Fourier Analysis, Stein and Shakarchi Basic Properties of Fourier Series

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1 Fourier Aalysis, Stei ad Shakarchi Chapter Basic Properties of Fourier Series Yug-Hsiag Huag 8.3. Abstract Notatio: T := [ π, π]. Note that we provide a proof for Big-O Tauberia theorem for Cesáro sum i Exercise 4 which is much easier tha the oe for Abel sum. Exercises. Trivial.. Easy. 3. Easy to see the Fourier series for the iitial coditio f i plucked strig problem coverge absolutely ad uiformly by Weierstrass M-test. So the uiqueess theorem implies the series equal to f everywhere (check page 7). 4. By computig the Fourier coefficiets of the π-periodic odd fuctio defied o [, π] by f(x) = x(π x) ad usig the covergece ad uiqueess theorem, oe ca show that f(x) = 8 π k, odd si kx k 3. Note that this result ad the Parseval s idetity ca be applied to compute the Riema zeta fuctio ζ(z) at z = 6. See Exercise 3.8(b). 5. As Exercise 4, oe ca show that the tet map f(x) = ( x /δ) + o [ π, π] equals to its the Fourier series, that is, f() = δ π + cos δ πδ cos x. Departmet of Math., Natioal Taiwa Uiversity. d4@tu.edu.tw

2 6. Similarly, oe ca show the followig idetities by cosiderig the Fourier series of f(x) = x, odd = π 8 ad = π Discrete summatio by parts formula ad the proofs for Dirichlet s ad Abel s test are stadard. 8. Verify that i e ix is the Fourier series of the π-periodic sawtooth fuctio illustrated i Figure 6, defied by f() =, f(x) = π x f(x) = π x if π < x <, ad if < x < π. Note that this fuctio is ot cotiuous. Show that evertheless, the series coverges for every x (by which we mea, as usual, that the symmetric partial sums of the series coverge). I particular, the value of the series at the origi, amely, is the average of the values of f(x) as x approaches the origi from the left ad the right. Proof. It s easy to compute f() = if ad f() =. The we ca apply the Abel- Dirichlet test to show the series coverges for every x. Note that for x =, the symmetric partial sum is always zero. Remark. There are two ways to compute the limit of series, see Problem 3 ad its remark. Also take a look at Problem 3. ad its remark. 9. Let f(x) = χ [a,b] (x) be the characteristic fuctio of the iterval [a, b] [ π, π]. (a) Show that the Fourier series of f is give by f(x) b a π + e ia e ib e ix. πi (b) Show that if a π or b π ad a b, the the Fourier series does ot coverge absolutely for ay x.

3 (c) However, prove that the Fourier series coverges at every poit x. happes if a = π ad b = π? What Proof. (a) is stadard, we omit it. (b) Stadard computatios shows that e ia e ib e ix = πi π > b a si( where θ := b a < π by the assumptio. So we ca fid c > such that π θ > si c. Note that for each k N, we ca fid a k N such that k θ ((k )π+si c, kπ si c) =: I k sice if mθ ((k )π si c, (k )π + si c), the (m + )θ I k, ad if mθ (kπ si c, kπ +si c), the (m )θ I k. (This is why we take si c istead of si c i the begiig.) Also ote that () I k I j =, so k j if k j; () si k θ c for all k; (3) k kπ θ. Hece e ia e ib e ix = for all x sice πi ), > si(θ ) k= si( k θ ) k k= c πk θ = θ cπ k= k =. (c) Note that the series is e i(x b) sum is πi πi e i(x a) e i(x b). If x {a, b}, the both e i(x a) are coverget by Abel-Dirichlet Test. If x = a, the the symmetric partial < N e i(a b), which coverges as N by Abel-Dirichlet Test sice a b. Similar argumet ca be applied to the case x = b. Fially, i this case a = π ad b = π, the coefficiets i series are. ad. Whe f is smoother, ˆf decays faster. (The big-o ca be improved to little-o by Riema- Lebesgue Lemma, see Exercise 3.3).. L (T) covergece of f k implies uiform covergece of ˆf k (i.e., C(Z)). The proof for coverget series is Cesàro summable is stadard. 3. It s ituitive to see that Abel summability is stroger tha the stadard or Cesàro methods of summatio because of the regularizatio effects of multiplyig r is much stroger tha takig the average of those partial sums. The purpose of this exercise is to prove this ituitio. (a) Show that c = σ implies that c = σ (Abel). 3

