D.8 Com­plete­ness of Fourier modes

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

$$\ldots, \frac{e^{-3{\rm i}x}}{\sqrt{2\pi}}, \frac{e^{-2{\... ...}x}}{\sqrt{2\pi}}, \frac{e^{3{\rm i}x}}{\sqrt{2\pi}}, \ldots$$

for de­scrib­ing func­tions that are pe­ri­odic of pe­riod $2\pi$. It is to be shown that all these func­tions can be writ­ten as com­bi­na­tions of the Fourier modes above. As­sume that $f(x)$ is any rea­son­able smooth func­tion that re­peats it­self af­ter a dis­tance $2\pi$, so that $f(x+2\pi)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(x)$. Then you can al­ways write it in the form

$$f(x) = \ldots + c_{-2} \frac{e^{-2{\rm i}x}}{\sqrt{2\pi}} ... ...sqrt{2\pi}} + c_3 \frac{e^{3{\rm i}x}}{\sqrt{2\pi}} + \ldots$$

or

$$f(x) = \sum_{k=-\infty}^\infty c_k \frac{e^{k{\rm i}x}}{\sqrt{2\pi}}$$

for short. Such a rep­re­sen­ta­tion of a pe­ri­odic func­tion is called a “Fourier se­ries.” The co­ef­fi­cients $c_k$ are called “Fourier co­ef­fi­cients.” The fac­tors 1$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ 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

$$\frac{1}{\sqrt{2\pi}}, \frac{\cos(x)}{\sqrt{\pi}}, \frac{\... ...ac{\cos(3x)}{\sqrt{\pi}}, \frac{\sin(3x)}{\sqrt{\pi}}, \ldots$$

so that $2\pi$-​pe­ri­odic func­tions may just as well be writ­ten as

$$f(x) = a_0 \frac{1}{\sqrt{2\pi}} + \sum_{k=1}^\infty a_k \... ...rt{\pi}} + \sum_{k=1}^\infty b_k \frac{\sin(kx)}{\sqrt{\pi}}.$$

The ex­ten­sion to func­tions that are pe­ri­odic of some other pe­riod than $2\pi$ is a triv­ial mat­ter of rescal­ing $x$. For a pe­riod $2\ell$, with $\ell$ any half pe­riod, the ex­po­nen­tial Fourier modes take the more gen­eral form

$$\ldots, \frac{e^{-k_2{\rm i}x}}{\sqrt{2\ell}}, \frac{e^{-... ... k_2 = \frac{2\pi}{\ell},\; k_3 = \frac{3\pi}{\ell}, \ldots$$

and sim­i­larly the real ver­sion of them be­comes

$$\frac{1}{\sqrt{2\ell}}, \frac{\cos(k_1x)}{\sqrt{\ell}}, \f... ...s(k_3x)}{\sqrt{\ell}}, \frac{\sin(k_3x)}{\sqrt{\ell}}, \ldots$$

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 $f(x)$ rel­e­vant to a prob­lem are de­fined only for 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\ell$ and must sat­isfy $f(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(\ell)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, then ex­tend them to the range $\vphantom{0}\raisebox{1.5pt}{$-$}$$\ell$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ 0 by set­ting $f(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-f(-x)$ and take the range $\vphantom{0}\raisebox{1.5pt}{$-$}$$\ell$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\ell$ to be the pe­riod of a $2\ell$-​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 $\ell$ be­comes in­fi­nite, the spac­ing be­tween the dis­crete $k$ val­ues be­comes zero and the sum over dis­crete $k$ val­ues turns into an in­te­gral over con­tin­u­ous $k$ 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 $c_k$ are now called the “Fourier trans­form” $F(k)$. 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 $f$ of pe­riod $2\pi$ that has con­tin­u­ous first and sec­ond or­der de­riv­a­tives can be writ­ten as

$$f(x) = \sum_{k=-\infty}^{k=\infty} c_k \frac{e^{k{\rm i}x}}{\sqrt{2\pi}},$$

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

First an ex­pres­sion for the val­ues of the Fourier co­ef­fi­cients $c_k$ is needed. It can be ob­tained from tak­ing the in­ner prod­uct $\big\langle{e}^{l{\rm i}{x}}/\sqrt{2\pi}\big\vert f(x)\big\rangle$ be­tween a generic eigen­func­tion $e^{l{\rm i}{x}}$$\raisebox{.5pt}{$/$}$​$\sqrt{2\pi}$ and the rep­re­sen­ta­tion for func­tion $f(x)$ above. Not­ing that all the in­ner prod­ucts with the ex­po­nen­tials rep­re­sent­ing $f(x)$ will be zero ex­cept the one for which $k$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$, if the Fourier rep­re­sen­ta­tion is in­deed cor­rect, the co­ef­fi­cients need to have the val­ues

$$c_l = \int_{x=0}^{2\pi} \frac{e^{-l{\rm i}x}}{\sqrt{2\pi}} f(x){ \rm d}x,$$

a re­quire­ment that was al­ready noted by Fourier. Note that $l$ and $x$ 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

$$c_k = \int_{\bar x=0}^{2\pi} \frac{e^{-k{\rm i}\bar x}}{\sqrt{2\pi}} f(\bar x){ \rm d}\bar x,$$

