Lecture 4: Fourier Transforms.

1 Definition. Lecture 4: Fourier Transforms. We now come to Fourier transforms, which we give in the form of a definition. First we define the spaces L 1 () and L 2 (). Definition 1.1 The space L 1 () is defined as follows: L 1 () {f : C L 2 () {f : C f(t) dt < }. f(t) 2 dt < }. The space L 1 () is called the space of absolutely integrable functions and L 2 () is called the space of square-integrable functions. We can now define the Fourier transform: Definition 1.2 The Fourier transform of f L 1 () is defined as F (ω) for ω, where the integral is defined as f(t)e iωt dt f(t)e iωt dt lim f(t)e iωt dt. The reason for requiring that f L 1 () is to ensure that the integral converges: it converges absolutely, and therefore it converges. In fact, it converges pointwise: we have where F (ω) f(t)e iωt dt. F (ω) F (ω) for each ω, Example 1.1 f(t) e t. It is easy to check that f L 1 (). First, we have 1

Then we see that F (ω) e t e iωt dt e t e iωt dt + e t e iωt dt + e (1+iω)t dt + 1 1 + iω + 1 2 1 + ω 2 e e t e iωt dt e t e iωt dt e (1 iω)t dt 1 iω e (1+iω) 1 + iω e (1 iω) ( ) 1 iω e iω 1 + iω + eiω. 1 iω Example 1.2 f(t) In this case we have { 1, t 1 F (ω) lim F (ω) 2 1 + ω 2., t > 1. F (ω) 1 1 e iωt dt { 2 sin ω ω ω 2, ω. 2 Continuity and differentiability. We now turn to the continuity and differentiability of the Fourier transform F (ω) of f L 1 (). To formulate the results, we need the concept of piecewise continuous functions on. Definition 2.1 A function f : C is said to be piecewise continuous on if f is piecewise continuous on every closed, bounded subinterval [a, b]. We denote the set of picewise continuous functions by G(). From the definition, we see that a piecewise continuous function on has only a finite number of discontinuities on each finite subinterval, but it may have an infinite number on the whole of the real line. As an example we take f(x) [x] which is defined as [x] n if n x < n + 1, where n Z. This is a picewise continuous function with discontinuities at every integer value for x. We now have the following result: 2

Theorem 2.1 If f L 1 () G() then its Fourier transform F (ω) has the following properties: 1. F (ω) is defined for all ω ; 2. F is continuous on ; 3. F (ω) as ω ±. Proof: The first part follows from the fact that f L 1 () and thus that the integral defining F (ω) is absolutely convergent for each ω. The second part uses the following result, which we quote without proof: Lemma 2.1 Suppose that {f h : h } is a family of piecewise continuous functions, and suppose further that (i) There is a function g so that f h (x) g(x) for all x and all h, (ii) g(x)dx <, (iii) lim h f h (x) f(x) for each x. Then we have lim f h (x)dx f(x)dx. h We then consider F (ω + h) F (ω) and show that this difference tends to zero as h. So, we have F (ω + h) F (ω) f(x)[e i(ω+h) x e iωx ]dx f(x)e iωx [e ihx 1]dx f h (x)dx, where f h (x) f(x)e iωx [e ihx 1]. Clearly we have for each x and we also have lim f h(x) h f h (x) f(x) e ihx 1 2 f(x), 3

and g(x) 2 f(x) satisfies the conditions of Lemma 2.1, so that we have That is lim h f(x)e iωx [e ihx 1]dx. and this proves the continuity of F (ω). lim[f (ω + h) F (ω)], h The third part, namely lim ω ± F (ω), can be found in standard textbooks, and we refer the reader to the course book. We shall also need a result on the differentiability of the Fourier transform F (ω) of f(x). To this end, we have first the M-test for integrals, which is the equivalent to the Weierstrass majorant theorem for functional series: Theorem 2.2 Let f(x, y) be a continuous function defined on 2 be such that (i) F (x) f(x, y)dy exists (ii) there is a function g(y) L 1 () so that f(x, y) g(y) for all x, y then converges uniformly on. That is, F (x) F (x) uniformly on as. This means that f(x, y)dy f(x, y)dy F (x) as. sup F (x) F (x) x Proof: First note that and we then have F (x) F (x) f(x, y)dy + f(x, y)dy F (x) F (x) f(x, y) dy + 4 g(y)dy + f(x, y) dy g(y)dy

