Pointwise Convergence of Fourier Series

Given a function f, we can construct its Fourier series

f(x)~ a0 + a1cos(x)+ b1sin(x) + a2cos(2x)+ b2sin(2x) + ... + ancos(nx)+ bnsin(nx) + ...   (real form)

or

f(x) ~ ∑n αneinx   (complex form),

where n runs over all integers. Under what conditions on f does the series converge pointwise to f?

The Dirichlet Kernel and Partial Sums

Assume that we have a 2π-periodic, piecewise continuous function f with Fourier series

f(x) ~ ∑n αneinx,  where αn = (1/(2π))∫π f(t)e-intdt.

The partial sum SN(x) is given by

SN(x) = ∑n=-NN αneinx = ∑n=-NN (1/(2π))∫π f(t)einxe-intdt.

Because the "integral of the sum is the sum of the integrals," we see that the partial sum SN is

SN(x) = ∫π f(t)DN(x-t)dt,

where DN is called the Dirichlet kernel, and is defined by

DN(u) := (1/(2π))∑n=-NNeinu.

We derived the properties of this important function in class. The first three listed below are easy to derive; the fourth is harder and we will establish it later.

Properties of DN

  1. DN(u) = 1/(2π) + (1/π)∑n=1N cos(nu)
  2. DN(u) is even and 2π periodic.
  3. π DN(u)du = 1 and ∫0π DN(u)du = ½
  4. DN(u) = sin((N+½)u)/(2π sin(u/2))

The properties given for DN help put each partial sum SN in a form that will make it easier to work with. First of all, in the form of SN above, we change variables to u = t-x and use the fact that DN is even to get

SN(x) = ∫-π-xπ-x f(u+x)DN(u)du.

Also, both DN and f(x+u) are 2π-periodic in u, and so by Lemma 1.3 (p. 43) in the text with c=-x,

SN(x) = ∫π f(u+x)DN(u)du.

We can change variables in the integral going from u to -u. The result is that

SN(x) = -∫π f(x-u)DN(-u)du = ∫π f(x-u)DN(u)du.

Add these two expressions together and divide by 2. The integrand in the result, (f(x+u)+f(x-u))DN(u)/2, is even. Using the fact that an integral of an even function over a symmetric interval is twice the integral of the function over half the interval, we get, after cancelling factors of 2, that

SN(x) = ∫0π (f(x+u)+f(x-u))DN(u)du.

The next ingredient in our proof is the Riemann-Lebesgue Lemma, which we now state.

Riemann-Lebesgue Lemma  If f is in L1[a,b], then

limk->∞abf(x)cos(kx)dx = limk->∞abf(x)sin(kx)dx = 0.

We first mention that k does not have to be an integer; it just has to go off to infinity. In any case, when k is an integer and f is in L2[− π,π], then the result is a consequence of Bessel's inequality.
Pointwise Convergence Theorem  If f is a 2π-periodic piecewise continuous function on the interval and if f has a right-hand derivative f′(x+) and left hand derivative f′(x-) at x, then when f is continuous at x

limN->∞SN(x) = f(x),

and when f has a jump discontinuity at x

limN->∞SN(x) = (f(x+)+f(x-))/2.

Proof: The trick here is to put the the error SN(x) - f(x) or SN(x) - (f(x+)+f(x-))/2 in the right form. Since the continuous case just has f(x+)=f(x-)=f(x), we can do both cases using

EN(x) = SN(x) - (f(x+)+f(x-))/2.

From the third property of DN listed above and the final form of SN derived earlier, we obtain

EN(x) = ∫0π (f(x+u) + f(x-u) - f(x+) - f(x-))DN(u)du.

Use property 4 of DN(u) to put the error in the form

EN(x) = ∫0π (f(x+u) + f(x-u) - f(x+) + f(x-)/(2π sin(u/2)) sin((N+½)u)du = ∫0π F(u)sin((N+½)u)du,

where

F(u) := (f(x+u) + f(x-u) - f(x+) - f(x-)/(2π sin(u/2)).

Since x is fixed, we can ignore the dependence of the expression above on x. L'Hospital's rule and the fact that f has right and left derivatives at x imply that

limu -> 0+ F(u) = (f'(x+) - f'(x-))/π,
so F has a right hand limit at u = 0. This and the piecewise continuity of f and 1/sin(u/2) on the open interval (0,π] mean that F(u) is piecewise continuous on [0,π]. From the Riemann Lebesgue Lemma we have that

limN -> ∞EN(x) = 0,

which is what we needed to show.

The Formula for the Dirichlet Kernel

We want to establish property 4 of DN(u). Let z = eiu, so zn = einu for n = -N, ..., N. We can write the Dirichlet kernel this way:
2π DN(u) = 1 + z + z2 + ... + zN + 1 + z-1 + ... + z-N - 1 .
From this, we get that
2π(z-1)DN = zN+1 -1 - z-N + z - z + 1 = zN+1 - z-N,
2π(eiu-1)DN = ei(N+1)u - e-iNu,
2i eiu/2 sin(u/2)(2π) DN = 2i eiu/2 sin((N+1/2)u) .
Dividing by the appropriate factors yields the formula that we want.