4 (b) However, show that there exist series which are Abel summable, but that do ot coverge. (c) Argue similarly to prove that if a series c = σ (Cesàro), the c = σ (Abel) [Hit: Note that ad assume σ =.] c r = ( r) σ r (d) Give a example of a series that is Abel summable but ot Cesàro summable. [Hit: Try c = ( ), which is highly oscillated. Note that if {c } is Cesàro summable, the c teds to.] Proof. Although (a) is a cosequece of Exercise ad (c), we provide a direct proof here. May assume σ =. Note that for each < r < ad N N, N c r = Passig N, we have N (s s )r = ( r) N s j r j + s N r N+ j= c r = ( r) s j r j Give ɛ >. Sice σ =, the series i the right-had side coverges absolutely ad hece j= M c r = ( r) s j r j + ( r) j= s j r j, where M is chose so that s j < ɛ for all j > M. Note that the absolute value of secod term is less tha ɛr M+ < ɛ. Ad the first term is less tha ɛ provided r is close to. M+ (b) The example is = ( ) which is ot coverget but coverges to (c) May assume σ =. Note that for each < r < ad N N, N c r = N (s s )r = i Abel s sese. N (σ ( )σ + ( )σ )r N = ( r) σ r Nσ N r N+ + (Nσ N (N )σ N )r N+ Passig N, we have (by L Hôpital s rule) c r = ( r) j= jσ j r j 4

5 Give ɛ >. Sice σ =, the series i the right-had side coverges absolutely ad hece c r = ( r) M j= jσ j r j + ( r) M+ jσ j r j, where M is chose so that σ j < ɛ for all j > M. Note that the absolute value of secod term is less tha ɛr M+ < ɛ. Ad the first term is less tha ɛ provided r is close to. (d) It s stadard to show that ( ) x = x (+x). We also ote that if {c } is Cesáro summable, the we have a cotradictio that ( ) = c = σ ( )σ = σ σ + σ, as. Exercise 3 says that Coverget Cesàro summable Abel summable. 4. This exercise deals with a theorem of Tauber which says that uder a additioal coditio o c, the above arrows ca be reversed. (a) If c = σ (Cesàro) ad c = o(/), the c = σ. (b) The above statemet holds if we replace Cesàro summable by Abel summable. [Hit: Estimate the differece betwee N c ad N c r where r = N.] (c)-(d) are from Rudi [4, Exercise 3.4]. Recall that s = j= c j ad σ = s + +s. (c) Ca it lim sup N s N = but c j = (Cesàro)? (d) Prove (a) uder a weaker assumptio c = O(/). Also see Problem 4.5. [Hit: Say, c M. If m <, the For these i, s σ = m + m (σ σ m ) + m s s i ( i)m i + j=m+ ( m )M. m + (s s j ). Fix ɛ > ad associate with each the iteger m that satisfies m ɛ + ɛ < m +. The (m + )/( m) /ɛ ad s s i < Mɛ. Hece Sice ɛ is arbitrary, lim s = σ.] lim sup s σ Mɛ. (e) We will also prove (b) uder a weaker assumptio c = O(/) i Problem 3. 5

