MATH 220: THE FOURIER TRANSFORM TEMPERED DISTRIBUTIONS. Beforehand we constructed distributions by taking the set Cc

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1 MATH 220: THE FOURIER TRANSFORM TEMPERED DISTRIBUTIONS Beforehand we constructed distributions by taking the set Cc ( ) as the set of very nice functions, and defined distributions as continuous linear maps u : Cc ( ) C (or into reals). While this was an appropriate class when studying just derivatives, we have seen that for the Fourier transform the set of very nice functions is that of Schwartz functions. Thus, we expect that the set S ( ) of corresponding distributions should consist of continuous linear maps u : S( ) C. In order to make this into a definition, we need a notion of convergence on S( ). Since Cc ( ) S( ), it is not unreasonable to expect that if a sequence {φ j } j=1 converges to some φ Cc ( ) inside the space Cc ( ) (i.e. in the sense of Cc ( )- convergence), then it should also converge to φ in the sense of S-convergence. We shall see that this is the case, which implies that every u S lies in D as well: for if φ j φ in Cc ( ) then φ j φ in S, hence u(φ j ) u(φ) by the continuity of u as an element of S. Thus, S D, i.e. S is a special class of distributions. In order to motivate the definition of S-convergence, recall that S = S( ) is the set of functions φ C ( ) with the property that for any multiindices α, β N n, x α β φ is bounded. Here we wrote x α = x α1 1 xα xαn n, and β = x β x βn n ; with xj = x j. With this in mind, convergence of a sequence φ m S, m N, to some φ S, in S is defined as follows. We say that φ m converges to φ in S if for all multiindices α, β, sup x α β (φ m φ) 0 as m, i.e. if x α β φ m converges to x α β φ uniformly. A tempered distribution u is defined as a continuous linear functional on S (this is written as u S ), i.e. as a map u : S C which is linear: u(aφ + bψ) = au(φ) + bu(ψ) for all a, b C, φ, ψ S, and which is continuous: if φ m converges to φ in S then lim m u(φ m ) = u(φ) (this is convergence of complex numbers). As mentioned already, any tempered distribution is a distribution, since φ Cc implies φ S, and convergence of a sequence in Cc implies that in S (recall that convergence of a sequence in Cc means that the supports stay inside a fixed compact set and the convergence of all derivatives is uniform). The converse is of course not true; e.g. any continuous function f on defines a distribution, but f(x)φ(x) dx will not converge for all φ S if f grows too fast at infinity; e.g. f(x) = e x 2 does not define a tempered distribution. On the other hand, any continuous function f satisfying an estimate f(x) C(1 + x ) N for some N and C defines a tempered distribution u = ι f via u(ψ) = ι f (ψ) = f(x)ψ(x) dx, ψ S, since f(x)ψ(x) dx CM (1 + x ) N (1 + x ) N n 1 dx <, M = sup ( (1 + x ) N+n+1 ψ ) <. 1

2 2 This is the reason for the tempered terminology: the growth of f is tempered at infinity. Another example of such a distribution u S is the delta distribution: u = δ a, given by δ a (φ) = φ(a) for φ S. A more extreme example is the following, on R for simplicity: Let a > 0, and let u = k Z δ ak, i.e. for φ S(R), let u(φ) = k Z φ(ak). This sum converges, since φ(x) C(1+ x 2 ) 1 as φ S, and k Z C(1+a2 k 2 ) 1 converges. It is also continuous since if φ j φ in S, then (1 + x 2 )φ j (1 + x 2 )φ uniformly on R, hence u(φ j ) u(φ) (1 + a 2 k 2 ) φ j (ak) φ(ax) (1 + a 2 k 2 ) 1 k Z ( sup (1 + x 2 ) φ j (x) φ(x) ) (1 + a 2 k 2 ) 1 x R k Z ( sup (1 + x 2 ) φ j (x) φ(x) ) (1 + a 2 k 2 ) 1, x R k Z and the last sum converges, so u(φ j ) u(φ) 0 as j. We defined the Fourier transform on S as (Fφ)() = ˆφ() = e ix φ(x) dx, and the inverse Fourier transform as (F 1 ψ)(x) = (2π) n e ix ψ() d. The Fourier transform satisfies the relation ˆφ()ψ() d = φ(x) ˆψ(x) dx, φ, ψ S. (Of course, we could have denoted the variable of integration by x on both sides.) Indeed, explicitly writing out the Fourier transforms, ( ) e ix φ(x) dx ψ() d = e ix φ(x)ψ() dx d ( ) = φ(x) e ix ψ() d dx, where the middle integral converges absolutely (since φ, ψ decrease rapidly at infinity), hence the order of integration can be changed. Of course, this argument does not really require φ, ψ S, it suffices if they decrease fast enough at infinity, e.g. φ(x) C(1 + x ) s for some s > n, and similarly for ψ. In the language of distributional pairing this just says that the tempered distribtions ι φ, resp. ι ˆφ, defined by φ, resp. ˆφ, satisfy ι ˆφ(ψ) = ι φ ( ˆψ), ψ S.

