Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is. e ikx f(x) dx. (1.

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1 Chapter 1 Fourier transforms 1.1 Introduction Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is ˆf(k) = e ikx f(x) dx. (1.1) It is a function on the (dual) real line R parameterized by k. The goal is to show that f has a representation as an inverse Fourier transform f(x) = e ikx ˆf(k) dk 2π. (1.2) There are two problems. One is to interpret the sense in which these integrals converge. The second is to show that the inversion formula actually holds. The simplest and most useful theory is in the context of Hilbert space. Let L 2 (R) be the space of all (Borel measurable) complex functions such that f 2 2 = f(x) 2 dx <. (1.3) Then L 2 (R) is a Hilbert space with inner product (f, g) = f(x)g(x) dx. (1.4) Let L 2 (R ) be the space of all (Borel measurable) complex functions such that h 2 2 = Then L 2 (R ) is a Hilbert space with inner product (h, u) = h(k)) 2 dk <. (1.5) 2π 1 h(k)u(k) dk 2π. (1.6)

2 2 CHAPTER 1. FOURIER TRANSFORMS We shall see that the correspondence between f and ˆf is a unitary map from L 2 (R) onto L 2 (R ). So this theory is simple and powerful. 1.2 L 1 theory First, we need to develop the L 1 theory. The space L 1 is a Banach space. Its dual space is L, the space of essentially bounded functions. An example of a function in the dual space is the exponential function φ k (x) = e ikx. The Fourier transform is then ˆf(k) = φ k, f = φ k (x)f(x) dx, (1.7) where φ k is in L and f is in L 1. Theorem. If f, g are in L 1 (R), then the convolution f g is another function in L 1 (R) defined by (f g)(x) = f(x y)g(y) dy. (1.8) Theorem. If f, g are in L 1 (R), then the Fourier transform of the convolution is the product of the Fourier transforms: (f g)(k) = ˆf(k)ĝ(k). (1.9) Theorem. Let f (x) = f( x). Then the Fourier transform of f is the complex conjugate of ˆf. Theorem. If f is in L 1 (R), then its Fourier transform ˆf is in L (R ) and satisfies ˆf f 1. Furthermore, ˆf is in C0 (R ), the space of bounded continuous functions that vanish at infinity. Theorem. If f is in L 1 and is also continuous and bounded, we have the inversion formula in the form where f(x) = lim ɛ 0 Proof: The inverse Fourier transform of this is It is easy to calculate that e ikxˆδɛ dk (k) ˆf(k) 2π, (1.10) ˆδ ɛ (k) = exp( ɛ k ). (1.11) δ ɛ (x) = 1 π ɛ x 2 + ɛ 2. (1.12) e ikxˆδɛ dk (k) ˆf(k) 2π = (δ ɛ f)(x). (1.13) However δ ɛ is an approximate delta function. The result follows by taking ɛ 0.

3 1.3. L 2 THEORY L 2 theory The space L 2 is its own dual space, and it is a Hilbert space. It is the setting for the most elegant and simple theory of the Fourier transform. Lemma. If f is in L 1 (R) and in L 2 (R), then ˆf is in L 2 (R ), and f 2 2 = ˆf 2 2. Proof. Let g = f f. Then g is in L 1 and is continuous and bounded. Furthermore, the Fourier transform of g is ˆf(k) 2. Thus f 2 2 dk 2 = g(0) = lim ˆδ ɛ (k) ˆf(k) ɛ 0 2π = 2 dk ˆf(k) 2π. (1.14) Theorem. Let f be in L 2 (R). For each a, let f a = 1 [ a,a] f. Then f a is in L 1 (R) and in L 2 (R), and f a f in L 2 (R) as a. Furthermore, there exists ˆf in L 2 (R ) such that ˆf a ˆf as a. Explicitly, this says that the Fourier transform ˆf(k) is characterized by a ˆf(k) e ikx f(x) dx 2 dk a 2π 0 (1.15) as a. These arguments show that the Fourier transformation F : L 2 (R) L 2 (R ) defined by F f = ˆf is well-defined and preserves norm. It is easy to see from the fact that it preserves norm that it also preserves inner product: (F f, F g) = (f, g). Define the inverse Fourier transform F in the same way, so that if h is in L 1 (R ) and in L 2 (R ), then F h is in L 2 (R) and is given by the usual inverse Fourier transform formula. Again we can extend the inverse transformation to F : L 2 (R ) L 2 (R) so that it preserves norm and inner product. Now it is easy to check that (F h, f) = (h, F f). Take h = F g. Then (F F g, f) = (F g, F f) = (g, f). That is F F g = g. Similarly, one may show that F F u = u. These equations show that F is unitary and that F = F 1 is the inverse of F. This proves the following result. Theorem. The Fourier transform F initially defined on L 1 (R) L 2 (R) extends by continuity to F : L 2 (R) L 2 (R ). The inverse Fourier transform F initially defined on L 1 (R ) L 2 (R ) extends by continuity to F : L 2 (R ) L 2 (R). These are unitary operators that preserve L 2 norm and preserve inner product. Furthermore, F is the inverse of F. 1.4 Absolute convergence We have seen that the Fourier transform gives a perfect correspondence between L 2 (R) and L 2 (R ). For the other spaces the situation is more complex. The map from a function to its Fourier transform gives a continuous map from L 1 (R) to part of C 0 (R ). That is, the Fourier transform of an integrable function is continuous and bounded (this is obvious) and approach zero

