### D.8 Com­plete­ness of Fourier modes

The pur­pose of this note is to show com­plete­ness of the Fourier modes

for de­scrib­ing func­tions that are pe­ri­odic of pe­riod . It is to be shown that all these func­tions can be writ­ten as com­bi­na­tions of the Fourier modes above. As­sume that is any rea­son­able smooth func­tion that re­peats it­self af­ter a dis­tance , so that . Then you can al­ways write it in the form

or

for short. Such a rep­re­sen­ta­tion of a pe­ri­odic func­tion is called a “Fourier se­ries.” The co­ef­fi­cients are called “Fourier co­ef­fi­cients.” The fac­tors 1 can be ab­sorbed in the de­f­i­n­i­tion of the Fourier co­ef­fi­cients, if you want.

Be­cause of the Euler for­mula, the set of ex­po­nen­tial Fourier modes above is com­pletely equiv­a­lent to the set of real Fourier modes

so that -​pe­ri­odic func­tions may just as well be writ­ten as

The ex­ten­sion to func­tions that are pe­ri­odic of some other pe­riod than is a triv­ial mat­ter of rescal­ing . For a pe­riod , with any half pe­riod, the ex­po­nen­tial Fourier modes take the more gen­eral form

and sim­i­larly the real ver­sion of them be­comes

See [41, p. 141] for de­tailed for­mu­lae.

Of­ten, the func­tions of in­ter­est are not pe­ri­odic, but are re­quired to be zero at the ends of the in­ter­val on which they are de­fined. Those func­tions can be han­dled too, by ex­tend­ing them to a pe­ri­odic func­tion. For ex­am­ple, if the func­tions rel­e­vant to a prob­lem are de­fined only for 0 and must sat­isfy 0, then ex­tend them to the range 0 by set­ting and take the range to be the pe­riod of a -​pe­ri­odic func­tion. It may be noted that for such a func­tion, the cosines dis­ap­pear in the real Fourier se­ries rep­re­sen­ta­tion, leav­ing only the sines. Sim­i­lar ex­ten­sions can be used for func­tions that sat­isfy sym­me­try or zero-de­riv­a­tive bound­ary con­di­tions at the ends of the in­ter­val on which they are de­fined. See again [41, p. 141] for more de­tailed for­mu­lae.

If the half pe­riod be­comes in­fi­nite, the spac­ing be­tween the dis­crete val­ues be­comes zero and the sum over dis­crete val­ues turns into an in­te­gral over con­tin­u­ous val­ues. This is ex­actly what hap­pens in quan­tum me­chan­ics for the eigen­func­tions of lin­ear mo­men­tum. The rep­re­sen­ta­tion is now no longer called a Fourier se­ries, but a “Fourier in­te­gral.” And the Fourier co­ef­fi­cients are now called the “Fourier trans­form” . The com­plete­ness of the eigen­func­tions is now called Fourier’s in­te­gral the­o­rem or in­ver­sion the­o­rem. See [41, pp. 190-191] for more.

The ba­sic com­plete­ness proof is a rather messy math­e­mat­i­cal de­riva­tion, so read the rest of this note at your own risk. The fact that the Fourier modes are or­thog­o­nal and nor­mal­ized was the sub­ject of var­i­ous ex­er­cises in chap­ter 2.6 and will be taken for granted here. See the so­lu­tion man­ual for the de­tails. What this note wants to show is that any ar­bi­trary pe­ri­odic func­tion of pe­riod that has con­tin­u­ous first and sec­ond or­der de­riv­a­tives can be writ­ten as

in other words, as a com­bi­na­tion of the set of Fourier modes.

First an ex­pres­sion for the val­ues of the Fourier co­ef­fi­cients is needed. It can be ob­tained from tak­ing the in­ner prod­uct be­tween a generic eigen­func­tion and the rep­re­sen­ta­tion for func­tion above. Not­ing that all the in­ner prod­ucts with the ex­po­nen­tials rep­re­sent­ing will be zero ex­cept the one for which , if the Fourier rep­re­sen­ta­tion is in­deed cor­rect, the co­ef­fi­cients need to have the val­ues

a re­quire­ment that was al­ready noted by Fourier. Note that and are just names for the eigen­func­tion num­ber and the in­te­gra­tion vari­able that you can change at will. There­fore, to avoid name con­flicts later, the ex­pres­sion will be reno­tated as