3 3 Motivated by this, we define the Fourier transform of an arbitrary tempered distribution u S by (Fu)(ψ) = u( ˆψ), ψ S. It is easy to check that û = Fu is indeed a tempered distribution, and as observed above, this definition is consistent with the original one if u is a tempered distribution given by a Schwartz function φ (or one with enough decay at infinity). It is also easy to see that the Fourier transform, when thus extended to a map F : S S, still has the standard properties, e.g. F(D xj u) = j Fu. Indeed, by definition, for all ψ S, (F(D xj u))(ψ) = (D xj u)(fψ) = u(d xj Fψ) = u(f( j ψ)) = (Fu)( j ψ) = ( j Fu)(ψ), finishing the proof. The inverse Fourier transform of a tempered distribution is defined analogously, and it satisfies F 1 u(ψ) = u(f 1 ψ), F 1 F = Id = FF 1 on tempered distributions as well. Again, this is an immediate consequence of the corresponding properties for S, for (F 1 Fu)(ψ) = Fu(F 1 ψ) = u(ff 1 ψ) = u(ψ). As an example, we find the Fourier transform of the distribution u = ι 1 given by the constant function 1. Namely, for all ψ S, û(ψ) = u( ˆψ) = ˆψ(x) dx = (2π) n F 1 ( ˆψ)(0) = (2π) n ψ(0) = (2π) n δ 0 (ψ). Here the first equality is from the definition of the Fourier transform of a tempered distribution, the second from the definition of u, the third by realizing that the integral of any function φ (in this case φ = ˆψ) is just (2π) n times its inverse Fourier transform evaluated at the origin (directly from the definition of F 1 as an integral), the fourth from F 1 F = Id on Schwartz functions, and the last from the definition of the delta distribution. Thus, Fu = (2π) n δ 0, which is often written as F1 = (2π) n δ 0. Similarly, the Fourier transform of the tempered distribution u given by the function f(x) = e ix a, where a is a fixed constant, is given by (2π) n δ a since û(ψ) = u( ˆψ) = e ix a ˆψ(x) dx = (2π) n F 1 ( ˆψ)(a) = (2π) n ψ(a) = (2π) n δ a (ψ), while its inverse Fourier transform is given by δ a since F 1 u(ψ) = u(f 1 ψ) = e ix a F 1 ψ(x) dx = F(F 1 ψ)( a) = ψ( a) = δ a (ψ). We can also perform analogous calculations on δ b, b : Fδ b (ψ) = δ b (Fψ) = (Fψ)(b) = e ix b ψ(x) dx, i.e. the Fourier transform of δ b is the tempered distribution given by the function f(x) = e ix b. With b = a, the previous calculations confirm what we knew anyway namely that FF 1 f = f (for this particular f).

4 4 We can now use tempered distributions to solve the wave equation on. Thus, consider the PDE ( 2 t c 2 )u = 0, u(0, x) = φ(x), u t (0, x) = ψ(x). Take the partial Fourier transform û of u in x to get ( 2 t + c 2 2 )û = 0, û(0, ) = Fφ(), û t (0, ) = Fψ(). For each this is an ODE that is easy to solve, with the result that Thus, û(t, ) = cos(c t)fφ() + sin(c t) Fψ(). c ( u = F 1 cos(c t)fφ() + sin(c t) ) Fψ(). c This can be rewritten in terms of convolutions, namely ( ) sin(c t) u(t, x) = F 1 (cos(c t)) x φ + F 1 x ψ(), c so it remains to evaluate the inverse Fourier transforms of these explicit functions. We only do this in R (i.e. n = 1). Here we need to be a little careful as we might be taking the convolution of two distributions in principle! However, any distribution can be convolved with elements of Cc ( ). Indeed, if f C( ), φ Cc ( ) then f φ(x) = f(y)φ(x y) dy = ι f (φ x ), where we write φ x (y) = φ(x y). Note that f φ C ( ) in fact, as differentiation under the integral sign shows (the derivatives fall on φ!). We make the consistent definition for u D ( ) that so u φ C ( ) since (u φ)(x) = u(φ x ), xj (u φ) = u( xj φ x ), in analogy with differentiation under the integral sign. As an example, δ a φ(x) = δ a (φ x ) = φ(x a). With some work one can even make sense of convolving distributions, as long as one of them has compact support, but we do not pursue this here, as we shall see directly that our formula makes sense for distributions even. We also note that for f(x) = H(a x ), a > 0, where H is the Heaviside step function (so H(s) = 1 for s 0, H(s) = 0 for s < 0), a x+a f φ(x) = H(a y )φ(x y) dx = φ(x y) dy = φ(s) ds, where we wrote s = x y. Returning to the actual transforms, (using that cos is even, sin is odd) a x a F 1 (cos(ct)) = 1 1 (F e ict + F 1 e ict ) = (δ ct + δ ct ),