Now the ques­tion is: sup­pose you com­pute the Fourier co­ef­fi­cients $c_k$ from this ex­pres­sion, and use them to sum many terms of the in­fi­nite sum for $f(x)$, say from some very large neg­a­tive value $\vphantom{0}\raisebox{1.5pt}{$-$}$$K$ for $k$ to the cor­re­spond­ing large pos­i­tive value $K$; in that case, is the re­sult you get, call it $f_K(x)$,

$$f_K(x) \equiv \sum_{k=-K}^{k=K} c_k \frac{e^{k{\rm i}x}}{\sqrt{2\pi}},$$

a valid ap­prox­i­ma­tion to the true func­tion $f(x)$? More specif­i­cally, if you sum more and more terms (make $K$ big­ger and big­ger), does $f_K(x)$ re­pro­duce the true value of $f(x)$ to any ar­bi­trary ac­cu­racy that you may want? If it does, then the eigen­func­tions are ca­pa­ble of re­pro­duc­ing $f(x)$. If the eigen­func­tions are not com­plete, a def­i­nite dif­fer­ence be­tween $f_K(x)$ and $f(x)$ will per­sist how­ever large you make $K$. In math­e­mat­i­cal terms, the ques­tion is whether $\lim_{K\to\infty}f_K(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(x)$.

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

$$f_K(x) = \frac{1}{2\pi} \int_{\bar x=0}^{2\pi} f(\bar x) \... ...m_{k=-K}^{k=K} e^{k{\rm i}(x-\bar x)}\right] { \rm d}\bar x.$$

The sum in the square brack­ets can be eval­u­ated, be­cause it is a geo­met­ric se­ries with start­ing value $e^{-K{\rm i}(x-\bar{x})}$ and ra­tio of terms $e^{{\rm i}(x-\bar{x})}$. Us­ing the for­mula from [41, item 21.4], mul­ti­ply­ing top and bot­tom with $e^{-{\rm i}(x-\bar{x})/2}$, and clean­ing up with, what else, the Euler for­mula, the sum is found to equal

$$\frac{\sin\Big((K+\frac12)(x-\bar x)\Big)} {\sin\Big(\frac12(x-\bar x)\Big)}.$$

This ex­pres­sion is called the Dirich­let ker­nel. You now have

$$f_K(x) = \int_{\bar x=0}^{2\pi} f(\bar x) \frac{\sin\Big((... ...Big)} {2\pi\sin\Big(\frac12(x-\bar x)\Big)} { \rm d}\bar x.$$

The sec­ond trick is to split the func­tion $f(\bar{x})$ be­ing in­te­grated into the two parts $f(x)$ and $f(\bar{x})-f(x)$. The sum of the parts is ob­vi­ously still $f(\bar{x})$, but the first part has the ad­van­tage that it is con­stant dur­ing the in­te­gra­tion over $\bar{x}$ and can be taken out, and the sec­ond part has the ad­van­tage that it be­comes zero at $\bar{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$. You get

$$\begin{eqnarray*} \lefteqn{f_K(x) = f(x) \int_{\bar x=0}^{2\pi} \frac{\sin\Bi... ...)\Big)} {2\pi\sin\Big(\frac12(x-\bar x)\Big)} { \rm d}\bar x. \end{eqnarray*}$$

Now if you back­track what hap­pens in the triv­ial case that $f(x)$ is just a con­stant, you find that $f_K(x)$ is ex­actly equal to $f(x)$ 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 $f(x)$, since the first in­te­gral above is still one, it makes the first term in the right hand side equal to the de­sired $f(x)$, and the sec­ond in­te­gral is then the er­ror in $f_K(x)$.

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

$$g(\bar x) = \frac{f(\bar x)-f(x)}{2\pi\sin\Big(\frac12(x-\bar x)\Big)}$$

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

$$f_K(x) = f(x) + \frac{1}{K+\frac12} \int_{\bar x=0}^{2\pi}... ...g((K+{\textstyle\frac{1}{2}})(x-\bar x)\Big) { \rm d}\bar x.$$

And since the in­te­grand of the fi­nal in­te­gral is con­tin­u­ous, it is bounded. That makes the er­ror in­versely pro­por­tional to $K+\frac12$, im­ply­ing that it does in­deed be­come ar­bi­trar­ily small for large $K$. 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 $x$. This can be ver­i­fied from Tay­lor se­ries with re­main­der. Also, the more con­tin­u­ous de­riv­a­tives the $2\pi$-​pe­ri­odic func­tion $f(x)$ has, the faster the rate of con­ver­gence, and the smaller the num­ber $2K+1$ of terms that you need to sum to get good ac­cu­racy is likely to be. For ex­am­ple, if $f(x)$ has three con­tin­u­ous de­riv­a­tives, you can do an­other in­te­gra­tion by parts to show that the con­ver­gence is pro­por­tional to 1$\raisebox{.5pt}{$/$}$​$(K+\frac12)^2$ rather than just 1/$(K+\frac12)$. 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 $f(x)$ 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 $f^2$ ex­ists, some­thing that will be true in quan­tum me­chan­ics ap­pli­ca­tions.