A continuous function with divergent Fourier series at a prescribed point

Let $X=\{f\in C([0,1],\mathbb{R}):f(0)=f(1)\}$ with the supremum norm. The interval is nonempty and compact, so a Cauchy sequence in $X$ has a continuous uniform limit by the completeness theorem. Its endpoint values remain equal, since $\mathrm{\mid }f(0)-f(1)\mathrm{\mid }\le 2\mathrm{\parallel }f-f_j{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every approximating member $f_j$. Thus $X$ is a real Banach space, identified isometrically with the real continuous periodic functions.

Define $T_N:X\to \mathbb{R}$ by $T_{N}f=S_{N}f(x_0)$. These maps are real-valued bounded linear functionals, and ${\sup }_N\mathrm{\parallel }{T}_N\mathrm{\parallel }=\mathrm{\infty }$. If all $f\in X$ had ${\sup }_N\mathrm{\mid }{T}_{N}f\mathrm{\mid }<\mathrm{\infty }$, uniform boundedness on this Banach space would make the operator norms uniformly bounded. Hence there exists a real $f\in X$ with ${\sup }_N\mathrm{\mid }{T}_{N}f\mathrm{\mid }=\mathrm{\infty }$. Every individual value is finite, so this sequence cannot converge.

For this witness, vanishing on any neighborhood of $x_0$ is impossible: if it vanished there, choose $0<\delta <1\mathrm{/}2$ within that neighborhood. The Dini integral with $s=0$ would be zero, giving $S_{N}f(x_0)\to 0$. DC supplies the countable choice assumed by that criterion. This contradicts the unboundedness in step 2.1 and proves the stated localization observation.