Now the ques­tion is: sup­pose you com­pute the Fourier co­ef­fi­cients from this ex­pres­sion, and use them to sum many terms of the in­fi­nite sum for , say from some very large neg­a­tive value for to the cor­re­spond­ing large pos­i­tive value ; in that case, is the re­sult you get, call it ,

a valid ap­prox­i­ma­tion to the true func­tion ? More specif­i­cally, if you sum more and more terms (make big­ger and big­ger), does re­pro­duce the true value of to any ar­bi­trary ac­cu­racy that you may want? If it does, then the eigen­func­tions are ca­pa­ble of re­pro­duc­ing . If the eigen­func­tions are not com­plete, a def­i­nite dif­fer­ence be­tween and will per­sist how­ever large you make . In math­e­mat­i­cal terms, the ques­tion is whether .

To find out, the trick is to sub­sti­tute the in­te­gral for the co­ef­fi­cients into the sum and then re­verse the or­der of in­te­gra­tion and sum­ma­tion to get:

The sum in the square brack­ets can be eval­u­ated, be­cause it is a geo­met­ric se­ries with start­ing value and ra­tio of terms . Us­ing the for­mula from [41, item 21.4], mul­ti­ply­ing top and bot­tom with , and clean­ing up with, what else, the Euler for­mula, the sum is found to equal

This ex­pres­sion is called the Dirich­let ker­nel. You now have

The sec­ond trick is to split the func­tion be­ing in­te­grated into the two parts and . The sum of the parts is ob­vi­ously still , but the first part has the ad­van­tage that it is con­stant dur­ing the in­te­gra­tion over and can be taken out, and the sec­ond part has the ad­van­tage that it be­comes zero at . You get

Now if you back­track what hap­pens in the triv­ial case that is just a con­stant, you find that is ex­actly equal to in that case, while the sec­ond in­te­gral above is zero. That makes the first in­te­gral above equal to one. Re­turn­ing to the case of gen­eral , since the first in­te­gral above is still one, it makes the first term in the right hand side equal to the de­sired , and the sec­ond in­te­gral is then the er­ror in .

To ma­nip­u­late this er­ror and show that it is in­deed small for large , it is con­ve­nient to re­name the -​in­de­pen­dent part of the in­te­grand to

Us­ing l’Hôpital's rule twice, it is seen that since by as­sump­tion has a con­tin­u­ous sec­ond de­riv­a­tive, has a con­tin­u­ous first de­riv­a­tive. So you can use one in­te­gra­tion by parts to get

And since the in­te­grand of the fi­nal in­te­gral is con­tin­u­ous, it is bounded. That makes the er­ror in­versely pro­por­tional to , im­ply­ing that it does in­deed be­come ar­bi­trar­ily small for large . Com­plete­ness has been proved.

It may be noted that un­der the stated con­di­tions, the con­ver­gence is uni­form; there is a guar­an­teed min­i­mum rate of con­ver­gence re­gard­less of the value of . This can be ver­i­fied from Tay­lor se­ries with re­main­der. Also, the more con­tin­u­ous de­riv­a­tives the -​pe­ri­odic func­tion has, the faster the rate of con­ver­gence, and the smaller the num­ber of terms that you need to sum to get good ac­cu­racy is likely to be. For ex­am­ple, if has three con­tin­u­ous de­riv­a­tives, you can do an­other in­te­gra­tion by parts to show that the con­ver­gence is pro­por­tional to 1 rather than just 1/. But watch the end points: if a de­riv­a­tive has dif­fer­ent val­ues at the start and end of the pe­riod, then that de­riv­a­tive is not con­tin­u­ous, it has a jump at the ends. (Such jumps can be in­cor­po­rated in the analy­sis, how­ever, and have less ef­fect than it may seem. You get a bet­ter prac­ti­cal es­ti­mate of the con­ver­gence rate by di­rectly look­ing at the in­te­gral for the Fourier co­ef­fi­cients.)

The con­di­tion for to have a con­tin­u­ous sec­ond de­riv­a­tive can be re­laxed with more work. If you are fa­mil­iar with the Lebesgue form of in­te­gra­tion, it is fairly easy to ex­tend the re­sult above to show that it suf­fices that the ab­solute in­te­gral of ex­ists, some­thing that will be true in quan­tum me­chan­ics ap­pli­ca­tions.