6 Proof. (a) (s σ ) = s j= s j = s j= (j (j ))s j = j= (j )s j j= js j = k= k(s k+ s k ) = k= kc k. Therefore give ɛ >, there is N such that ka k < ɛ if k > N. So there is some N (ɛ) such that wheever > N (ɛ). s σ = kc k N kc k + k= k= k=n + kc k ɛ, (b) Give ɛ >, there is N = N(ɛ) such that c ɛ wheever > N. Cosider r = as. Note that sup N c =: M <, so for sufficetly large, s A r N c j ( r) j + j= j=n+ r j c j c j ( ( )j ) + ɛ j= N c j j + ɛ ( )N+ ( ) MN + ɛ ɛ. j= j=n+ r j j (c) A example is s = if N ad s m = m wheever m N. The lim sup s =. However σ < m ( (m+)m + ) if m < m+. So σ. (d) We first prove a idetity similar to the hit: for > m, So s m σ m = σ mσ m = j=m+ ( j + )a j j= = ( m) ( j + )a j m (m j + )a j j= m a j = ( m)s m j= j=m+ ) (σ m σ m m j=m+ ( j + )a j =: I II. ( j + )a j Now we estimate II as itegral test, that is, (with sup N c =: M < ) II M m M( + ) m j=m+ m ( + j) j dt t M. ( So s m σ m σ + m σ m + M ). log m m = M( + ) m j=m+ j + Give ɛ >, by Cesáro summability, there is N ɛ such that σ σ m < ɛ if, m > N ɛ. Note that ( ɛ)ɛ m m ɛ ( + ɛ ɛ )m ( + ɛ)m 6

7 So < + = + m m+ + ( + ) < + +ɛ if m > m m m m ( ɛ)ɛ m ( ɛ)ɛ ɛ wheever m > max( ɛ, N ɛ). ad hece s m σ m ( + + ɛ ( ( ɛ)ɛ )ɛ + M ( + + ɛ ) ( ɛ)ɛ ) log( + ɛ) ɛ + ɛ + M(ɛ + + ɛ ɛ ) = ɛ + ( + M + ɛ )ɛ, Remark. Aother proof for (d), via delay meas, is give i Problem We omit the basic computatio for showig the Fejér kerel equals to F N (x) = N si (Nx/) si (x/). This result ca also be predict from the graph of F N, that is, F N = c D M D M for some c, M. I leared this perspective from [5, page 9]. 6. Prove the Weierstrauss approximatio theorem by Fejér s theorem (Corollary 5.4). Proof. Exted f from [a, b] to [a, c] cotiuously so that f(a) = f(c). So there is Q(x) = N k=m a ke ikx such that Q f < ɛ. Note that those fiitely may eikx ca be approximate by polyomial uiformly. 7. I Sectio 5.4 we proved that the Abel meas of f coverge to f at all poits of cotiuity, that is, lim r wheever f is cotiuous at θ. Ar(f)(θ) = lim (P r f)(θ) = f(θ) r I this exercise, we will study the behavior of A r (f)(θ) at certai poits of discotiuity. A itegrable fuctio is said to have a jump discotiuity at θ if the two limits exist. lim f(θ + h) = f(θ+) ad lim h + f(θ + h) = f(θ ) h (a) Prove that if f has a jump discotiuity at θ, the lim A r(f)(θ) = r f(θ+) + f(θ ). (b) Usig a similar argumet, show that if f has a jump discotiuity at θ, the Fourier series of f at θ is Cesáro summable to f(θ+)+f(θ ). 7

8 Proof. (a) Sice P r (t) = P r ( t) ad π π P r =, we have π P r = π P r = for all r <. Give ɛ >, there is δ = δ(ɛ) > such that f(θ t) f(θ ) < ɛ ad f(θ + t) f(θ+) < ɛ for all < t < δ. So π f(θ+) + f(θ ) A r (f)(θ) = f(θ t)p r (t) dt π π f(θ+) + f(θ ) = f(θ t)p r (t) dt = = ( π π π δ [f(θ + t) f(θ+)]p r (t) dt + + δ =: I + II + III + IV π )[f(θ + t) f(θ+)]p r (t) dt + ( f(θ+) + f(θ ) [f(θ t) f(θ )]P r (t) dt δ π + δ )[f(θ t) f(θ )]P r (t) dt It s easy to see II, IV < ɛ for all r <. Sice Poisso kerel o the uit disc is a good kerel, there is R = R(δ) = R(ɛ) < such that I, III < ɛ if r > R. So A r (f)(θ) f(θ+)+f(θ ) < 4ɛ if r > R(ɛ). (b) Same proof as (a) except P r is replaced by F N. Remark 3. See Problem 3 ad its remark to derive the Fourier series coverges to f(θ+)+f(θ ). 8. This is a example metioed i Remark of Theorem 5.7. If P r (θ) deotes the Poisso kerel, show that the fuctio defied for r < ad θ R, satisfies: u(r, θ) = P r θ, (i) u = i the disc. (ii) lim r u(r, θ) = for each θ. (iii) u does ot coverges to uiformly as r. However, u is ot idetically zero. Proof. Direct computatio. Note u(x, y) = y(x +y ) (( x) +y ). So u( ɛ, ɛ) as ɛ Solve Laplace s equatio u = i the semi ifiite strip S = {(x, y) : < x <, < y}, subject to the followig boudary coditios u(, y) = = u(, y) whe y, u(x, ) = f(x) whe x, 8

