Divergence and Almost Everywhere Convergence of Fourier Series — Examples

Real continuous periodic functions identify isometrically with $X=\{f\in C([0,1],\mathbb{R}):f(0)=f(1)\}$. A Cauchy sequence has a continuous uniform limit by completeness on the nonempty compact interval. Endpoint equality passes to that limit because evaluation at either endpoint changes by at most the uniform error. Hence $X$ is Banach.

The family $T_N:X\to \mathbb{R}$ consists of bounded linear maps, with unbounded operator norms. The bounded-norm alternative in the dichotomy is therefore excluded; its other alternative says precisely that $\mathcal{U}$ is dense and $G_{\delta }$ in $X$.

Explicitly, $\mathcal{U}={\bigcap }_{r=1}^{\mathrm{\infty }}{\bigcup }_{N=0}^{\mathrm{\infty }}\{f\in X:\mathrm{\mid }{T}_{N}f\mathrm{\mid }>r\}$. Each inner set is open by continuity of $T_N$, and the displayed membership condition is exactly unboundedness of the sequence of finite values. Thus the topology in this conclusion is topology on a space of functions; it asserts no full-measure set of points of the torus for any fixed function.

Apply the prescribed-point witness with $x_0=0$. It supplies a real continuous periodic $f$ whose finite partial-sum values at zero form an unbounded sequence. Such an $f$ is nonzero, and this sequence cannot converge to a finite value.

Regard the same real $f$ as complex-valued. It meets the continuity and periodicity hypotheses of uniform Fejer convergence, so $\mathrm{\parallel }{\sigma }_{N}f-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Thus the Cesaro averages ${\sigma }_{N}f=(N+1{)}^{-1}{\sum }_{j=0}^N{S}_{j}f$ converge uniformly while the ordinary sums at zero diverge unboundedly. Both properties hold for the single function from step 1.1, refuting the claimed implication.

Choose an integer $N\ge 1$ with $\mathrm{\parallel }{D}_N{\mathrm{\parallel }}_1>A+1$, possible from the logarithmic lower bound. For each $K\ge 0$, the function $F_K$ is a real nonnegative trigonometric polynomial of $L^1$ norm one.

Expanding the square formula for $F_K$ gives coefficient $1-\mathrm{\mid }k\mathrm{\mid }\mathrm{/}(K+1)$ for $\mathrm{\mid }k\mathrm{\mid }\le K$ and zero otherwise. Orthogonality of finite characters therefore gives $S_N{F}_K=F_K\ast D_N={\sigma }_K{D}_N$: either side has coefficient $1-\mathrm{\mid }k\mathrm{\mid }\mathrm{/}(K+1)$ for $\mathrm{\mid }k\mathrm{\mid }\le \min \{K,N\}$ and zero elsewhere. Since $D_N$ is continuous, $\mathrm{\parallel }{S}_N{F}_K-D_N{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ as $K\to \mathrm{\infty }$, with $N$ fixed.

For the additional consequence, let $h$ be a real integrable periodic representative. The convolution formula, after a periodic change of variables, gives $\mathrm{\mid }{S}_{N}h(x)\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }h(u)\mathrm{\mid }\mathrm{\mid }{D}_N(x-u)\mathrm{\mid }du$. The integrand is product-measurable: $h(u)$ depends on one coordinate and $D_N(x-u)$ is continuous. Both measure spaces are finite, so Tonelli and translation invariance give $\mathrm{\parallel }{S}_{N}h{\mathrm{\parallel }}_1\le {\int }_0^1\mathrm{\mid }h(u)\mathrm{\mid }{\int }_0^1\mathrm{\mid }{D}_N(x-u)\mathrm{\mid }dxdu=\mathrm{\parallel }{D}_N{\mathrm{\parallel }}_1\mathrm{\parallel }h{\mathrm{\parallel }}_1$. Thus $S_N$ is a bounded real linear operator on real $L^1$.

Choose $K$ large enough that this uniform error is less than one. The measure has mass one, so $\mathrm{\parallel }{S}_N{F}_K{\mathrm{\parallel }}_1\ge \mathrm{\parallel }{D}_N{\mathrm{\parallel }}_1-\mathrm{\parallel }{S}_N{F}_K-D_N{\mathrm{\parallel }}_1>A$. Since $C{F}_K\ge \mathrm{\mid }{S}_N{F}_K\mathrm{\mid }$, the polynomial $f=F_K$ is the required witness. Its maximal function is finite and bounded: when the partial-sum index is at least $K$, the partial sum equals $F_K$, so only finitely many distinct continuous polynomials enter its supremum. Thus the strict norm inequality is an ordinary finite integral, not an artifact of an infinite value.

For each fixed $N$, step 1.2 and the unit-norm tests $F_K$ imply $\mathrm{\parallel }{S}_N:L^1(\mathbb{T},\mathbb{R})\to L^1(\mathbb{T},\mathbb{R})\mathrm{\parallel }\ge {\lim }_{K\to \mathrm{\infty }}\mathrm{\parallel }{S}_N{F}_K{\mathrm{\parallel }}_1=\mathrm{\parallel }{D}_N{\mathrm{\parallel }}_1$. These operator norms are consequently unbounded.

Now assume DC. Real $L^1$ is Banach by completeness. If ${\sup }_N\mathrm{\parallel }{S}_{N}h{\mathrm{\parallel }}_1$ were finite for each $h$, uniform boundedness would contradict step 2.2. Therefore some real integrable $h$ has unbounded $L^1$ norms of its partial sums. This is a norm-divergence conclusion and does not assert almost-everywhere divergence.