schwartz

Math 395 (Fall 2009). Honors Analysis I 36.Schwartz functions, Fourier inversion, Plancherel formula 36.1. Goal. We would like to discuss Fourier transform on some classes of functions other thanL1(R); in particular, we would like to extend it to L2(R). As a key ingredient we will study the space of Schwartz functions on R(which is contained in Lp(R) for every 1p1 , and is dense in Lp(R) when 1p<1). As a by-product of our discussion we will also obtain a proof of the Fourier inversion theorem, which was stated earlier. 36.2. Recollections. Let us recall the denitions of the Fourier and the inverse Fourier transforms. If f=f(x)2L1(R), the Fourier transform offis dened to be the function bf() = (Ff)() =Z Reixf(x)dm(x); 2R: Ifg=g()2L1(R), the inverse Fourier transform ofgis dened to be the function (F1g)(x) =1 2Z Reixg()dm(); x2R: As I mentioned earlier, F1isnotthe inverse map to Fin the strict sense: for example, the domain ofF1(which is the space L1(R), by denition) is not equal to the image of F(which is hard to describe explicitly, but certainly contains many functions that are not integrable on R). So for the time being it is better to think of FandF1as two separate maps, even though they are \inverse to each other in some sense." 36.3. Motivation. As a motivation behind the notion of a Schwartz function, let us consider the following question. Is there a convenient subspace SL1(R) that is invariant under the Fourier transform, in the sense that if f2S, thenbf2S? The adjective \convenient" (which is not a mathematical term) means that Sshould have a fairly explicit description, and that it should be dense inL1(R) with respect to the L1norm. For example, the 1-dimensional space spanned by the function f(x) =ex2=2is invariant underF, because, as we saw in an earlier lecture, bf() =p 2
e2=2. However, this space is not dense in L1(R), and hence is not a good candidate. Note also that the space L1(R) itself is not going to work because there are many functionsf2L1(R) such thatbf62L1(R), an easy example being f(x) = 1[0;1](x). 36.4. Smooth functions with compact support. As a next candidate, let us try the space of innitely dierentiable functions with compact support on R. By denition, this is the space C1 0(R) consisting of innitely dierentiable functions f:R!Cfor which there exists a compact subset KR(depending on f) such that1f(x) = 0 for all x62K. 1In general, the support of a function fis dened as the closure of the set of points xwhere f(x)6= 0. Thus, for a function on Rn, having compact support is equivalent to having bounded support, by the Heine-Borel theorem. 1 2 By this week's homework, C1 0(R) is dense in L1(R) (in fact, it is dense in Lp(R) for every 1p <1). However, this space is notinvariant under the Fourier transform. Even worse: if f2C1 0(R) andf60, thenbfcannever have compact support on R. The reason is as follows. One can easily check that if f2C1 0(R), thenbf() is dened for all 2C, andbfis an entire function, that is, bfis holomorphic (() analytic) on the whole complex plane. This implies (by a general result of complex analysis) that if bf60, then the set of zeroes of bfin the complex plane is discrete , and therefore at most countable. In particular, the restriction of bftoRcertainly cannot have compact support. 36.5. Schwartz functions. It turns out that there is a space S(R) that \sits between" C1 0(R) andL1(R) that is invariant under F. This is precisely the space of Schwartz functions, which we will dene soon. As motivation for the denition, let us consider the following two statements, which are part of this week's homework. Lemma 36.1. Letf:R!Cbe a measurable function such that both f(x)andx
f(x) are integrable. Then bf()is dierentiable w.r.t. , and, in fact, dbf d=F ixf(x)
: Using this lemma, one immediately checks by induction that if f:R!Cis a measurable function such that xn