9 where f is a give C α fuctio (α >, see Exercise 3.6), with f() = f() =. Write f(x) = a si(πx) ad expad the geeral solutio i terms of the special solutios give by u (x, y) = e πy si(πx). Express u as a itegral ivolvig f, aalogous to the Poisso itegral formula. Proof. Note that for y > ad x [, ], the series e πy si(πx) si(πz) coverges uiformly o z [, ] by Weierstrass M-Test. So u(x, y) := = a e πy si(πx) = e πy si(πx) ( ) f(z) e πy si(πx) si(πz) dz =: First we ote that u(x, y) did solve the problem. Secod we have f(z) si(πz) dz f(z)φ y (x, z) dz Φ y (x, z) = } e {(e πy iπ(x+z) + e iπ(x+z) ) (e iπ(x z) + e iπ(x z)) ) 4 = { e + π[ y+i(x+z)] e π[ y i(x+z)] e } π[ y+i(x z)] e π[ y i(x z)] = { e πy (e iπ(x+z) + e iπ(x+z) ) e πy (e iπ(x+z) + e iπ(x+z) ) + e e πy (e iπ(x z) + e iπ(x z) ) } πy e πy (e iπ(x z) + e iπ(x z) ) + e πy e πy cos π(x z) = e πy cos π(x z) + e e πy cos π(x + z) πy e πy cos π(x + z) + e. πy Remark 4. Note that this computatio is the same as the oe for computig P r (θ) if oe use odd extesio from (, ) to (, ). Also ote there is o zero-th order ω = i this result which exists i P r, this is why the deomiator e πy cos t.. Cosider the Dirichlet problem i the aulus defied by {(r, θ) : ρ < r < }, where < ρ < is the ier radius. The problem is to solve u r + u r r + u r θ = subject to the boudary coditios u(, θ) = f(θ), u(ρ, θ) = g(θ), where f ad g are give cotiuous fuctios. 9

10 Arguig as we have previously for the Dirichlet problem i the disc, we ca hope to write u(r, θ) = c (r)e iθ with c (r) = A r + B r,. Set f(θ) a e iθ ad g(θ) b e iθ. We wat c () = a ad c (ρ) = b. This leads to the solutio u(r, θ) := ( ) [((ρ/r) (r/ρ) )a ρ ρ + (r r )b ]e iθ + a + (b a ) log r log ρ. Show that as a result we have u(r, θ) (P r f)(θ) as r uiformly i θ, ad u(r, θ) (P ρ/r g)(θ) as r ρ uiformly i θ. Proof. First we cofirm the absolute covergece of the series u (r, θ) := ( ( ρ r ) ( ρ ) r ) a ρ ρ e iθ ad u ρ (r, θ) := ( r r ) b ρ ρ e iθ so that u =: u + u ρ + u is cofirmed to be well-defied. This is a cosequece that ad for each ( ρ r ) ( ρ r ) ρ ρ ( ( ρ = r r ) ρ { } sup a, b ( ) π max f L (T), g L (T) ) = r ( ρ )( r + ( ρ ρ r ) + + ( ρ ) r ) ( ρ r ) r r () r r ( ρ ) ( r ) ( ρ ) ρ ρ = ( r )( + r + r r ) ρ ) r = ( r ρ r ρ r ρ. ()

