Page:EB1911 - Volume 10.djvu/777

Rh applies to every continuous portion of ƒ(x), for which the functions (x), (x) are formed in the same manner; we now take ƒ1(x) = (x) + ƒ( + 0) + C, ƒ2(x) = (x) + C, where C is constant between consecutive discontinuities, but may have different values in the next interval between discontinuities; the C can be so chosen that neither ƒ1(x) nor ƒ2(x) diminishes as x increases through a value for which ƒ(x) is discontinuous. We thus see that ƒ(x) = ƒ1(x) − ƒ2(x), where ƒ1(x), ƒ2(x) never diminish as x increases from a to b, and are discontinuous only where ƒ(x) is so. The function ƒ(x) is a particular case of a class of functions defined and discussed by Jordan, under the name “functions with limited variation” (fonctions à variation bornée); in general such functions have not necessarily only a finite number of maxima and minima.

Proof of the Convergence of Fourier’s Series.—It will now be assumed that a function ƒ(x) arbitrarily given between the values − and +, has the following properties:—

(a) The function is everywhere numerically less than some fixed positive number, and continuous except for a finite number of values of the variable, for which it may be ordinarily discontinuous.

(b) The function only changes from increasing to diminishing or vice versa, a finite number of times within the interval; this is usually expressed by saying that the number of maxima and minima is finite.

These limitations on the nature of the function are known as Dirichlet’s conditions; it follows from them that the function is integrable throughout the interval.

On these assumptions, we can investigate the limiting value of Dirichlet’s integral; it will be necessary to consider only the case of a function F(z) which does not diminish as z increases from 0 to, since it has been shown that in the general case the difference of two such functions may be taken. The following lemmas will be required:

1. Since $$\int_{0}^{\frac{\pi}{2}} \frac{\text{sin } mz}{\text{sin } z} dz=\int_{0}^{\frac{\pi}{2}}\{1+2 \text{cos } 2z+2 \text{cos } 4z +...+2 \text{cos } 2nz\}dz=\frac{\pi}{2};$$

this result holds however large the odd integer m may be.

2. If 0 < <  ≦ $undefined⁄2$,

where <  <, hence

a precisely similar proof shows that |∫  $sin mz⁄z$ dz | < $4⁄m$,

hence the integrals ∫  $sin mz⁄sin z$dz, ∫   $sin mz⁄z$dz, converge to the limit zero, as m is indefinitely increased.

3. If > 0, |∫ ∞  $1⁄sin$ d | cannot exceed. For by the mean-value theorem |∫ h $1⁄sin$d | < $1⁄2$ + $2⁄h$,

hence | Lh = ∞ ∫ h $1⁄sin$ d | ≦ 2/

in particular if ≧  |∫ ∞  $1⁄sin$d | ≦ $1⁄2$ < $undefined⁄2$.

Again $d⁄d$ ∫ ∞ $1⁄sin$d = − $1⁄sin$,  > 0,

therefore ∫ ∞ $1⁄sin$d increases as  diminishes, when  <  < ;

but lim =0 ∫ ∞ $1⁄sin$d = $undefined⁄2$, hence |∫ ∞  $1⁄sin$) d | < $undefined⁄2$,

where <, and < $1⁄2$ where  ≧. It follows that

To find the limit of ∫ $undefined⁄2$ 0 F(z) $sin mz⁄sin z$dz, we observe that it may be written in the form

where is a fixed number as small as we please; hence if we use lemma (1), and apply the second mean-value theorem,

when ¹ lies between and. When m is indefinitely increased, the two last integrals have the limit zero in virtue of lemma (2). To evaluate the first integral on the right-hand side, let G(z) = {F(z) − F(0)} (z / sin z), and observe that G(z) increases as z increases from 0 to, hence if we apply the mean value theorem

where 0 < <, since G(z) has the limit zero when z = 0. If be an arbitrarily chosen positive number, a fixed value of  may be so chosen that G < , and thus that |∫  0 G(z) $undefined⁄2$ dz | <. When has been so fixed, m may now be so chosen that

It has now been shown that when m is indefinitely increased ∫ $undefined⁄2$ 0 F(z) $undefined⁄2$ dz − $undefined⁄2$ F(0) has the limit zero.

Returning to the form (4), we now see that the limiting value of

hence the sum of n + 1 terms of the series

converges to the value {ƒ(x + 0) + ƒ(x − 0)}, or to ƒ(x) at a point where ƒ(x) is continuous, provided ƒ(x) satisfies Dirichlet’s conditions for the interval from −l to l.

Proof that Fourier’s Series is in General Uniformly Convergent.—To prove that Fourier’s Series converges uniformly to its sum for all values of x, provided that the immediate neighbourhoods of the points of discontinuity of ƒ(x) are excluded, we have

Using this inequality and the corresponding one for F(−z), we have

|S2n+1(x) − ƒ(x)| < cosec  [|ƒ(x + 2) − ƒ(x)| + |ƒ(x − 2) − ƒ(x)|] + A|m cosec ,

where A is some fixed number independent of m. In any interval (a, b) in which ƒ(x) is continuous, a value 1 of can be chosen such that, for every value of x in (a, b), |ƒ(x + 2) − ƒ(x)|, number, provided = 1. Also a value 2 of can be so chosen that 2 cosec 2 <, where  is an arbitrarily assigned positive number. Take for the lesser of the numbers 1, 2, then |S2n+1 − ƒ(x)| <  + A|m cosec for every value of x in (a, b). It follows that, since and m are independent of x, |S2n+1 − ƒ(x)| < 2, provided n is greater than some fixed value n1 dependent only on. Therefore S2n+1 converges to ƒ(x) uniformly in the interval (a, b).
 * ƒ(x − 2) − ƒ(x)| are less than an arbitrarily prescribed positive

Case of a Function with Infinities.—The limitation that ƒ(x) must be numerically less than a fixed positive number throughout the interval may, under a certain restriction, be removed. Suppose F(z) is indefinitely great in the neighbourhood of the point z = c, and is such that the limits of the two integrals ∫ c± c F(z) dz are both zero, as is indefinitely diminished, then

∫ $sin mx⁄z$ 0 F(z) $undefined⁄2$dz denotes the limit when = 0, ¹ = 0 of ∫ c− 0 F(z) $sin mz⁄sin z$dz + ∫ $undefined⁄2$ c+¹ F(z) $undefined⁄2$dz, both these limits existing; the first of these integrals has F(+0) for its limiting value when m is indefinitely increased, and the second has zero for its limit. The theorem therefore holds if F(z) has an infinity up to which it is absolutely integrable; this will, for example, be the case if F(z) near the point C is of the form x(z)(z − c) − + (z), where (c), (c) are finite, and 0 < < 1. It is thus seen that ƒ(x) may have a finite number of infinities within the given interval, provided the function is integrable through any one of these points; the function is in that case still representable by Fourier’s Series.

The Ultimate Values of the Coefficients in Fourier’s Series.—If ƒ(x) is everywhere finite within the given interval − to +, it can be shown that an, bn, the coefficients of cos nx, sin nx in the series which represent the function, are such that nan, nbn, however