f(x) is integrable for every integer n0, then the Fourier transform bf:R!Cis an innitely dierentiable function. Lemma 36.2. Letf:R!Cbe a continuously dierentiable function such that both f(x)andf0(x)are integrable, and such that limx!1f(x) = 0 . Then lim!1
bf() = 0 . In the situation of the last lemma, one can check (using integration by parts) that 
bf() =i
F(f0(x)): By induction, we see that if f:R!Cis innitely dierentiable and f(m)(x) is integrable for everym0 (wheref(m)denotes the m-th derivative of f), thenm
bf()!0 as !1 for eachm0. Now, one should view the property that xn
f(x) is integrable for each n0 as a kind of \fast decay at innity" condition. In particular, if a measurable function f(x) is bounded and satises lim x!1xn
f(x) = 0 for all n0, it follows that xn
f(x) is integrable for all n0, by observing that xn
f(x) = (xn+2
f(x))
x2for allx6= 0. The upshot of the discussion above is that if a function decays very quickly at innity, then its Fourier transform is innitely dierentiable. Conversely, if a function is innitely dierentiable and all of its derivatives are integrable, then the Fourier transform of the function decays very quickly at innity. It is therefore not surprising that by imposing a combination of these two properties, we obtain a space of functions that is invariant under the Fourier transform: 3 Denition 36.3.A function f:R!Cis said to be a Schwartz function iffis innitely dierentiable and for all integers m;n0, we have lim x!1xm
f(n)(x) = 0. The space of all Schwartz functions on Ris denoted byS(R) (it is a vector space over C). Examples 36.4.There are essentially two main types of examples of Schwartz functions. On the one hand, C1 0(R)S(R). On the other hand, the function f(x) =ex2is also a Schwartz function (but it does not have compact support). As far as I know, most interesting examples of Schwartz functions that do not have compact support are obtained as simple variations of the example ex2. 36.6. Fourier transform of Schwartz functions. We can now prove Lemma 36.5. Iff2S(R), thenbf2S(R). Proof. The comments in x36.5 above imply that if f2 S(R), thenbf() is innitely dierentiable. They also imply that m
bf()!0 as!1 for everym0. It remains to check that the same remains true if bf() is replaced with bf(n)() for anyn1. However, this also follows by induction. Indeed, it is easy to see that if f2S(R), then x
f(x)2S(R) (and, in fact, xn
f(x)2S(R) for eachn1). We already saw that bf(n)() is the Fourier transform of ( ix)n
f(x) for eachn0, which completes the proof.  36.7. A key observation. The following observation will play an important role in the proofs of the results of this section. We claim that: (36.1)Z Rf(x)bg(x)dm(x) =Z Rg(y)bf(y)dm(y)8f;g2L1(R): We note that both sides are well dened. Indeed, we know that the function bg(x) is continuous (in particular, measurable) and bounded. Hence the product f
bgis integrable (becausefis integrable). Similarly, g
bfis integrable. Identity (36.1) is very easy to prove. First one checks, using Fubini's theorem, that the function F(x;y) =eixyf(x)g(y) is integrable with respect to the Lebesgue measure onR2. Next one uses Fubini's theorem to evaluateR R2Fdm in two dierent ways, which implies thatR R2Fdm is equal to both the left hand side and the right hand side of (36.1). A very similar argument also shows that (36.2)Z Rf(x)
(F1g)(x)dm(x) =Z Rg(y)
(F1f)(y)dm(y)8f;g2L1(R): 36.8. Fourier inversion theorem. Our next goal is to sketch a proof of the Fourier inversion theorem for L1functions. We begin with a version for Schwartz functions: Theorem 36.6. Letf2S(R). Then for every x2R, we have (36.3) f(x) =1 2Z Reixbf()dm()def= (F1bf)(x): 4 Note that by Lemma 36.5, we have bf2S(R), and, in particular, bfis integrable, so the inverse Fourier transform of bfis dened. In addition, there is no need to say \for almost everyx2R" in Theorem 36.6 because Schwartz functions are continuous. Proof of Theorem 36.6. First, using a linear change of variables in (36.1), one shows that for allf;g2L1(R) and every number >0, we have (36.4)Z Rfx 
bg(x)dm(x) =Z Rgy 
