Chapter 1. Fourier Series. 1.1 Introduction.

Transcription

1 Chapter Fourier Series. Introduction. We will consider comple valued periodic functions on R with period. We can view them as functions defined on the circumference of the unit circle in the comple plane or equivalently as function f defined on [,π] with f() = f(π). In that case, smoothness requires matching the derivatives as well at ±π. he Fourier Coefficcients of a periodic function f L [,π] are defined by a n = π f()e in d = f()e in d (.) If a function f has the representation as a Fourier Series f() a n e in (.) <n< with <n< a n <, since e im e in d = δ n,m we see that the coefficients a n can be recovered from f by formula (.). If we assume that f L [] then clearly a n is well defined and a n f() d

2 CHAPER. FOURIER SERIES Itisnotclear thattheseriesonrighthandsideofequation(.)converges and even if it does, it is not clear that the sum of the series is actually equal tothethefunctionf(). It isrelatively easyto findconditionsonf( )so that the series (.) is convergent. If f() is assumed to be k times continuously differentiable on, integrating by parts k times one gets, for n, a n = n k f (k) ()e in d n sup f {k} () (.3) k From the estimate (.3) it is easily seen that the series is convergent if f is twice continuously differentiable. An important but elementary fact is the Riemann-Lebesgue theorem. heorem. For every f L [], lim a n = (.4) n ± Proof. Let f L [] and ǫ > be given. Since smooth functions are dense in L, given ǫ >, we can approimate f by a function g ǫ such that f() g ǫ () d ǫ and g ǫ is continuously differentiable. hen a n g ǫ e in d + f() g ǫ () d g ǫ n () d+ f() g ǫ () d n sup g ǫ () +ǫ and limsup n ± a n ǫ. Since ǫ > is arbitrary limsup n ± a n =.. Convergence of Fourier Series. Let us define the partial sums (s N f)() = s N (f,) = a n e in (.5) n N

3 .. CONVERGENCE OF FOURIER SERIES. 3 and the Fejér sum We can calculate where (S N f)() = S N (f,) = N + (s n f)() = e ij j n e ijy f(y)dy n N s n (f,) (.6) = f(y)[ e ij( y) ]dy j n = f(y) e in( y) (e i(n+)( y) ) dy e i( y) = f(y)k n ( y)dy (.7) = (f k n )() (.8) k n (z) = e inz (e i(n+)z ) e iz = sin(n+ )z sin z (.9) and the convolution f g of two functions f,g in L [] is defined as (f g)() = f(y)g( y)dy = f( y)g(y)dy (.) where A similar calculation reveals (S N f)() = f(y)k N ( y)dy = (f K N )() (.) K N (z) = (N +) sin z = (N +) n N [ sin (N+)z sin z [sin(n+ )z] ] (.)

4 4 CHAPER. FOURIER SERIES Since k n (z) = a j e ij j n K N (z) = N k n (z) N + = ( n N + )a ne in n N Notice that for every N, k N (z)dz = K N (z)dz = (.3) he following observations are now easy to make.. Nonnegativity. K N (z). For any δ >, lim sup K N (z) = z δ 3. herefore It is now an easy eercise to prove lim K N (z)dz = z δ heorem. For any f that is bounded and continuous on lim sup S N (f,) f() = Proof. Let δ > be given. hen S N (f,) f() = [f( z) f()]k N (z)dz f( z) f() K N (z)dz + f( z) f() K N (z)dz z δ z δ sup sup f( z) f() +sup f() K N (z)dz z δ z δ

5 .. CONVERGENCE OF FOURIER SERIES. 5 If we let N and then δ as δ. lim sup sup S N (f,) f() sup sup f( z) f() z δ heorem 3. For p < and f L p [,π] and therefore S N (f, ) p f p lim S Nf f p = Proof. By Hölder s inequality for any S N (f,) p f(z) p K N ( z)dz and integrating with respect to we obtain the first part of the theorem. For any f L p and ǫ > we can find g such that it is continuous and f g p ǫ. S N f f p S N f S N g p + S N g g p + g f p S N g g p +ǫ S N g g p sup S N (g,) g() asn andǫ > isarbitrary. hebehaviorofs N (f,)ismorecomplicated. Itiseasyenoughtoobserve that for f C (), lim sup s N (f,) f() = heseriesconvergesandsos N (f, )hasauniformlimitg. hecesàroaverage S N (f, ) has the same limit, which has just been shown to be f. herefore f = g. he following theorem is fairly easy. heorem 4. If f L satisfies at some, f(y) f() c y α for some α > and c <, then at that, lim s N(f,) = f()

