Math 597. Homework 13 Bryden Cais

(1) (a) It is clear that for f, g ∈ L^{1}(T) and 2π periodic the convolution f ∗g is also 2π periodic: indeed, we have

f ∗ g(x + 2π) = ∫_{T} f(y)g(x + 2π − y) dy
= ∫_{T} f(y)g(x − y) dy

since g is 2π periodic,

= f ∗ g(x).

Moreover, observe that |f(y)g(x − y)| is nonnegative and measurable. Since Lebesgue measure is complete and σ-finite, it follows from the Tonelli Theorem that

∫_{T} |f ∗ g(x)| dx ≤ ∫_{T} ∫_{T} |f(y)||g(x − y)| dy dx
= ∫_{T} ∫_{T} |f(y)||g(x − y)| dx dy
= ∫_{T} ∫_{T} |f(y)||g(x)| dx dy

since g is 2π periodic,

= ∥f∥_{L^{1}(T)}∥g∥_{L^{1}(T)},

so that

∥f ∗ g∥_{L^{1}(T)} ≤ ∥f∥_{L^{1}(T)}∥g∥_{L^{1}(T)}.

(b) Let ϵ > 0, and let g be a continuous 2π periodic function with ∥f − g∥ < ϵ/3. Moreover, since g ∗ K_{n} → g uniformly (since g is continuous), let N be such that for all n > N we have |g ∗ K_{n}(x) − g(x)| < ϵ/6π for all x ∈ T. Then we have

∥f ∗ K_{n} − f∥ = ∥f ∗ K_{n} − g ∗ K_{n} + g ∗ K_{n} − g + g − f∥
≤ ∥(f − g) ∗ K_{n}∥ + ∥g ∗ K_{n} − g∥ + ∥g − f∥
≤ ∥f − g∥∥K_{n}∥ + ∥g ∗ K_{n} − g∥ + ∥g − f∥

by part (i),

≤ \frac{ϵ}{3} + ∫_{T} \frac{ϵ}{6π} dx + \frac{ϵ}{3}

where we have used the fact that ∥K_{n}∥ = 1 since K_{n} > 0 and ∫_{T} K_{n} = 1,

≤ ϵ.

It follows that the n^{th} Casàro partial sum of the Fourier Series of f tends to f in L^{1} as n → ∞.

(2) (a) Suppose that f ∈ C^{k}(T) has Fourier Series ∑_{j∈Z} c_{j}e^{ijx}. Then we have c_{j} = \frac{1}{2π} ∫_{T} f(x)e^{−ijx} dx. Now, if l < k then f ∈ C^{l}(T) and f^{(l+1)} exists and is continuous. It follows that

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

= 0 since f^{(l)} is 2π periodic. Thus we have for l < k

∫_{T} f^{(l+1)}(x)e^{−ijx} dx = (ij) ∫_{T} f^{(l)}(x)e^{−ijx} dx.

It then follows by induction that f^{(k)} has Fourier Series

∑_{j∈Z}(ij)^{k}c_{j}e^{ijx}.

(b) Suppose that f ∈ C^{k}(T). Then since T is compact and f, f^{(k)} are continuous, they are both bounded in absolute value on T; it follows that ∥f∥_{2}^{2} and ∥f^{(k)}∥_{2}^{2} are finite. On the other hand, by part (i) and Parseval’s identity, we have

∥f∥_{2}^{2} = ∑_{j∈Z} |c_{j}|^{2},

and

∥f^{(k)}∥_{2}^{2} = ∑_{j∈Z} |j^{k}c_{j}|^{2}.

Since ∥f∥_{2}^{2}, ∥f^{(k)}∥_{2}^{2} < ∞ we see that

∥f∥_{2}^{2} + ∥f^{(k)}∥_{2}^{2} = ∑_{j∈Z}(1 + |j|^{2k})|c_{j}|^{2} = ∥f∥_{H^{k}}^{2} < ∞

where we are able to rearrange the sum however we like since all the terms are positive.