from which we have sup F (x) F (x) g(y)dy + x as since g(y) L 1 (). This is given in the following result: g(y)dy Theorem 2.3 Suppose that f(x, y) and f x(x, y) are continuous on 2 and that we also have: (i) F (x) (ii) G(x) Then uniformly as. f(x, y)dy converges for each x, f x(x, y)dy converges uniformly on, that is G (x) F (x) f x(x, y)dy G(x) f x(x, y)dy. Proof: For the proof of this result, see pp 16-18 in the notes for Lecture 1, and in particular Theorem 2.2. As an example of Theorem 2.3 we take f(x) e x2 and calculate its Fourier transform. First put f(ω, x) e x2 e iωx. Then as is easily seen. Secondly F (ω) f(ω, x)dx, G(ω) f ω(ω, x)dx ( ix)f(ω, x)dx converges uniformly (Theorem 2.2, see also the notes for Lecture 1). Hence, by Theorem 2.3, we have 5

since we have as is easily verified. F (ω) i i d 2 dx i [ d 2 dx ω 2 ω 2 F (ω), xe x2 e iωx dx ( e x2) e iωx dx (e x2 e iωx ) dx + iω e x2 e iωx dx d ) (e x2 e iωx dx, dx ] e x2 e iωx dx Consequently, we find that F (ω) satisfies the differential equation which has general solution F (ω) ω 2 F (ω), F (ω) Ce ω2 /4, where C is an arbitrary real constant. We then note that F () e x2 dx π. Then C π and we have that the Fourier transform of e x2 is 3 ules for computation. F (ω) πe ω2 /4. We now turn to consider some useful rules for simplifying the calculation of Fourier transforms. In the following we will also use the notation F[f] for the Fourier transform of a function f. The first rule we have is quite obvious: Theorem 3.1 If f, g G() L 1 () and a, b C, then F[af + bg](ω) af (ω) + bg(ω). One of the uses of the Fourier transform is to transform differential equations into algebraic equations, which are (usually) easier to solve. The following result is then of great utility: 6

Theorem 3.2 If f is continuous and f, f G() L 1 () then F[f ](ω) iωf (ω). Proof: The proof relies on the fact that, under the given conditions, f(x) as x ±. First we have f(x) f() + and since f G() L 1 (), we have that x f (t)dt. exists, so that x lim x ± f (t)dt lim f(x) x ± exists. Now, if f(x) a as x we have f(x) a > as x. Thus there is an > so that f(x) > a /2 for x > and therefore f(x) dx diverges, contradicting the fact that f G() L 1 (). A similar argument shows that f(x) as x. Having established these limits, we then obtain: because F[f ](ω) lim f (x)e iωx dx lim iω f (x)e iωx dx [f(x)e iωx iωf (ω), f(x)e iωx dx ] f(x)( iω)e iωx dx lim f(x)e iωx lim [f()e iω f()e iω ] since f(x) x ±. In the same way we can prove: Theorem 3.3 If f, f, f (2),..., f (n 1) are continuous and f, f, f (2),..., f (n) L 1 (), then G() F[f (n) ](ω) (iω) n F (ω). 7

Proof: Proof by induction. To deal with differential equations with variable (polynomial) coefficients we need the following results: Theorem 3.4 If f, xf G() L 1 () then F[xf(x)](ω) if (ω). Also, if we have f, x n f G() for some natural number n, then F[x n f(x)](ω) i n F (n) (ω). Proof: For a proof, consult the course textbook. Finally, we come to the scaling law: Theorem 3.5 If f G() L 1 () and A, B with A, then F[f(Ax + B)](ω) 1 B A ei A ω F ( ω A ). Proof: The proof is a simple calculation, using a change of variable t Ax + B. Note that there are two cases. A > and A <. For A > we have For A < we have F[f(Ax + B)](ω) f(ax + B)e iωx dx 1 f(t)e A 1 B A ei A ω F ( ω A ). t B iω A dt This proves the result. F[f(Ax + B)](ω) f(ax + B)e iωx dx 1 f(t)e A 1 B ( A) ei A ω F ( ω A ) 1 B A ei A ω F ( ω A ). t B iω A In order to use Fourier transforms as a tool in calculations, we need to be able to find the inverse transform. That is, we need to be able to calculate the function which has a given expression as its Fourier transform. To this end we define the inverse Fourier transform as follows: 8 dt

Definition 3.1 The inverse Fourier transform of F (ω) where F F[f] for some function f G() L 1 () is defined as F 1 1 [F ](x) lim F (ω)e ixω dω. 2π Note that we may write the right-hand side as at least formally. 1 F[F ]( x) 2π We end this lecture with the following result, which we shall return to in the following lecture: Theorem 3.6 Suppose f G() L 1 () has Fourier transform F (ω). Then we have F 1 1 [F ](x) lim F (ω)e ixω dω f(x ) + f(x + ) 2π 2 at each x where the right- and left-hand derivatives of f(x) exist. Note the similarity of this with the theorem on the convergence of Fourier series. 9