{j}∥_{p}∥g_{j} − g∥_{q} + ∥g∥_{q}∥f_{j} − f∥_{p},

where we have used Minkowski’s inequality. However, since f_{j} → f and g_{j} → g in L^{p}, L^{q} respectively, we have ∥f_{j}∥_{p} ≤ C and ∥g∥ ≤ D for all sufficiently large j and some constants C, D (since ∥f − f_{j}∥_{p} can be made arbitrarily small by making j suff. large and since ∥g∥ < ∞). Moreover, for any ϵ > 0 there exists N such that for all j > N we have ∥g − g_{j}∥_{q} < ϵ/C and ∥f_{j} − f∥ < ϵ/D so that for all sufficiently large j we have

|∫ f_{j}g_{j} dμ − ∫ fg dμ| < ϵ

as required.

(2) (a) Let (X, A, μ) be a finite measure space and f ∈ L^{2}(X, μ). Let f_{1} = \frac{1}{\sqrt{μ(X)}}χ_{X} (which is well

defined and in L^{2} since X is finite) and observe that ⟨f_{1}, f_{1}⟩ = 1. From class, we have the formula

∥f − αf_{1}∥^{2} = ∥f∥^{2} − ∥⟨f, f_{1}⟩∥^{2} + ∥⟨f, f_{1}⟩ − α∥

so that ∥f − αf_{1}∥ = ∥f − \frac{α}{\sqrt{μ(X)}}∥ := ∥f − k∥ is minimized when α = ⟨f, f_{1}⟩, that is, when

k = \frac{1}{μ(X)} ∫_{X} f dμ.

(b) Let f_{j} = \frac{1}{2^{−n/2}} χ_{I_{j}} where I_{j} is the j^{th} dyadic interval of length 2^{−n} contained in [0, 1]. It is clear that {f_{j}} is orthonormal, so that we may apply the result from class as above to see that

∥f − ∑_{j=1}^{2^{n}} α_{j}f_{j}∥

is minimized when α_{j} = ⟨f, f_{j}⟩. We thus conclude that

∥f − ∑_{j=1}^{2^{n}} c_{j}χ_{I_{j}}∥

is minimized when

c_{j} = 2^{n} ∫_{I_{j}} f dμ.

(c) Observe that the span of the first 2^{n} Haar functions is exactly the span of the χ_{I_{j}} for I_{j} the dyadic intervals in [0, 1] of length 2^{−n}. Now, from class, we know that

∑_{j=1}^{2^{n}}⟨f, f_{j}⟩f_{j}

is the best approximation to f (with respect to ∥˙∥) that we can obtain using functions in the span of the first 2^{n} f_{j}. But in part (b) we showed that

∑_{j=1}^{2^{n}} c_{j}χ_{I_{j}}

is the best approximation to f we can obtain from functions in the span of the 2^{n} χ_{I_{j}}. Since the span of the χ_{I_{j}} and the span of the first 2^{n} f_{j} agree, it follows that the sums

∑_{j=1}^{2^{n}} c_{j}χ_{I_{j}} and ∑_{j=1}^{2^{n}}⟨f, f_{j}⟩f_{j}

are equal (as functions on L^{2}([0, 1], μ_{Leb})).

(3) Write

nχ_{[0,1/n]} = ∑_{j∈Z} a_{n}e^{ijx}.

Then we easily find that

a_{j} = \frac{1}{2πi} ∫_{0}^{2π} nχ_{[0,1/n]}e^{−ijx} dx

= \frac{n}{2π} ∫_{0}^{1/n} e^{−ijx} dx

= \frac{n}{2πij} (1 − e^{−ij/n}),

if j ≠ 0 and

a_{0} = \frac{n}{2π} ∫_{0}^{1/n} dx = \frac{1}{2π} ,

so that

nχ_{[0,1/n]}(x) = \frac{1}{2π} (1 + n ∑_{j≠0} \frac{1 − e^{−ij/n}}{ij} e^{−ijx}) .

Observe that for all n we have ∫_{R} nχ_{[0,1/n]} = 1 and

nχ_{[0,1/n]} = \begin{cases}n & if x ∈ [0, 1/n] \\ 0 & otherwise \end{cases} ,

so intuitively speaking, the limit function as n → ∞ is just the dirac delta function δ(x) satisfying δ(x) = 0 for all x ≠ 0 and ∫_{R} δ = 1.

(4) Let A denote the annulus {z ∈ C : r_{1} < |z| < r_{2}} with 0 < r_{1} < 1 < r_{2}, and H the region {z ∈ C : − log(r_{2}) < ℑ(z) < − log(r_{1})}. Then the map z ↦ exp(iz) is a conformal map from H to A. Any holomorphic function h on A pulls back to a holomorphic function f = h ∘ exp on H. Further, any such h has a finite tailed laurent expansion converging uniformly to h in the annulus; this gives rise to a finite tailed laurent expansion (now in the uniformizer w = e^{iz}) for f converging uniformly in H. Observe that H contains the real axis. It follows that the fourier series of f converges uniformly to f on R.

(5) (a) Suppose that f has fourier series ∑_{j∈Z} c_{j}e^{ijx} so that

\frac{1}{2π} ∫_{0}^{2π} f(x)e^{−ijx} dx = c_{j}

. Then we have

\frac{1}{2π} ∫_{0}^{2π} f_{α}(x)e^{−ijx} dx = \frac{1}{2π} ∫_{0}^{2π} f(x + α)e^{−ijx} dx

= \frac{1}{2π} ∫_{α}^{2π+α} f(x)e^{−ij(x−α)} dx

= \frac{1}{2π} (∫_{2π}^{2π+α} f(x)e^{−ij(x−α)} dx + ∫_{α}^{2π} f(x)e^{−ij(x−α)} dx)

= \frac{1}{2π} (∫_{0}^{α} f(x)e^{−ij(x−α)} dx + ∫_{α}^{2π} f(x)e^{−ij(x−α)} dx)

since f has period 2π,

= \frac{e^{ijα}}{2π} ∫_{0}^{2π} f(x)e^{−ijx} dx

= c_{j}e^{ijα},

from which it follows that

f_{α}(x) = ∑_{j∈Z} c_{j}e^{ijα}e^{ijx}.

(b) Observe that e^{ijx} ∈ L^{2}([0, 2π)) for all j ∈ Z and that these functions are linearly independent. Now if α = πp/q ∈ πQ for p, q ∈ Z then e^{2qijx} = e^{2qij(x+α)} as functions on [0, 2π). It follows that if α is a nonzero rational multiple of π then the dimension of {f ∈ L^{2} : f = f_{α}} is infinite. On the other hand, suppose that α/π is not rational. Then e^{ijx} ≠ e^{ij(x+α)} for any j ∈ Z (for this would require e^{ijα} = 1, or equivalently, α/π ∈ Q). Since e^{ijx} for j ∈ Z are dense in L^{2}([0, 2π)), we see that the dimension of {f ∈ L^{2} : f = f_{α}} is 0. Finally, if α = 0 it is clear that the above space is infinite dimensional.