bf(y)dm(y): Next suppose that f;g2S(R). Then for a xed x2R, we havef(x=)!f(0) as !+1becausefis continuous. Moreover, fis also bounded, so there is a constant C > 0 such thatjf(x=)jCfor allx2Rand all>0. Hence the product fx 

bg(x) is bounded by C
jbg(x)j, and we already know that bgis integrable because bgis a Schwartz function. Hence the Dominated Convergence Theorem allows us to interchange the limit as! 1 with the integral on the left hand side of (36.4). Similarly, we can also interchange the limit with the integral on the right hand side of (36.4). Therefore f(0)
Z Rbg(x)dm(x) =g(0)
Z Rbf(y)dm(y)8f;g2S(R): Let us now substitute the function g(y) =ey2=2into the identity above. We saw earlier thatbg(x) =p 2
ex2=2, and henceR Rbg(x)dm(x) = 2. We conclude that f(0) =1 2Z Rbf(y)dm(y)8f2S(R) (which is in fact a special case of the Fourier inversion formula). Finally, to complete the proof, lett2Rbe arbitrary and consider the function ft(x) =f(x+t). It is clear that ft2S(R), and one checks using the change of variables x+t$xthatbft(y) =eity
bf(y). Moreover,ft(0) =f(t), so replacing fwithftin the identity above yields (36.3).  Corollary 36.7. The Fourier transform F:S(R)!S (R)is a linear isomorphism. Moreover, if f2S(R), thenF2(f)def=F(F(f)) = 2
f, where f(x) =f(x). Proof. By Theorem 36.6, we have f=F1(F(f)) for allf2S(R). Moreover,FandF1 are related to each other via ( Ff)(y) = 2(F1f)(y). It follows thatF(F(f)) = 2
f, which in turn implies that F:S(R)!S (R) is bijective, because the map f(x)7!f(x) is obviously bijective. (In fact, FFFF equals 42times the identity map.)  Theorem 36.8. Letf2L1(R)be such thatbf2L1(R)as well. Then f=F1(bf)a.e. Proof. Let us write f0=F1(bf) for the time being. We will rst check thatR Rf
hdm =R Rf0
hdm for everyh2S(R). Givenh2S(R), we know by Theorem 36.6 and Corollary 36.7 thath=bg, whereg:=F1(h)2S(R). ThereforeZ fhdm =Z fbgdm =Z gbfdm =Z (F1h)bfdm =Z F1(bf)hdm; 5 where in the second equality we used (36.1) and in the fourth one we used (36.2). So we see thatR R(ff0)hdm = 0 for every h2S(R). In particular, this holds for every h2C1 0(R). Now the lemma below shows that f=f0a.e. and completes the proof.  Lemma 36.9. Iff2L1(R)andR Rf
'dm = 0 for all'2C1 0(R), thenf= 0 a.e. A slightly more general statement is proved in the next section of these lecture notes. 36.9. Plancherel's formula. Our last topic is Plancherel's formula and the extension of the Fourier transform to the space L2(R). Let us recall (from Problem Set 7) that L2(R) is a Hilbert space with respect to the inner product f;g =Z Rf(x)
g(x)dm(x); where the bar denotes complex conjugation. Theorem 36.10 (Plancherel's formula) .Iff;g2S(R), then bf;bg = 2
f;g . Proof. Using (36.1), we obtain bf;bg =R Rbf
bgdm =R Rf
F bg
. Next, one easily checks that for any h2L1(R), we haveF(h) = 2
F1(h). SinceF1(bg) =gby Theorem 36.6, we nd that bf;bg = 2
f;g , as claimed.  In particular, we see that (2 )1=2
F:S(R)'!S (R) is an isometry with respect to theL2norm: for any f2S(R), we havebfp 2 2=jjfjj2. Moreover, by this week's homework, C1 0(R) is dense in L2(R), and hence, a fortiori ,S(R) is also dense in L2(R). By homework problem 10.1, it follows that (2 )1=2
Fextends uniquely to an isometry L2(R)'!L2(R). Thus we can now dene Fas a linear map from L2(R) toL2(R). Remarks 36.11 .(1) Earlier we had a denition of F=Foldfor functions f2L1(R). Now we have another Fourier transform, F=Fnew:L2(R)!L2(R). By construction, Fold(f) =Fnew(f) for every f2S(R). In fact, one can show that the same is true wheneverf2L1(R)\L2(R). Thus, the notation Fremains unambiguous: if F(f) can be dened for a given function fin two dierent ways, the results are the same. (2) However, there is an important dierence between FoldandFnew. For anyf2L1(R), the valuesbf() are unambiguously dened for every 2R. On the other hand, if f2L2(R), thenF(f) is only well dened as an element of L2(R). In particular, it does not make sense to speak of the value of F(f) at a given point of R, becauseF(f) is only dened up to modifying it on a set of measure zero.