11 Note that the above method to derive the estimates also imply u is harmoic o the aual. We omit the details. To check it did satisfy the boudary coditio, it s eough to show the rest assertios. To arrive this, we first show u ρ ad u + u P r f as r uiformly i θ. Note that the u ρ part is a cosequece of (). For δ < r < ad, ( ρ r ) ( ρ r ) r = r ρ [r ] ρ ρ ρ ρ r ρ ρ ( δ). = r ρ ρ ( r )[ + r + r 4 + r ] I particular we pick δ > ρ so that the majorat series of u + u P r f coverges (by ratio test) ad hece bouded by a multiple (idepedet of θ) of r. This proves that u + u P r f as r uiformly i θ. Fially, we show that u ad u ρ + u P ρ/r f as r ρ uiformly i θ. Note that the u part is a cosequece of (). For ρ < r < ad, r r ( ρ ) ρ ρ (ρ r ρ r ) = = ρ r [ ρ ( r ρ ρ r ) + ( ρ r ) + ( ρ r )4 + ( ρ ] r ) ρ r ρ ρ. So the majorat series of u + u P ρ/r f coverges (by ratio test) ad hece bouded by a multiple (idepedet of θ) of ρ r. This proves that u + u P ρ/r f as r ρ uiformly i θ. Problems. Oe ca costruct Riema itegrable fuctios o [, ] that have a dese set of discotiuities as follows. (a) Let f(x) = whe x < ad f(x) = if x. sequece {r } i [, ]. The, show that the fuctio Choose a coutable dese F (x) = f(x r ) is itegrable ad discotiuous precisely at each r. [Hit: F is mootoic ad bouded.]

12 (b) Cosider ext F (x) = 3 g(x r ) where g(x) = si(/x) whe x, ad g() =. The F is itegrable, discotiuous precisely at each x = r, ad fails to be mootoic i ay subiterval of [, ]. [Hit: Use the fact that 3 k > >k 3.] (c) The origial example of Riema is the fuctio F (x) = where (x) = x for x ( /, /] ad exteded to R by periodicity, that is, (x + ) = (x). It ca be show that F is discotiuous wheever x D := { k+ k, m Z, m } m ad cotiuous at R D. Moreover, F is Riema-itegrable o every bouded iterval. (d) If the above (x) is replace by {x}, where {x} is the fractioal part of x (that is, {z} := z if z [, ) ad the exted periodically), the we have F is discotiuous precisely at Q. (x) Remark 5. Are the fuctios F i (c)(d) mootoe? Proof. We may assume all r are distict (why?) (a) For every x [, ], the series coverges ad is less tha <, so F is well-defied ad bouded. The itegrability ca be easily proved from the obvious mootoicity of F. The cotiuity o [, ] \ {r } N ad discotiuity o {r } N ca be proved similarly to the cases for the popcor fuctio (ote that F (r +) F (r ) ). Remark 6. Aother proof for discotiuity at r is a cotradictio proof, that is, suppose F is cotiuous at r, the with the advatage of uiform covergece we kow the fuctio F = j f(x r j j ) is cotiuous at r. So f(x r ) = F (x) F (x) is cotiuous at r, which is a cotradictio. This method ca be applied to (b) too. (b) Sice the Riema itegrability, cotiuity ad discotiuity of F ca be proved as (a), we oly prove the last assertio here. Give I [, ], by the desity of {r j } j there is some r k ad a iterval I k such that r k I k I k ad r, r,, r k I k. So F k(x) = k 3 g(x r ) is cotiuous o I k, ad hece F k 3 k i some small eighboorhood J k of r k.

