MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

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1 MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x, hen () g() d c+ g() d for all real numbers c. c () g( + a) d g() d. for all real numbers a. (3) g( ) d g() d. EXERCISE 6.. Verify ha he above inegraion formulas are correc. Wha can Fourier really prove? sounded. Here s a firs sep: I s no quie as wonderful as i may have THEOREM 6.. Le f be a square-inegrable, periodic funcion, suppose ha f is differeniable a a poin x. Then f(x) n f(n)e πinx, where of course f(n) c n f()e πin d. REMARK. So, Fourier s Theorem may no hold for all funcions all poins. For insance, we are assuming here ha f is in fac differeniable a he poin in quesion, clearly no every funcion is necessarily differeniable a every poin. PROOF. We know from Secion 5 ha he parial sum S N (x) n N c n e πinx of he Fourier series for f is given in an inegral form by S N (x) f()d N (x ) d, where D N is he Dirichle kernel. Now, using he inegraion formulas of Exercise 6., we see ha also Hence, S N (x) S N (x) S N (x) f(x + )D N () d f(x )D N () d. f(x + ) + f(x ) D N () d.

2 And hen, since he inegr here is an even funcion, we ge ha S N (x) f(x + ) + f(x ) D N () d (f(x + ) + f(x ))D N () d. Now, f(x) f(x) D N () d f(x)d N () d. To show ha f(x) is he sum of is Fourier series, we mus show ha f(x) is he limi of he parial sums of he Fourier series. Tha means we mus show ha lim N (f(x) S N (x)). We have f(x) S N (x) [f(x) f(x + ) f(x )]D N () d f(x) f(x + ) f(x ) sin(π) sin(π(n + )) d, so, by he Riemann-Lebesgue Lemma, lim N f(x) S N (x) will be if he funcion g defined by f(x) f(x + ) f(x ) g() sin(π) is inegrable. Now, because f is assumed o be differeniable a x, here exiss a δ > such ha he differenial quoien (f(x) f(x + ))/ is close o f (x), implying ha his quoien is bounded by a number M, for all < < δ. We will prove ha he funcion g() is inegrable by showing ha i is inegrable on he inerval [, δ] also inegrable on he inerval [δ, /]. On he inerval [δ, /], he hree erms in he numeraor of he funcion g, f(x), f(x + ), f(x ) are all inegrable funcions of, so he sum is also inegrable. And, on ha inerval [δ, /], he denominaor sin(π) is bounded away from, so ha he reciprocal / sin(π) is bounded. Hence, he enire funcion g, on he inerval [δ, /] mus be inegrable, being he produc of an inegrable funcion a bounded funcion. Now, on he inerval [, δ], wrie g() f(x) f(x + ) f(x ) f(x) f(x + ) ( + sin(π) f(x) f(x ) ) sin(π). Recall ha he funcion / sin(π) is bounded on he inerval [, δ], (Why is ha?), since boh he differenial quoiens are bounded by M on ha inerval, he funcion g mus be bounded on he inerval [, δ) hence inegrable. This complees he proof. EXERCISE 6.. (a) Le f be he periodic funcion defined by f(x) x on [, ). Compue he Fourier coefficiens c n f(n) for f, hen, for < x <, derive he formula: x πn sin(πnx). n

3 (Consul your compuaions in Exercise 4.8.) Check his equaion ou for x / for x /4. Derive he formula π 4 ( ) k k +. k (b) Now le f be he periodic funcion defined by f(x) x. Compue he Fourier coefficiens for f, derive a formula analogous o he one in par (a). (c) Le h be he periodic funcion defined by h(x) for x < /, h(x) for / x <. Compue he fourier coefficiens for h, examine he corresponding formulas. Wha happens o his series for he disconinuiy poins x / or x? REMARK. Fourier s Theorem cerainly would imply ha he Fourier ransform T has an inverse. The heorem says ha we can recover he funcion f from he ransform f. Theorem 6. esablishes Fourier s Theorem for cerain funcions, bu we don ye really know ha he Fourier ransform has an inverse. However, we can use Theorem 6. o prove his. THEOREM 6.. The Fourier ransform T is - on L ([, )). Tha is, i has an inverse. PROOF. Since T is a linear ransformaion from one vecor space ino anoher, we can prove i is - by showing ha is kernel N is rivial, i.e., he only elemen of N is he funcion. Thus, suppose f is a funcion ha saisfies T (f), le us show ha f mus be he zero funcion. We know hen ha f(n) for every n. Wrie F for he funcion defined by F (x) f() d. Then, i is known ha F is coninuous on [, ] differeniable (almos everywhere), F (x) f(x). Also, noe ha F () F () f() d f(), so ha F can be exended o a coninuous periodic funcion. Now Theorem 6. applies o he funcion F, so ha for almos every x F (x) n F (n)e πinx. Bu, by he exercise below, we know ha F (n) for all n. So, F (x) F () for all x, i.e., F is a consan funcion. Bu hen, f mus be he funcion, since i is he derivaive of he consan funcion F. This complees he proof. EXERCISE 6.3. Le f F be as in he preceding heorem. Use inegraion by pars o show ha, for all nonzero n, F (n) πin f(n). 3

