By Gustaf Gripenberg

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**Additional info for Fourier analysis**

**Example text**

E. ˆ Here fˆ(t) = the Fourier transform of fˆ evaluated at the point t. Proof. e. 43 CHAPTER 2. 18. fˆ(t) = f (t) (If we repeat the Fourier transform 4 times, then we get back the original function). (True at least if f ∈ L1 (R) and fˆ ∈ L1 (R). 19. Let f ∈ L1 (R) and g ∈ L1 (R). Then ˆ f(s)g(s)ds f (t)ˆ g (t)dt = R R Proof. f (t)ˆ g (t)dt = e−2πits g(s)dsdt (Fubini) f (t) t∈R R s∈R f (t)e−2πist dt g(s)ds = s∈R t∈R ˆ f(s)g(s)ds. = s∈R ˆ ∈ L1 (R). 20. Let f ∈ L1 (R), h ∈ L1 (R) and h ˆ fˆ(ω)h(ω)dω.

1. 2. C0 (R) = “continuous functions f (t) satisfying f (t) → 0 as t → ±∞”. The norm in C0 is f C0 (R) = max|f (t)| (= sup|f (t)|). t∈R t∈R Compare this to c0 (Z). 3. , 36 CHAPTER 2. FOURIER INTEGRALS 37 i) fˆ is continuous ii) fˆ(ω) → 0 as ω → ±∞ iii) |fˆ(ω)| ≤ ∞ |f (t)|dt, −∞ ω ∈ R. Note: Part ii) is again the Riemann-Lesbesgue lemma. Proof. 4 i). 4 ii), (replace n by ω, and prove this first in the special case where f is continuously differentiable and vanishes outside of some finite interval).

Suppose that gˆ(n) = f(n) for all n. Define h(t) = f (t) − g(t). Then ˆ h(n) = fˆ(n) − gˆ(n) = 0, n ∈ Z. e. e. 37. Suppose that f ∈ L1 (T) and that series ∞ ∞ ˆ n=−∞ |f(n)| < ∞. e. Proof. 14. e. 36. The following theorem is much more surprising. It says that not every sequence {an }n∈Z is the set of Fourier coefficients of some f ∈ L1 (T). 30 CHAPTER 1. 38. e. fˆ(n) is an odd function). Then ∞ i) n=1 ii) 1ˆ f (n) < ∞ n ∞ 1ˆ | f(n)| < ∞. n n=−∞ n=0 Proof. Second half easy: Since fˆ is odd, 1 1 1 | fˆ(n)| + | fˆ(−n)| | fˆ(n)| = n n n n>0 n<0 n=0 n∈Z = 2 ∞ 1 | fˆ(n)| < ∞ n n=1 if i) holds.