(c) Suppose that f ∈ H^{s}(T) has Fourier Series ∑_{j∈Z} c_{j}e^{ijx}. Then we have

∑_{j∈Z} |c_{j}| = ∑_{j∈Z} |c_{j}|\sqrt{1 + |j|^{2s}} \frac{1}{\sqrt{1 + |j|^{2s}}}

≤ (∑_{j∈Z}(1 + |j|^{2s})|c_{j}|^{2})^{1/2} (∑_{j∈Z} \frac{1}{1 + |j|^{2s}} )^{1/2}

by Minkowski’s identity,

≤ ∥f∥_{H^{s}} (∑_{j∈Z} \frac{1}{1 + |j|^{2s}} )^{1/2}

< ∞,

since f ∈ H^{s}(T) and we have used the fact the s > 1/2 so the sum ∑_{j∈Z} \frac{1}{1+|j|^{2s}} converges.

(d) Let f ∈ H^{s}(T) with t > 1/2. Then by part (c) the Fourier Series for f converges uniformly to f on T; this follows from the Weirstrass M-test and the fact that ∑ |c_{j}| = ∑ |c_{j}e^{ijx}| < ∞. Thus, the Fourier Series of f converges to a continuous function that agrees with f a.e., as required.

(e) Since f ∈ H^{l+s}(T) for any l ≤ k, we have

∑_{j∈Z} |j^{l}c_{j}| = ∑_{j∈Z} |j^{l}c_{j}|\sqrt{1/|j|^{2l} + |j|^{2s}} \frac{1}{\sqrt{1/|j|^{2l} + |j|^{2s}}}

≤ (∑_{j∈Z}(1 + |j|^{2(s+l)})|c_{j}|^{2})^{1/2} (∑_{j∈Z} \frac{1}{1/|j|^{2l} + |j|^{2s}} )^{1/2}

by Minkowski’s identity,

≤ ∥f∥_{H^{l+s}} (∑_{j∈Z} \frac{1}{1/|j|^{2l} + |j|^{2s}} )^{1/2}

< ∞,

as in part (c). We now proceed by induction. Suppose that for l < k f ∈ C^{l}(T) and set f_{n} = ∑_{−n}^{n} c_{j}e^{ijx}. Then our calculations above (via the Weierstrass M-test) show that f_{n}^{(l)} converges uniformly to f^{(l)} and that f^{(l+1)} = lim_{n→∞} f_{n}^{(l+1)} exists a.e. (See, for example, [1][p. 152]). In fact, it is not difficult to see, as before, that f_{n}^{(l+1)} converges uniformly, so that f ∈ C^{l+1}(T). The base case is part (d) so that f ∈ C^{k}(T) by induction.

(f) Let f ∈ C^{∞} have Fourier Series ∑_{j∈Z} c_{j}e^{ijx}. Then by part (a), we find that Sf has Fourier series ∑_{j∈Z} d_{j}e^{ijx} where

d_{j} = c_{j} ∑_{l=0}^{k}(ij)^{l}a_{l},

where for convenience we set a_{k} = 1. Now set p(z) = ∑_{l=0}^{k} a_{k}(iz)^{k}. Let g ∈ C^{∞} have Fourier Series ∑_{j∈Z} b_{j}e^{ijx}. As long as p(j) ≠ 0 for any integer j, the Fourier Series ∑_{j∈Z} \frac{b_{j}}{p(j)}e^{ijx} defines a C^{∞} function f with Sf = g so that S in this case is surjective (the fact that f ∈ C^{∞}(T) follows from the fact that for large j we have |p(j)| > 1 so that ∑_{j∈Z} |j^{k}b_{j}/p(j)| converges for all k > 0 since the tail end tends to 0 since g ∈ C^{∞}(T)). Conversely, if p(j) = 0 for some integer j, then e^{ijx} ∈ C^{∞}(T) is not in the image of S (since the coefficient of e^{ijx} in the Fourier Series of Sf for any f is 0 as we calculated above). Thus, S is surjective if and only if p(z) has no integer roots.

References

[1] Rudin, W. Principles of Mathematical Analysis. McGraw-Hill, Inc. 1976.