13 Note that 3 k = j=k+ 3 j ad F (x) = F k (x) + 3 k g(x r k ) + j=k+ 3 j g(x r j ). So for ay eighborhood J J k of r k, we ca fid z < w < z such that g(z r k ) = = g(z r k ) ad = g(w r k ) so F (z) 3 k + 3 k j=k+ 3 j > 3 k 3 k + j=k+ 3 j F (w) ad similarly F (z ) > F (w). We prove (d) first to explai our idea clearly. Sice F N (x) = N {x} is Riema itegrable for each N N ad coverges to F (x) uiformly o every bouded iterval (by Weierstrass M-test), F is Riema-itegrable o every bouded iterval. R Q will be a cosequece of the cotiuity of F N ad hece of {x} which ca be easily proved. Also, the cotiuity of F o o R Q for each N Now we examie that F is discotiuous at ay z Q, say z = p, where gcd(p, q) =. Note q that the covergece of the series F (x) ad f r (x) := {(kq+r)x} k= (kq+r) q F (x) = f r (x). r= for each r q imply We are doe if we prove f r is cotiuous at p q if r < q ad f q has a discotiuity at p q. For every x ( p q 4q, p q ), that is, qx (p 4, p), we have {qx} > 3 4 ad hece f q (x) = j= Therefore f q is discotiuous at p q. {jqx} j q > 3 4q q j= j = q (7 4 π 6 ) > = f q( p q ). For r < q, oe otes that for each k the map x {(kq + r)x} is cotiuous at p q sice (kq + r) p q = kp + rp q Z. Hece f r is cotiuous at p q by the uiform covergece of its series. Fially, i (c), we chage the otatios ( ) to [ ] ad F (x) to G(x) for otatioal coveiece. For G(x) := [x] the Riema-itegrability ad cotiuity are proved as (d). Now we examie that G is discotiuous at ay z D, say z = p, where gcd(p, q) =, p is q odd. Note that the absolute covergece of the series G(x) ad g r (x) := [(kq+r)x] k= (kq+r) q G(x) = g r (x). We are doe if we prove g r is cotiuous at p q if r < q ad g q has a discotiuity at p q. For every x ( p, + p ), that is, qx ( p, p+ ), we have [qx] < sice p is odd. Hece q q q g q (x) = j= [jqx] j q < q j= r= j = q (π 6 ) < q (π 6 π 6 ) = q ( 3 j= j j= imply (j) ) = g q( p q ).

14 Therefore g q is discotiuous at p q. For r < q, oe otes that for each k the map x [(kq + r)x] is cotiuous at p q sice (kq +r) p Z+ (if (kq +r) p Z+ rp, the kp+ q q q Hece g r is cotiuous at p q = (kq +r) p q by the uiform covergece of its series. Z which is a cotradictio). Remark 7. Aother costructio us the so-called popcor fuctio. More iterestigly, there is o fuctios which are precisely cotiuous at the ratioal. Oe ca prove this by usig Baire Category Theory, see Exercise 4.6 of Book IV.. Let D N deote the Dirichlet kerel D N (θ) = N k= N e ikθ = si((n + )θ, si(θ/) ad defie (a) Prove that L N = π D N (θ) dθ. π π L N c log N for some costat c >. A more careful estimate gives L N = 4 log N + O(). π (b) Prove the followig as a cosequece: for each, there exists a cotiuous fuctio f such that f ad S (f )() c log. Proof. (a) π π L N = si(n + )θ dθ = si(n + )θ dθ + si(n + ( π si(θ/) π θ/ π )θ si(θ/) ) dθ θ/ = (N+ )π si z dz + O() = N (j+ )π ( si z π z π (j )π jπ + z ) dz + O() jπ = 4 π N j= j= j + O() = 4 log N + O(). π (b) Sice D has fiite zeros, F (t) := sgd (t) has fiitely may discotiuities. Note that S (F )() = L ad we ca approximate F by cotiuous fuctio i L orm sice [, ] has fiite Lebesgue measure. The desired result follows by otig D () for all N so D have a uiform boud i every eighborhood of discotiuities of every F. π 4

