Page:EB1911 - Volume 10.djvu/777

applies to every continuous portion of ƒ(x), for which the functions ψ(x), χ(x) are formed in the same manner; we now take ƒ₁(x) = ψ(x) + ƒ(α + 0) + C, ƒ₂(x) = χ(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 −π and +π, 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 1/2π, 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 < α < β ≦ π/2,

$\int _{\alpha }^{\beta }$	sin mz	dz =	1	∫ γ α sin mz dz +	1	∫ β γ sin mz dz
	sin z		sin α		sin β

where α < γ < β, hence

| $\int _{\alpha }^{\beta }$	sin mz	dz | <	2	(	1	+	1	) <	4	;
	sin z		m		sin α		sin β		m sin α

a precisely similar proof shows that |∫ β α sin mz/z dz | < 4/mα,

hence the integrals ∫ β α sin mz/sin zdz, ∫ β α sin mz/zdz, converge to the limit zero, as m is indefinitely increased.

3. If α > 0, |∫ ∞ α sin θ/θ dθ | cannot exceed 1/2π. For by the mean-value theorem |∫ h α sin θ/θdθ | < 2/α + 2/h,

hence | Lh = ∞ ∫ h α sin θ/θ dθ | ≦ 2/α

in particular if α ≧ π |∫ ∞ α sin θ/θdθ | ≦ 2/π < π/2.

Again d/dα ∫ ∞ α sin θ/θdθ = − sin α/α, α > 0,

therefore ∫ ∞ α sin θ/θdθ increases as α diminishes, when θ < α < π;

but lim α=0∫ ∞ α sin θ/θdθ = π/2, hence |∫ ∞ α sin θ/θ) dθ | < π/2,

where α < π, and < 2/π where α ≧ π. It follows that

| $\int _{\alpha }^{\beta }$	sinθ	dθ | ≦ π, provided 0 ≦ α < β.
	θ

To find the limit of ∫ π/2 0 F(z) sin mz/sin zdz, we observe that it may be written in the form

F(0)   ∫ π/2 0	sin mz	dz + ∫ μ 0 {F(z) − F(0)}	sin mz	dz + ∫ π/2 μ {F(z) − F(0)}	sin mz	dz
	sin z		sin z		sin z

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

∫ π/2 0 F(z)	sin mz	dz −	π	F(0) = ∫ μ 0 {F(z) − F(0)}	z		sin mz	dz
	sin z		2		sin z		z

+ {F(μ + 0) − F(0)} ∫ ξ¹ μ	sin mz	dz + [F (1/2π − 0) − F(0)] ∫ π/2 ξ¹	sin mz	dz
	sin z		sin z

when ξ¹ lies between μ and 1/2π. 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 μ, hence if we apply the mean value theorem

|∫ μ 0 G(μ)	sin mz	dz | = | G(μ) ∫ μ ξ	sin mz	dz | = | G(μ) ∫ mμ mξ	sin θ	dθ | < πG(μ),
	z		z		θ

where 0 < ξ < μ, since G(z) has the limit zero when z = 0. If ε be an arbitrarily chosen positive number, a fixed value of μ may be so chosen that πG(μ) < 1/2 ε, and thus that |∫ μ 0 G(z) sin mx/z dz | < 1/2 ε. When μ has been so fixed, m may now be so chosen that

|∫ 1/2π 0 F(z)	sin mz	dz −	π	F(0) | < ε.
	sin z		2

It has now been shown that when m is indefinitely increased ∫ π/2 0 F(z) sin mz/sin z dz − π/2 F(0) has the limit zero.

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

1	∫ π/2 0 F(z)	sin mz	+	1	∫ π/2 0 F(−z)	sin mz	dz is 1/2 {F(+0) + F(−0)};
π		sin z		π		sin z

hence the sum of n + 1 terms of the series

1	∫ l −l ƒ(x) dx +	1	Σ ∫ l −l ƒ(x¹) cos	nπ(x − x¹)	dx
2l		l		l

converges to the value 1/2 {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

|∫ π/2 F(z)	sin mz	dz −	π	F(0) | < πG(μ) +	4	{F(μ + 0) − F(0)} +	4	{F(1/2π − 0) − F(0)}
	sin z		2		m sin μ		m sin ξ¹

<	πμ	{ƒ(x + 2μ) − ƒ(x)} +	4	{ƒ(x + 2μ) − ƒ(x)} +	4	{ƒ(x + π) − ƒ(x)}.
	sin μ		m sin μ		m sin ξ¹

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

|S_2n+1(x) − ƒ(x)| < μ cosec μ [|ƒ(x + 2μ) − ƒ(x)| + |ƒ(x − 2μ) − ƒ(x)|] + A|m cosec μ,

where A is some fixed number independent of m. In any interval (a, b) in which ƒ(x) is continuous, a value μ₁ of μ can be chosen such that, for every value of x in (a, b), |ƒ(x + 2μ) − ƒ(x)|, |ƒ(x − 2μ) − ƒ(x)| are less than an arbitrarily prescribed positive number ε, provided μ = μ₁. Also a value μ₂ of μ can be so chosen that εμ₂ cosec μ₂ < 1/2 η, where η is an arbitrarily assigned positive number. Take for μ the lesser of the numbers μ₁, μ₂, then |S_2n+1 − ƒ(x)| < η + A|m cosec μ for every value of x in (a, b). It follows that, since η and m are independent of x, |S_2n+1 − ƒ(x)| < 2ε, provided n is greater than some fixed value n₁ dependent only on ε. Therefore S_2n+1 converges to ƒ(x) uniformly in the interval (a, b).

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±ε c F(z) dz are both zero, as ε is indefinitely diminished, then

∫ π/2 0 F(z) sin mz/sin zdz denotes the limit when ε = 0, ε¹ = 0 of ∫ c−ε 0 F(z) sin mz/sin zdz + ∫ π/2 c+ε¹ F(z) sin mz/sin zdz, both these limits existing; the first of these integrals has 1/2πF(+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)−μ + ψ(z), where χ(c), ψ(c) are finite, and 0 < μ < 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 −π to +π, it can be shown that a_n, b_n, the coefficients of cos nx, sin nx in the series which represent the function, are such that na_n, nb_n, however