5 5 while (note that 1 sin(ct) is continuous, indeed C, at = 0!) from the homework so In summary, F x H(ct x ) = 2 sin(ct), F 1 (c 1 1 sin(ct)) = 1 H(ct x ). 2c u(t, x) = 1 2 (δ ct + δ ct )) x φ + 1 2c H(ct x ) x ψ = 1 2 (φ(x ct) + φ(x + ct)) + 1 2c x+ct x ct ψ(y) dy, so we recover d Alembert s formula. In order to have more examples of tempered distributions, note that any distribution that vanishes at infinity should be tempered. Of course, we need to make sense of this. We recall that the support of a continuous function f, denoted by supp f, is the closure of the set {x : f(x) 0}. Thus, x / supp f if and only if there exists a neighborhood U of x such that y U implies f(y) = 0. Thus, supp f is closed by definition; so for continuous functions on, it is compact if and only if it is bounded. The support of a distribution u is defined similarly. One says that x / supp u if there exists a neighborhood U of x, such that on U, u is given by the zero function. That is, x / supp u if there exists U as above such that for all φ Cc with supp φ U, u(φ) = 0. For example, if u = δ a is the delta distribution at a, then supp u = {a}, since u(φ) = φ(a), so if x a, taking U as a neighborhood of x that is disjoint from a, u(φ) = 0 follows for all φ Cc with supp φ U. With this definition of the support one in fact concludes that if u D, φ Cc and supp u supp φ = then u(φ) = 0. Note that if u is a distribution and supp u is compact, u, which is a priori a map u : Cc C, extends to a map u : C ( ) C, i.e. u(φ) is naturally defined if φ is just smooth, and does not have compact support. To see this, let f Cc be identically one in a neighborhood of supp u, and for φ C ( ) define u(φ) = u(fφ), noting that fφ Cc. If u = ι g is given by integration against a continuous function g of compact support, this just says that we defined for φ C ( ) ι g (φ) = g(x)f(x)φ(x) dx = g(x)φ(x) dx, which is of course the standard definition if φ had compact support. Note that the second equality above holds since we are assuming that f is identically 1 on supp g, i.e. wherever f is not 1, g necessarily vanishes. We should still check that the definition of the extension of u does not depend on the choice of f (which follows from the above calculation if u is given by a continuous function g). But this can be checked easily, for if f 0 is another function in Cc which is identically one on supp u, then we need to make sure that u(fφ) = u(f 0 φ) for all φ C ( ), i.e. that u((f f 0 )φ) = 0 for all φ C ( ). But f = f 0 = 1 on a neighborhood of supp u, so (f f 0 )φ vanishes there, hence u((f f 0 )φ) = 0 indeed. Moreover, any distribution u of compact support, e.g. δ a for a, is tempered. Indeed, ψ S certainly implies that ψ C ( ), so u(ψ) is defined, and it is easy

6 6 to check that this gives a tempered distribution. In particular, δ a (ψ) = ψ(a), is a tempered distribution as we have already seen. Note that the Fourier transform of a compactly supported distribution can be calculated directly. Indeed, g (x) = e ix is a C function (of x), and compactly supported distributions can be evaluated on these. Thus, we can define Fu as the tempered distribution given by the function u(g ). For example, if u = δ b, then Fu is given by the function δ b (g ) = g (b) = e i b in accordance with our previous calculation. Of course, if u is given by a continuous function f of compact support, then u(g ) = f(x)g (x) dx = e ix f(x) dx = (Ff)() indeed, this motivated the definition of Fu. This definition is also consistent with the general one for tempered distributions, as we have seen on the particular example of delta distributions. Indeed, in general, ( ) u(g )φ() d = u g φ() d = u(fφ) = (Fu)(φ). The fact that for compactly supported distributions u, Fu is given by u(g ) shows directly that for such u, Fu is given by a C function: u(g ) = u(e ix ), and differentiating this with respect to simply differentiates g, i.e. simply gives another exponential (times a linear function), which is still C.