f(x)~ a0 + a1cos(x)+ b1sin(x) + a2cos(2x)+ b2sin(2x) + ... + ancos(nx)+ bnsin(nx) + ... (real form)
orf(x) ~ ∑n αneinx (complex form),
where n runs over all integers. Under what conditions on f does the series converge pointwise to f?
f(x) ~ ∑n αneinx, where αn = (1/(2π))∫-ππ f(t)e-intdt.
The partial sum SN(x) is given by
SN(x) = ∑n=-NN αneinx = ∑n=-NN (1/(2π))∫-ππ 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) = ∫-ππ 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.
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) = ∫-ππ 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-π f(x-u)DN(-u)du = ∫-ππ 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], thenWe 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.limk->∞∫abf(x)cos(kx)dx = limk->∞∫abf(x)sin(kx)dx = 0.
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 xProof: 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 usinglimN->∞SN(x) = f(x),
and when f has a jump discontinuity at x
limN->∞SN(x) = (f(x+)+f(x-))/2.
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.