15 Remark 8. I Book IV, Chapter 4, oe ca combie this problem with Uiform Boudedess Priciple to show the existece of cotiuous fuctio whose Fourier series diverges at a give poit, a result will also be proved by a explicit costructio of P. du Bois-Reymod (873) i Sectio 3... O the other had, oe ca combie this with Ope Mappig Theorem to show the mappig f L [ π, π] { ˆf()} C (Z) is ot surjective. 3. Littlewood provided a refiemet of Tauber s theorem: (a) If c is Abel summable to s ad c M for all, the c coverges to s. (b) As a cosequece of Exercise 3(c), we kow that if c is Cesàro summable to s ad c M for all, the c coverges to s. Note that aother easier proof is also give i Exercise 4(c). These results may be applied to Fourier series. By Exercise 7, they imply that if f is a itegrable fuctio that satisfies ˆf(ν) = O(/ ν ), the: (i) If f is cotiuous at θ, the S N (f)(θ) f(θ) as N (ii) If f has a jump discotiuity at θ, the S N (f)(θ) f(θ+ ) + f(θ ) as N (iii) If f is cotiuous o [ π, π], the S N (f) f uiformly. Other proofs for the simpler assertio (b), hece proofs of (i),(ii),ad (iii), ca be foud i my Exercise.4 ad Problem 4.5 Remark 9. The proof by Wieladt s method is take from [3, Sectio.], I thik this book is a great survey of Tauber s theorem. Remark. Aother similar result (weaker hypothesis ad weaker coclusio) is the Jorda s test: lim N S N (f)(θ ) = f(θ +)+f(θ ) for f is oly mootoe (ad hece for f is of bouded variatio) o a eighborhood of θ, which ca be proved by Riema-Lebesgue lemma ad secod mea-value theorem. See [, Theorem..]. Note that the cotrol of Dirichlet kerel is much difficult tha the Fejér kerel, cf. [, Theorem 3.4.]. Proof. For every polyomial P (x) = m k= b kx k, m m c P (x ) = b k x k = = = c k= k= b k = c x k m b k s = P ()s (3) k= 5

16 as x. (The reader should be able to prove the secod idetity i (3) by ɛ N argumet.) Defie g(x) = χ [e,](x), that is, the characteristic fuctio of [e, ] so that for s N = N c = = c g(e /N ); To complete the proof that s N s, we will show below that the coclusio i (3) is also valid with g istead of P : s lim if x c g(x ) lim sup c g(x ) s. x To the last iequality of (3), we look for a suitable polyomial majorat Equivaletly, P g for which P () = g() =, P () = g() =. P (t) t t( t) must be a polyomial Q(t) h(t) = g(t) t t( t). (4) The piecewise cotiuous fuctio h o [, ] looks like the followig graph. For < x < it follows from (4) ad the coditio c M that c g(x ) c P (x ) = c {P (x ) g(x )} M {P (x ) g(x )} = = x M x {P (x ) g(x )} = M( x) φ(x P (t) g(t) ), where φ(t) =. (5) t By (3), the secod term i the first member of (5) has limit s as x. Note that the fial member i (5) has limit (6) sice it ca serves as a Riema sum of that itegral uder the partitio P x = {, x, x, }. (The itegrability of φ(t) t Q(t) h(t). O the other had, the detail for reducig P x is easy to see from ito a fiite set as a regular partitio is left to the reader, this is ot trivial, you eed to observe some uiformity or mootoicity.) M φ(t) dt t = M P (t) t (g(t) t) t( t) dt = M {Q(t) h(t)} dt. (6) Now for give ɛ >, Weierstrass approximatio theorem (WAT) makes it possible to costruct a polyomial Q h such that (Q h) ɛ. A costructio is give i the figure below. (Note that the graph is ot accurate due to the author s poor drawig techique ad softwares) 6

17 For such a Q ad the correspodig P determied by (4), it follows from (5), (3) ad (6) that lim sup x = c g(x ) lim x c P (x ) + lim M(l x) φ(x ) x = = P ()s + M Sice ɛ was arbitrary, our lim sup does ot exceed s. = {φ(t)/t} dt s + Mɛ. To the first iequality of (3), the method is the same with lookig some polyomial P g, P () = g() = ad P () = g() =. We omit it. Refereces [] Duoadikoetxea: Fourier Aalysis. Vol. 9. America Mathematical Soc., [] Grafakos, Loukas: Classical Fourier Aalysis. 3rd ed., Spriger, 4. [3] Korevaar, Jacob: Tauberia theory: A cetury of developmets. Vol. 39. Spriger, 4. [4] Rudi, Walter: Priciple of Mathematical Aalysis, 3rd editio, 976. [5] Muscalu, Camil, ad Wilhelm Schlag: Classical ad multiliear harmoic aalysis. Vol.. Cambridge Uiversity Press, 3. 7