Let \(f:\RR\to\RR\) be the \(2\pi\)-periodic function satisfying \(f(x)=x\) for \(-\pi\le x<\pi\).
- Prove that its Fourier series is \[ \sum_{n=1}^{\infty}\frac{2(-1)^{n+1}\sin(nx)}{n}. \]
- Prove that this series does not converge uniformly.
- For each \(x\in\RR\), find the sum of the series.
Solution
1. The cosine Fourier coefficients of \(f\) vanish, and its sinecoefficients are\[ b_n=\frac{2(-1)^{n+1}}{n} \qquad(n\ge1). \]
Changing a function at the endpoint \(\pi\) does not affect its Fourier coefficients, so the coefficients may be computed from \(f(x)=x\) on \([-\pi,\pi]\). Since \(x\) is odd, \[ a_0 = \frac1\pi\int_{-\pi}^{\pi}x\,dx = 0 \] and, for \(n\ge1\), \[ a_n = \frac1\pi\int_{-\pi}^{\pi}x\cos(nx)\,dx = 0, \] because \(x\cos(nx)\) is odd. Also \(x\sin(nx)\) is even, so \[ \begin{aligned} b_n &= \frac1\pi\int_{-\pi}^{\pi}x\sin(nx)\,dx\\ &= \frac2\pi\int_0^\pi x\sin(nx)\,dx\\ &= \frac2\pi \left[ -\frac{x\cos(nx)}n+\frac{\sin(nx)}{n^2} \right]_{0}^{\pi}\\ &= -\frac{2(-1)^n}{n} = \frac{2(-1)^{n+1}}n. \end{aligned} \]
2. Therefore the Fourier series of \(f\) is\[ \boxed{ \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}\sin(nx)}{n}. } \]
Insert the coefficients from step 1 into the real Fourier-series formula.
3. The series in step 2 does not satisfy the uniform Cauchycriterion on \(\RR\).
For an integer \(N\ge1\), set \[ x_N\coloneqq\pi-\frac1N. \] For \(N\le n\le2N\), \[ \sin(nx_N) = \sin\left(n\pi-\frac nN\right) = (-1)^{n+1}\sin\left(\frac nN\right). \] Hence the corresponding block of the series is \[ \sum_{n=N}^{2N} \frac{2(-1)^{n+1}\sin(nx_N)}n = 2\sum_{n=N}^{2N} \frac{\sin(n/N)}n. \] Let \[ c\coloneqq\min_{1\le t\le2}\sin t. \] Since \([1,2]\subset(0,\pi)\), one has \(c>0\). Thus \[ 2\sum_{n=N}^{2N}\frac{\sin(n/N)}n \ge 2c\sum_{n=N}^{2N}\frac1n \ge 2c(N+1)\frac1{2N} \ge c. \] Therefore tails with arbitrarily large starting index have supremum at least \(c\), so the uniform Cauchy criterion fails.
4. The Fourier series does not converge uniformly.
Uniform convergence of a series of functions implies the uniform Cauchy criterion. Step 3 shows that criterion fails.
5. At every point where \(f\) is continuous, the Fourier series convergesto \(f(x)\); at each jump point\[ x=(2k+1)\pi, \qquad k\in\ZZ, \]it converges to \(0\).
The periodic function \(f\) is piecewise \(C^1\). By the Dirichlet convergence theorem, its Fourier series converges at each \(x\) to \[ \frac{f(x^-)+f(x^+)}2. \] Away from the odd multiples of \(\pi\), the function is continuous, so this value is \(f(x)\). At \(x=(2k+1)\pi\), periodicity gives one-sided limits \[ f(x^-)=\pi, \qquad f(x^+)=-\pi, \] whose average is \(0\).
6. Explicitly, the sum \(S\) of the series is\[ \boxed{ S(x)= \begin{cases} x-2k\pi, & (2k-1)\pi<x<(2k+1)\pi\text{ for some }k\in\ZZ,\\ 0, &x=(2k+1)\pi\text{ for some }k\in\ZZ. \end{cases} } \]
On the open interval \[ ((2k-1)\pi,(2k+1)\pi), \] periodicity and the defining formula on \([-\pi,\pi)\) give \[ f(x)=x-2k\pi. \] Combine this with step 5.
7. Q.E.D.
Step 2 proves part (1), step 4 proves part (2), and step 6 answers part (3).