The Uniform Boundedness Theorem showed that such a function exists. We now construct one explicitly. The idea is to build a trigonometric polynomial whose terms cancel at zero when the whole polynomial is summed, but whose first half has a large sum at zero.

1. Start with a symmetric sum.

For each positive integer \(N\), define

$$ f_N(x)=\sum_{1\le |n|\le N}\frac{e^{inx}}{n}. $$

Pairing the terms at frequencies \(n\) and \(-n\) gives

$$ f_N(x) =\sum_{n=1}^{N}\frac{e^{inx}-e^{-inx}}{n} =2i\sum_{n=1}^{N}\frac{\sin(nx)}{n}. $$

At \(x=0\) each pair cancels, so \(f_N(0)=0\). If we keep only the negative frequencies, however, their sum at zero is

$$ \sum_{n=-N}^{-1}\frac{1}{n} =-\sum_{n=1}^{N}\frac{1}{n} =-H_N, $$

where \(H_N\) is the \(N\)th harmonic number. Its magnitude grows at least as fast as \(\log(N+1)\).

Lemma
There is a constant \(C\) such that \(|f_N(x)|\le C\) for every \(N\ge 1\) and every \(x\in[-\pi,\pi]\).

To see why the bound is uniform even near zero, take \(0\lt x\le\pi\) and set \(m=\min\{N,\lfloor 1/x\rfloor\}\). For the first \(m\) terms, \(|\sin(nx)|\le nx\) gives

$$ \sum_{n=1}^{m}\frac{|\sin(nx)|}{n}\le mx\le 1. $$

For the remaining terms, the finite geometric sum gives \(\left|\sum_{n=m+1}^{r}\sin(nx)\right|\le 1/\sin(x/2)\). Summation by parts with the decreasing weights \(1/n\) therefore gives, when \(m\lt N\),

$$ \left|\sum_{n=m+1}^{N}\frac{\sin(nx)}{n}\right| \le \frac{1}{(m+1)\sin(x/2)} \le \pi. $$

The last inequality uses \(m+1>1/x\) and \(\sin(x/2)\ge x/\pi\). Oddness gives the same bound for negative \(x\), and \(f_N(0)=0\). Thus we may take \(C=2(1+\pi)\). The same summation-by-parts estimate also gives uniform convergence of the sine series away from any neighborhood of zero in \([-\pi,\pi]\).

2. Shift the frequencies so a partial sum can capture just the first half.

Define

$$ P_N(x)=e^{2iNx}f_N(x) =\sum_{1\le |n|\le N}\frac{e^{i(2N+n)x}}{n}. $$

Multiplication by \(e^{2iNx}\) shifts each frequency \(n\) to \(2N+n\). The original negative frequencies now run from \(N\) to \(2N-1\). The original positive frequencies run from \(2N+1\) to \(3N\). Thus a Fourier partial sum with cutoff \(2N\) includes exactly the first half of this polynomial.

The shift preserves the uniform bound because \(|e^{2iNx}|=1\). It also preserves the value at zero. Hence

$$ \|P_N\|_\infty\le C, \qquad P_N(0)=0, \qquad S_{2N}P_N(0)=-H_N. $$

3. Combine the shifted polynomials into one continuous function.

Choose positive integers \(N_k\) and positive weights \(\alpha_k\) so that

$$ N_{k+1}>3N_k, \qquad \sum_{k=1}^{\infty}\alpha_k\lt\infty, \qquad \alpha_k\log N_k\to\infty. $$

For example, take \(N_k=4^{4^k}\) and \(\alpha_k=1/k^2\). The first condition keeps the frequency intervals \([N_k,3N_k]\) disjoint. Define

$$ f(x)=\sum_{k=1}^{\infty}\alpha_k P_{N_k}(x). $$

Since \(|\alpha_kP_{N_k}(x)|\le C\alpha_k\) and the weights are summable, the Weierstrass M-test gives uniform convergence. Each summand is continuous and \(2\pi\)-periodic, so \(f\in C(\mathbb{T})\). Also, \(f(0)=0\) because every summand vanishes at zero.

4. Evaluate the Fourier partial sums at zero.

Take the cutoff \(2N_k\). Every earlier polynomial is included in full and contributes zero at the origin. Every later polynomial has all its frequencies above the cutoff. Only the first half of the \(k\)th polynomial contributes. Uniform convergence allows us to compute the Fourier coefficients term by term, so

$$ S_{2N_k}f(0) =\alpha_k\sum_{n=-N_k}^{-1}\frac{1}{n} =-\alpha_k H_{N_k}. $$

Since \(H_{N_k}\ge\log(N_k+1)\) and \(\alpha_k\log N_k\to\infty\), these partial sums tend to \(-\infty\). At the cutoff \(3N_k\), the entire \(k\)th polynomial is included, so the cancellation is restored and \(S_{3N_k}f(0)=0\). Thus the Fourier series of this continuous function diverges at \(x=0\).

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi