Our main topic for today is the second question, which is essentially the reverse of the first. Given a "formal" trigonometric series, when does it actually define a function? In the course of discussing it, we will also introduce the idea of uniform convergence of a Fourier series. Before continuing in this section, the reader would be wise to review the sections in calculus text concerning convergence and absolute convergence of series.
Proof of the Riemann-Lebesgue Lemma We will restrict ourselves to functions f(x) that are piecewise continuous and have piecewise continuous derivatives on a finite interval [a,b]. This is a stronger condition than the lemma requires. On the other hand, we will derive an important formula in the process.
List the points in [a,b] where f or f′ is discontinuous, including, possibly, the end points a and b:
c0 = a < c1 < c2 < ... < cM < b = cM+1 .
We will first integrate by parts over the interval [cj,cj+1],
∫cjcj+1 f(x) cos(kx)dx = k-1(f(cj+1-) sin(kcj+1) - f(cj+)sin(kcj)) - k-1 ∫cjcj+1 f′(x) sin(kx)dx,
where f(cj+) and f(cj+1-) indicate the right and left hand limits of f at cj and cj+1. Since we have
∫ab f(x) cos(kx)dx = ∑j=0M ∫cjcj+1 f(x) cos(kx)dx,
we can replace the integrals on the right by the expressions we found for them above and then simplify to arrive at this formula,
∫ab f(x) cos(kx)dx = k-1 (f(b-)sin(kb) - f(a+)sin(ka)) - ∑j=1MΔj k-1sin(kcj) - k-1∫ab f′(x) sin(kx)dx ,
where Δj = f(cj+) - f(cj-). A similar formula holds for ∫ab f(x) sin(kx)dx, namely,
∫ab f(x) sin(kx)dx = k-1 (f(a+)cos(ka) - f(b-)cos(kb)) + ∑j=1MΔj k-1cos(kcj) + k-1∫ab f′(x) cos(kx)dx.
The version of the Riemann-Lebesgue Lemma stated earlier is now easy to get, if we recall the triangle inequality for numbers and integrals,
|z1 + z2 +... + zn| < |z1| + |z2| +...+ |zn| and |∫ab g(x) dx| < ∫ab |g(x)| dx.
Applying these to the formulas we just obtained, we see that
|∫ab f(x) cos(kx)dx| < k-1(|f(a+)|+|f(b-)| + ∑j=1M|Δj| + ∫ab |f′(x)|dx) = Ck-1,
and
|∫ab f(x) sin(kx)dx| < Ck-1,
where C, which is independent of k, is the same as in the previous formula. Now, since Ck-1 -> 0 as k -> ∞, by the "squeeze" or "pinching" theorem from calculus, both integrals go to zero as well. This proves the Riemann-Lebesgue Lemma in the special case where f and f′ are both piecewise continuous. ∎
Applications to Fourier Coefficients The formulas and inequalities for ∫ab f(x) cos(kx)dx and ∫ab f(x) sin(kx)dx derived above can be applied to Fourier coefficents. Let k = n be an integer, and let a = -π and b = π. If we let an, bn be the Fourier coefficients for f, and An, Bn be the ones for f′, then, after adjusting for factors of 2π and using sin(nπ) = 0, we have
an = - ∑j=1M(2πn)-1Δj sin(ncj) - n-1Bn
Similarly,
bn = (2πn)-1(f(-π+)-f(π-))cos(nπ) + ∑j=1M(2πn)-1Δj cos(ncj) + n-1An
In particular, if f is continuous and f(-π) = f(π), then, for all n > 0,
an = - n-1Bn and bn = n-1An.
For f and f′ piecewise continuous, the inequalities we obtained in proving the Riemann-Lebesgue Lemma hold, so in that case for all n > 0,
|an| < Cn-1 and |bn| < Cn-1
Now, suppose that f is continuous, with f(-π) = f(π), and that f′ and f" are piecewise continuous. The assumptions on f′ and f" imply that
|An| < Cn-1 and |Bn| < Cn-1,
and the assumptions on f imply that an = - n-1Bn and bn = n-1An. Evidently, we have that if f is continuous, with f(-π) = f(π), and that f′ and f" are piecewise continuous, then for n > 0,
|an| < Cn-2 and |bn| < Cn-2,
where C is independent of n.
Convergence of Trigonometric Series A simple, but very important, condition that guarantees a formal series converges is the following one, which we state as a proposition here.
Proposition. If the series ∑n=1∞ |an|+|bn| is convergent, then the trigonometric seriesa0 + ∑n=1∞ ancos(nx)+ bnsin(nx)
converges to a 2π periodic, continuous function f(x). Moreover, the function f(x) has the trigonometric series as its Fourier series. Finally, the error from truncating the series at the Nth term can be estimated as follows:
|SN(x) - f(x)| < ∑n=N+1∞ |an|+|bn|
Proof. Because each of the terms ancos(nx)+ bnsin(nx) satisfies
|ancos(nx)+ bnsin(nx)| < |an|+|bn|,
where the series ∑n=1∞ |an|+|bn| is convergent, the comparison test implies that the trigonometric series above is absolutely convergent for each x. Consequently, it defines a function f(x). Now, we also have that
|SN(x) - f(x)| = |∑n=N+1∞ ancos(nx)+ bnsin(nx)|,
so by the triangle inequality we get the error estimate,
|SN(x) - f(x)| < ∑n=N+1∞ |an|+|bn| .
Changing x to x+2π leaves each term in the series unchanged; thus, f(x) = f(x+2π), and so f is 2π periodic. The continuity is true, but requires more analysis than we have time for. The fact the trigonometric series is the Fourier series for f follows from the argument we made in Lecture 2, plus a little more work to make it absolutely precise. ∎
Let's connect this up with what we said earlier about Fourier coefficients of functions.
Corollary. If f is continuous, with f(-π) = f(π), and that f′ and f" are piecewise continuous, then ∑n=1∞ |an|+|bn|, is convergent and the conclusions of the proposition above apply.
Proof. From the discussion earlier, the Fourier coefficnets of f satisfy |an| < Cn-2 and |bn| < Cn-2, for n >0, where C is independent of n. Hence,
|an|+|bn| < 2n-2,
so by the p-series test, with p=2, the series in question is convergent. ∎
The important thing about the truncation error estimate,
|SN(x) - f(x)| < ∑n=N+1∞ |an|+|bn|,
is what the right hand side of the inequality does not depend on; namely, x. We say that the error is estimated uniformly in x. In fact, the maximum value of |f(x) -S2N-1| over all x goes to 0 as N -> ∞. We will talk about what this means next lecture.