6 6 CHAPER. FOURIER SERIES Proof. Wecanassume without lossofgenerality that = andlet f() = a. We need to show that lim f(y) sin(n + )y sin y dy = a (.4) Because f(y) a is integrable, (.4) is a consequence of the Riemann-Lebesgue sin y heorem, i.e. heorem. Let us now assume that f is a function of bounded variation on which has left and right limits a l and a r at. It is easy to check that sin( y ) y C y and it follows from Riemann-Lebesgue theorem that π [ lim f(y)sinλy λ sin( y ) ] ( y ) dy = By change of variables one can reduce the calculation of to calculating If we denote by lim then G( ) = and G() = π. a r (λ) = π λπ = a r + π a r λπ lim λ π λπ G(y) = f(y) sin(n + )y sin y dy f ( y)siny λ y dy f ( y)siny λ y dy = π λπ y sin d λπ G(y)df ( y) = λ a r + π f ( y) dg(y) λ π G(λy)df(y) by the bounded convergence theorem. his establishes the following

7 .3. SPECIAL CASE P = 7 heorem 5. If f is of bounded variation on, for every, lim s N(f,) = f(+)+f( ) On the other hand the behavior of s N (f,) for f in C(), the space of continuous functions on or in L p [,π] for p < is more comple. For eample one can ask. Does s N (f,) f() in L p? How about for almost all? Let us define the linear operator ( λ f)() = π f(+y) sinλy sin y dy (.5) on smooth functions f. It is more convenient to think of f as a periodic function of period defined on R. If s N (f,) were to converge uniformly to f for every bounded continuous function it would follow by the uniform boundedness principle that sup ( λ f)() Csup f() with a constant independent of f as well as λ, at least for λ = N + where N is a positive integer. Let us show that this is false. he best possible bound C = C λ is seen to be C λ = π sinλy sin y dy and because is sin y y integrable on [,π], C λ differs from π sinλy dy = y π λπ λπ siny dy y siny by a uniformly bounded amount. he divergence of dy implies y that C λ as λ. By duality this means that λ f is not uniformly bounded as an operator from L [,π] into itself either. Again because of uniform boundedness principle one cannot epect that s N (f, ) tends to f( ) in L [,π] for every f L [,π]..3 Special case p = When p = we have a Hilbert Space L () with the inner product f,g = f gd

8 8 CHAPER. FOURIER SERIES and f = f() d We have taken the normalized Lebesgue measure dµ = d so that µ() =. he functions {e n ( ) = e in : n Z} form a complete orthonormal basis and the Fourier series is the epansion f() = n= which converges in L () with a n given by a n e in a n = e n,f = e in f()d he Plancherel-Perseval identities state n= a n = f() d and n= ā n b n = f()g()d.4 Higher dimensions. If we have periodic functions f() = f(,..., d ) of d variables with period in every variable then the Fourier transforms are defined on Z d. If n = (n,...,n d ) then a n = e i n, d () d d While most of the one dimensional results carry over to d dimensions, one needs to be careful about the partial sums. Results that depend on the eplicit form of the kernels k N and K N have to reeamined. While partial sumsoversetsoftheform i { n i N}oreven i { n i N i }canbehandled, it is hard to analyze partial sums over sets of the form {n : i n i N}.

9 .5. SINGULAR INEGRALS. 9.5 Singular Integrals. We start with a useful covering lemma known as Vitali covering lemma. Lemma 6. Let K S be a compact subset of R and {I α } be a collection of intervals covering K. hen there is a finite sub-collection {I j } such that. {I j } are disjoint.. he intervals {3I j } that have the same midpoints as {I j } but three times the lenghth cover K. Proof. We first choose a finite subcover. From the finite subcover we pick the largest interval. In case of a tie pick any of the competing ones. hen, at any stage, of the remaining intervals from our finite subcollection we pick the largest one that is disjoint from the ones already picked. We stop when we cannot pick any more. he collection that we end up with is clearly disjoint and finite. Let K. his is covered by one of the intervals I from our finite subcollection covering K. If I was picked there is nothing to prove. If I was not picked it must intersect some I j already picked. Let us look at the first such interval and call it I. I is disjoint from all the previously picked ones and I was passed over when we picked I. herefore in addition to intersecting I, I is not larger than I. herefore 3I I. he lemma is used in proving maimal inequalities. For instance, for the Hardy-Littlewood maimal function we have heorem 7. Let f L (). Define M f () = sup <r< π r y <r f(y) dy (.6) µ[ : M f () > l] 3 f(y) dy l Proof. Let us denote by E l the set (.7) E l = { : M f () > l} and let K E l be an arbitrary compact set. For each K there is an interval I such that I f(y) dy lµ(i )

