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For each integer $m\ge 1$, the set of continuous functions satisfying $|S_N(f,0)|\le m$ for every $N$ is closed in the uniform norm. If it contained a ball of radius $r$, subtracting two members of that ball would bound every partial-sum functional norm by $4m/r$. The Dirichlet kernel harmonic lower bound excludes this, so these sets are nowhere dense. By the Baire category theorem, their complement is a dense $G_{\delta }$ set. Every function in that complement has unbounded Fourier partial sums at zero. Translation gives the same conclusion at any prescribed point. This is a direct Uniform boundedness principle argument on the real Banach space of continuous functions on the circle.

First prove the Baire category theorem in its complete-metric form. Let $X$ be a nonempty complete metric space, let $G_n$ be open dense sets, and let $V$ be any nonempty open set. Choose a closed ball $\overline{B}(x_1,r_1)\subset V\cap G_1$ with $0<r_1<1/2$. Inductively, density and openness permit the choiceFor $m>n$, $x_m$ belongs to the $n$th closed ball, so $d(x_m,x_n)\le r_n$. The centers form a Cauchy sequence and converge to some $x\in X$ by completeness. Each closed ball contains all later centers and therefore contains their limit. Hence $x\in V\cap {\bigcap }_n{G}_n$. Since this holds for every $V$, every countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of closed sets with empty interior; this is the form needed below.

For the Banach-Steinhaus theorem, let $X$ be a Banach space, $Y$ a normed vector space, and $\mathcal{T}$ a family of bounded linear operators from $X$ to $Y$ such that ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x$. DefineEach $E_m$ is closed, as an intersection of inverse images of closed balls under continuous maps, and pointwise boundedness gives $X={\bigcup }_m{E}_m$. The Baire category theorem gives some $E_m$ with nonempty interior. Choose $B(x_0,r)\subset E_m$. If $\Vert h\Vert \le r/2$, both $x_0+h$ and $x_0$ lie in $E_m$, so $\Vert Th\Vert \le 2m$ for every $T$. Applying this to $h=(r/2)u$ with $\Vert u\Vert \le 1$ provesThis proves the Uniform boundedness principle; completeness is required for the domain, but not for $Y$.

Now fix the prescribed point ${\theta }_0$ on the unit circle, parametrized by angles modulo $2\pi$. On the real Banach space $C(\mathbb{T};\mathbb{R})$ with the supremum norm, consider the continuous linear functionals $T_{N}f=S_{N}f({\theta }_0)$, where $S_N$ is the symmetric Fourier partial sum. Summing the finite geometric series gives its Dirichlet kernel representationThe continuous extension at $t=0$ is $2N+1$. The kernel bound gives $\Vert T_N\Vert \le {\Lambda }_N=(2\pi {)}^{-1}\int |D_N|$. Equality holds: the continuous periodic functions defined byhave supremum norm at most one, and dominated convergence gives $T_N{f}_{\delta }\to {\Lambda }_N$ as $\delta \downarrow 0$.

These norms are unbounded. Indeed, evenness and $\sin (t/2)\le t/2$ for $0<t\le \pi$ giveFor the second inequality, integrate over $[(j-1)\pi ,j\pi ]$, where $1/u\ge 1/(j\pi )$ and the integral of $|\sin u|$ is two. The harmonic series diverges, proving the Dirichlet kernel harmonic lower bound. If every continuous $f$ had bounded partial sums at ${\theta }_0$, the Banach-Steinhaus theorem would bound all the operator norms, a contradiction. Consequently there exists a continuous real function $f$ with ${\sup }_N|S_{N}f({\theta }_0)|=\infty$, and its Fourier series diverges at the prescribed point. The function depends on the chosen point; the argument requires no claim of simultaneous divergence everywhere.

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized measure $dt/(2\pi )$. The Fourier partial sum at zero is the real linear functionalwhere the Dirichlet kernel isThe quotient follows by summing the finite geometric progression, with its removable value $D_N(0)=2N+1$. It is a real continuous periodic function. Ifthen $|L_{N}f|\le {\Lambda }_N\Vert f{\Vert }_{\infty }$. This bound is its exact operator norm even on the real continuous functions. Indeed, for $\delta >0$ setIt is real, continuous and periodic with norm at most one, andby dominated convergence. At zeros of the kernel the integrand is zero, and elsewhere its limit is $|D_N|$.

For $0<t\le \pi$, $\sin (t/2)\le t/2$. Using evenness and putting $v=(N+1/2)t$ givesFor the last inequality, split into the first $N$ complete intervals $[(j-1)\pi ,j\pi ]$: their sine integrals are two and $1/v\ge 1/(j\pi )$. The harmonic sum tends to infinity. This proves the Dirichlet kernel harmonic lower bound.

Choose $N$ with ${\Lambda }_N>K$, then choose $\delta$ sufficiently small that $L_N{f}_{N,\delta }>K$. ThusThis constructs continuous near-maximizers, rather than using the discontinuous sign of the kernel as the desired function.