4 4 Convergence of Fourier Series a Jump Disconinuiies Theorem 6. assers ha he Fourier series for a funcion f converges a each poin x where f is differeniable. Somehing analogous occurs if f is differeniable near a poin x, bu has a jump disconinuiy exacly a x. THEOREM 6.3. Suppose f is a square-inegrable, periodic funcion, suppose ha f saisfies he following addiional assumpions. () There exiss a poin x such ha boh a lef a righ limi as approaches x exis: r lim f() l lim f(). x+ x () f has boh a lef a righ derivae a he poin x; i.e., f(x + h) r lim h + h f(x + h) l lim h h exiss, exiss. Then, he Fourier series for f a x converges o he average of he righ lef limis a x : lim S N(x) f(n)e πinx r + l N. n PROOF. Jus as in he proof of Theorem 6., we have ha S N (x) (f(x + ) + f(x ))D N () d. Now, So, r r l l D N () d rd N () d D N () d ld N () d. l + r S N (x) [ l + r f(x + ) f(x )]D N () d l + r f(x + ) f(x ) sin(π) sin(π(n + )) d, so, by he Riemann-Lebesgue Lemma, (l + r)/ will equal he limi of S N (x) if he funcion g defined by g() l + r f(x + ) f(x ) sin(π)

5 is inegrable on he inerval [, ]. Now, from he assumpions abou he exisence of lef righ derivaes of f a x, we know ha here mus exis a δ > such ha he wo quoiens (r f(x+))/ (l f(x )/ are bounded by a number M for all < < δ. We will prove ha he funcion g() is inegrable by showing ha i is inegrable on he inerval [, δ] also inegrable on he inerval [δ, /]. On he inerval [δ, /], he four erms in he numeraor of g, l, r, f(x + ), f(x ) are all inegrable funcions of, so he sum is also inegrable. And, on ha inerval [δ, /], he denominaor sin(π) is bounded away from, so ha he reciprocal / sin(π) is bounded. Hence, he enire funcion g, on he inerval [δ, /] mus be inegrable, being he produc of an inegrable funcion a bounded funcion. Now, on he inerval [, δ], wrie g() l + r f(x + ) f(x ) r f(x + ) ( + l f(x ) ) sin(π) sin(π). So, since he funcion / sin(π) is bounded on he inerval [, δ], since boh of he quoiens in he above expression are bounded by M on ha inerval, he funcion g mus be bounded on ha inerval hence inegrable. This complees he proof. The Gibb s Phenomenon We begin our discussion of he Gibb s Phenomenon wih a special example. EXERCISE 6.4. Le f be he periodic funcion defined on [, ) by f() /. (a) Use he resuls from Exercise 6. o wrie down a formula for he parial sums S N of he Fourier series for f. You should ge S N (x) n (b) Jusify he following compuaions: n sin(πnx) πn n (c) Use pars (a) (b) o conclude ha S N (x) sin(πnx). nπ cos(πn) d cos(πn) d n (D N () ) d D N () d x. D N () d x, 5

6 6 where S N (x) is he Nh parial sum of he Fourier series for he funcion f(x) / x. (d) Verify ha boh he hypoheses he conclusion of Theorem 6.3 hold x for his funcion. Tha is, show ha lim N D N() d x x for every x [, ). REMARK. The Gibb s Phenomenon concerns he manner in which he parial sums S N (x) of a Fourier series converge as he poin x varies. For insance, how does he rae of convergence depend on he poin x. In he funcion of he preceding exercise, f(x) / x, we see ha he funcion values f(x) are always beween / +/. One would expec ha he values of he parial sums S N (x) would end, in he limi a leas, also o be beween / +/. The fac ha his is no uniformly he case near a jump disconinuiy poin is an example of wha s called he Gibb s Phenomenon. The nex heorem demonsraes his explicily for he funcion / x he disconinuiy poin x. Firs, some preliminary calculaions esimaes: EXERCISE 6.5. (a) Esimae he number π sin()/ d. Show ha i is approximaely.8. (b) Define he Gibb s consan G by G π π sin() d. Show ha G is approximaely.8. THEOREM 6.4. Le f be he periodic funcion of Exercise 6.5. Then lim S N ( N N + ) G G >.8, lim S N ( N N + ) G G <.8, where G is he Gibb s consan. REMARK. Wha his shows is ha no maer how large N ges, he values of he parial sums of his Fourier series are no uniformly squeezing down o he inerval [ /, /), even hough he funcion / x iself is confined o ha inerval. PROOF. We will make use of he funcion h() sin(π) π

7 7 from Secion 5. Using he resuls from Exercises , we have S N ( N + ) N+ N+ N+ + π D N () d N + sin(π(n + )) sin(π) d sin(π(n + )) d π N+ sin() π N + sin(π(n + ))h() d N + N+ d + sin(π(n + ))h() d N +. Now he firs erm is exacly equal o G/, he limis of he second hird erms are boh. (Why?) This proves he firs asserion of he heorem. The second asserion is easy now, because he parial sums S N are periodic also are odd funcions. S N ( N + ) S N( N + ) S N ( N + ), so ha lim S N( N N + ) lim S N ( N N + ) G. EXERCISE 6.6. Skech he graph of he funcion f(x) / x, hen skech he graphs of he parial sums S N (x). Convince yourself ha, as x approaches he jump disconinuiy, he graphs of he parial sums overshoo or undershoo he proper values. Here is he general saemen of he Gibb s Phenomenon. THEOREM 6.5. Suppose f is a square-inegrable, periodic funcion, suppose ha f saisfies he following addiional assumpions. () There exiss a poin x such ha boh a lef a righ limi as approaches x exis: r lim f() l lim f(). x+ x () f has boh a lef a righ derivae a he poin x; i.e., f(x + h) r lim h + h exiss, f(x + h) l lim h h exiss.

8 8 Then, he Fourier series for f a x converges o he average of he righ lef limis a x : lim S N(x) f(n)e πinx r + l N. n Moreover, here exis wo sequences {x N } {y N } converging o x for which lim S N(x N ) l + r + G l r N lim S N(y N ) l + r G l r. N