Why can't the Fourier partial sums of a continuous function tend to infinity at a point?

Question

I know that there exists a $2\pi$ periodic continuous function whose Fourier series diverges at a point.

However, I heard that for any $2\pi$ periodic continuous function $f(x)$, it is wrong that $\mid S_N(f,0)\mid \to \infty$ as $N$ goes to $\infty$.

Here $S_N(f,x)=1/2\pi\int_{-\pi}^{\pi}[\sin(N+1/2)(x-\theta)/\sin((x-\theta)/2)]f(\theta)d\theta$.

I tried with the $L^2$ convergence of $S_Nf$ to $f$. But this only shows that there exists 'some points' on which $S_Nf$ has a subsequence converging to $f$. Those points are not guaranteed to include $0$. So I am just stymied.

Answer 1

I don't know the answer in general. But if $f$ is real-valued it's very simple.

Easy Lemma. Suppose $(x_n)$ is a sequence of real numbers such that $|x_n|\to\infty$ and $|x_{n+1}-x_n|$ is bounded. Then $\left|\frac{x_1+\dots+x_n}n\right|\to\infty$.

(Hint: The hypothesis implies that either $x_n\to+\infty$ or $x_n\to-\infty$.)

Now set $$\sigma_n(f)=\frac{s_1(f)+\dots+s_n(f)}n$$as usual. If $f$ is a real-valued continuous function and $|s_n(f,0)|\to\infty$ then $|\sigma_n(f,0)|\to\infty$, which is impossible since $\sigma_n(f)=f*K_n\to f$ uniformly. (Where of course $K_n$ is the Fejer kernel.)

But the lemma is false for sequences of complex numbers...

(Say $C_j$ is the circle of radius $j$ centered at the origin. Let $(z_n)$ be a sequence consisting of one point on $C_1$, then two points evenly spaced on $C_2$, then three on $C_3$, etc. Then $(z_1+\dots+z_n)/n\to0$.)