Using this inequality and the corresponding one for F(−z), we have
|S2n+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 < ½ η, where η is an arbitrarily assigned positive number. Take for μ the lesser of the numbers μ1, μ2, then |S2n+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, |S2n+1 − ƒ(x)| < 2ε, provided n is greater than some fixed value n1 dependent only on ε. Therefore S2n+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 z) dz denotes the limit when ε = 0, ε¹ = 0 of ∫ c-ε 0 F(z) (sin mz / sin z) dz + ∫ π/2 c+ε¹ F(z) (sin mz / sin z) dz, both these limits existing; the first of these integrals has ½ πF(+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 an, bn, the coefficients of cos nx, sin nx in the series which represent the function, are such that nan, nbn, however great n is, are each less than a fixed finite quantity. For writing ƒ(x) = ƒ1(x) − ƒ2(x), we have
∫ π −π ƒ1(x) cos nxdx = ƒ1(−π + 0) ∫ ξ −π cos nxdx + ƒ1(π − 0) ∫ π ξ cos nxdx
hence
| ∫ π −π ƒ1(x) cos nxdx = ƒ1(−π + 0) | sin nξ | + ƒ1(π − 0) | sin nξ |
| n | n |
with a similar expression, with ƒ2(x) for ƒ1(x), ξ being between π and −π; the result then follows at once, and is obtained similarly for the other coefficient.
If ƒ(x) is infinite at x = c, and is of the form φ(x) / (x − c)K near the point c, where 0 < K < 1, the integral ∫ π −π ƒ(x)cos nxdx contains portions of the form ∫ ε+ε c [φ(x) / (x − c)K] cos nxdx ∫ c c−ε [φ(x) / (x − c)K] cos nxdx; consider the first of these, and put x = c + u, it thus becomes ∫ ε 0 [φ(c + u) / uK] cos n(c + u) du, which is of the form φ(c + θε) ∫ ε 0 [cos n(c + u) / uK] du; now let nu = v, the integral becomes