4 4 CHAPTER 1. FOURIER TRANSFORMS (Riemann-Lebesgue lemma). Furthermore, this map is one-to-one. That is, the Fourier transform determines the function. The inverse Fourier transform gives a continuous map from L 1 (R ) to C 0 (R). This is also a one-to-one transformation. One useful fact is that if f is in L 1 (R) and g is in L 2 (R), then the convolution f g is in L 2 (R). Furthermore, f g(k) = ˆf(k)ĝ(k) is the product of a bounded function with an L 2 (R ) function, and therefore is in L 2 (R ). However the same pattern of the product of a bounded function with an L 2 function can arise in other ways. For instance, consider the translate f a of a function f in L 2 (R) defined by f a (x) = f(x a). Then f a (k) = exp( ika) ˆf(k). This is also the product of a bounded function with an L 2 (R ) function. One can think of this last example as a limiting case of a convolution. Let δ ɛ be an approximate δ function. Then (δ ɛ ) a f has Fourier transform exp( ika)ˆδ ɛ (k) ˆf(k). Now let ɛ 0. Then (δ ɛ ) a f f a, while exp( ika)ˆδ ɛ (k) ˆf(k) exp( ika) ˆf(k). Theorem. If f is in L 2 (R) and if f exists (in the sense that f is an integral of f) and if f is also in L 2 (R), then the Fourier transform is in L 1 (R ). As a consequence f is is C 0 (R). Proof: ˆf(k) = (1/ 1 + k2 ) 1 + k 2 ˆf(k). Since f is in L 2 (R), it follows that ˆf(k) is in L 2 (R). Since f is in L 2 (R), it follows that k ˆf(k) is in L 2 (R ). Hence 1 + k 2 ˆf(k) is in L 2 (R ). Since 1/ 1 + k 2 is also in L 2 (R ), it follows from the Schwarz inequality that ˆf(k) is in L 1 (R ). 1.5 Fourier transform pairs There are some famous Fourier transforms. Fix σ > 0. Consider the Gaussian Its Fourier transform is g σ (x) = 1 2πσ 2 x2 exp( ). (1.16) 2σ2 ĝ σ (k) = exp( σ2 k 2 ). (1.17) 2 Proof: Define the Fourier transform ĝ σ (k) by the usual formula. Check that This proves that ( ) d dk + σ2 k ĝ σ (k) = 0. (1.18) ĝ σ (k) = C exp( σ2 k 2 ). (1.19) 2 Now apply the equality of L 2 norms. This implies that C 2 = 1. By looking at the case k = 0 it becomes obvious that C = 1.

5 1.6. PROBLEMS 5 Let ɛ > 0. Introduce the Heaviside function H(k) that is 1 for k > 0 and 0 for k < 0. The two basic Fourier transform pairs are and its complex conjugate f ɛ (x) = 1 x iɛ (1.20) ˆf ɛ (k) = 2πiH( k)e ɛk. (1.21) f ɛ (x) = 1 x + iɛ (1.22) ˆf ɛ ( k) = 2πiH(k)e ɛk. (1.23) These may be checked by computing the inverse Fourier transform. Notice that f ɛ and its conjugate are not in L 1 (R). Take 1/π times the imaginary part. This gives the approximate delta function δ ɛ (x) = 1 ɛ π x 2 + ɛ 2. (1.24) ˆδ ɛ (k) = e ɛ k. (1.25) Take the real part. This gives the approximate principal value of 1/x function x p ɛ (x) = x 2 + ɛ 2 (1.26) 1.6 Problems ˆp ɛ (k) = πi[h(k)e ɛk H( k)e ɛk ]. (1.27) 1. Let f(x) = 1/(2a) for a x a and be zero elsewhere. Find the L 1 (R), L 2 (R), and L (R) norms of f, and compare them. 2. Find the Fourier transform of f. 3. Find the L (R ), L 2 (R ), and L 1 (R ) norms of the Fourier transform, and compare them. 4. Compare the L (R ) and L 1 (R) norms for this problem. Compare the L (R) and L 1 (R ) norms for this problem. 5. Use the pointwise convergence at x = 0 to evaluate an improper integral. 6. Calculate the convolution of f with itself. 7. Find the Fourier transform of the convolution of f with itself. Verify in this case that the Fourier transform of the convolution is the product of the Fourier transforms.

6 6 CHAPTER 1. FOURIER TRANSFORMS 1.7 Poisson summation formula Theorem: Let f be in L 1 (R) with ˆf in L 1 (R ) and such that k ˆf(k) <. Then 2π f(2πn) = ˆf(k). (1.28) n k Proof: Let S(t) = n f(2πn + t). (1.29) Since S(t) is 2π periodic, we can expand S(t) = k a k e ikt. (1.30) It is easy to compute that a k = 1 2π 2π 0 S(t)e ikt dt = 1 2π ˆf(k). (1.31) So the Fourier series of S(t) is absolutely summable. In particular S(0) = k a k. (1.32)