10 CHAPER. FOURIER SERIES Clearly {I } is a covering of K and by lemma we get a finite disjoint sub collection {I j } such that {3I j } covers K. Adding them up f(y) dy l µ(i j ) = l µ(3i j ) l 3 3 µ(k) j Sine K E l is arbitrary we are done. here is no problem in replacing { : M f () > l} by { : M f () l}. Replace l by l ǫ and let ǫ. his theorem can be used to prove the Labesgue diffrentiability theorem. heorem 8. For any f L (S), lim f(y) f() dy = a.e. (.8) h h y h Proof. It is sufficient to prove that for any δ > µ[ : limsup f(y) f() dy δ] = h h y h Given ǫ > we can write f = f +g with f continuous and g ǫ and µ[ : limsup h f(y) f() dy δ] h y h = µ[ : limsup h µ[ : sup h> 3 g δ h 3ǫ δ Since ǫ > is arbitrary we are done. h j y h y h g(y) g() dy δ] g(y) g() dy δ] In other words the maimal inequality is useful to prove almost sure convergence. ypically almost sure convergence will be obvious for a dense set and the maimal inequality will be used to interchange limits in the approimation.

11 .5. SINGULAR INEGRALS. Another summability method, like the Fejer sum that is often considered is the Poisson sum S(ρ,) = n a n ρ n e in and the kernel corresponding to it is the Poisson kernel p(ρ,z) = n ρ n e inz = ρ ( ρcosz +ρ ) (.9) so that P(ρ,) = f(y)p(ρ, y)dy It is left as an eercise to prove that for for p <, every f L p P(ρ, ) f( ) in L p as ρ. We will prove a maimal inequality for the Poissonsum, sothatasaconsequence wewillgetthealmostsureconvergence of P(ρ,) to f for every f in L. heorem 9. For every f in L µ[ : sup P(ρ,) l] C f ρ< l (.) Proof. he proof consists of estimating the Poisson maimal function in terms of the Hardy-Littlewood maimal function M f (). We begin with some simple estimates for the Poisson kernel p(ρ, z). p(ρ,z) = ρ ( ρ) +ρ( cosz) +ρ ρ π ρ = ρ ( ρ) he problem therefore is only as ρ. Lets us assume that ρ. Lemma. For any symmetric function φ(z) the integral

12 CHAPER. FOURIER SERIES π f(z)φ(z)dz = = π π π π M f () [f(z)+f( z)]φ(z)dz φ(z)[ d dz φ (z)[ z z z z zφ (z) z [ π For the Poisson kernel f(y)dy]dz f(y)dy]dz + φ(π) z π f(y) dy]dz + φ(π) f(z)dz π z zφ (z) dz + φ(π) M f () f(z) dz and z d dz p(ρ,z) = ρ ( ρcosz +ρ ) ρ zsinz ( ρ)z π( ρ) 4 +( cosz) C ( ρ)z ( ρ) 4 +z 4 π uniformly in ρ. z d π dz p(ρ,z) dz C = π ρ π ρ ( ρ)z ( ρ) 4 +z 4dz z +z 4dz z +z 4dz C

13 .6. EXERCISES. 3.6 Eercises.. Instead of Fejér sum if we use (W N f)() = a n w(n,n)e in with w(n,n) as N what simple additional conditions will ensure the convergence of W N f to f? In L [] or L []. What about w(n,n) = e n N? 3. What about w(n,n) = e n N? 4. What about w(n,n) = e n N e n N 5. Can you formulate higher dimensional analogs? 6. Use lemma to complete the proof of heorem For an function f defined on, the harmonic etension inside the circle is given by U(f,r,α) = f(θ) ( rcos(θ α)+r ) dθ Show that for f L [] for almost all α limu(f,r,α) = f(α) r