A sequence of kernels $K_n(x)$ on $[-\pi, \pi]$ is called *good* if
- $\frac{1}{2\pi} \int_{-\pi}^\pi K_n(x) \, dx = 1$,
- $\frac{1}{2\pi} \int_{-\pi}^\pi |K_n(x)| \, dx$ is bounded, and
- $\frac{1}{2\pi} \int_{|x| \geq c} |K_n(x)| \, dx \to 0$ as $n \to \infty$ for each $c> 0$.
Theorem. Suppose $|f| \leq B$ on $[-\pi, \pi]$, $K_n$ are good kernels and $f$ is continuous at $x_0$, then $(K_n\star f)(x_0) \to f(x)$, where $\star$ represents convolution.
Proof. We can choose some bound $\int_{-\pi}^\pi |K_n(y) |\, dy \leq C$ uniform in all $n$ by the second property. Pick $\epsilon > 0$. By continuity, there is some $\delta > 0$ such that $|y \, – x| < \delta$ implies $|f(y \, – x) – f(x)| < \epsilon$. We have
$$
\begin{eqnarray}
2\pi \left| (K_n \star f)(x_0) – f(x_0) \right| &=& \left| \int_{-\pi}^\pi K_n(y)f(x_0 – y) \, dy \, – \int_{-\pi}^{\pi} K_n(y)f(x_0) \, dy \right| \\
&=& \left| \int_{-\pi}^\pi K_n(y)(f(x_0 – y) – f(y)) \, dy \right| \\
&\leq& \int_{-\pi}^\pi |K_n(y)| |f(x_0 – y) – f(y)| \, dy \\
&\leq& \int_{y < \delta} |K_n(y)| |f(x_0 – y) – f(y)| \, dy +
\int_{y \geq \delta} |K_n(y)| |f(x_0 – y) – f(y)| \, dy \\
&\leq& \epsilon \int_{y < \delta} |K_n(x)| \, dy + 2B \int_{y \geq \delta} |K_n(y)| \, dy \\
&\leq& \epsilon C + 2B \int_{y \geq \delta} |K_n(y)| \, dy.
\end{eqnarray}
$$
By the third property, the second term converges to zero. So the original quantity can be made arbitarily small. $\blacksquare$
Theorem. If $f$ is continuous on $[-\pi, \pi]$, then this convergence is uniform.
Proof. If $f$ is continuous on a closed interval, it’s uniformly continuous there, so the $\delta$ used in the above proof does not depend on $x_0$. $\blacksquare$
Pointwise Convergence
The Dirichlet kernel $D_n$ can be convolved with $f$ to yield the partial sums $S_n$ of the Fourier series of $f$. While it fails to obey property 2. above, we can still use it to show pointwise convergence of the Fourier series at any point where $f$ is differentiable:
$$
\begin{eqnarray}
2\pi |(D_n \star f)(x) – f(x)|
&=& \left| \int_{-\pi}^{\pi} D_n(t)f(x-t) dt -\int_{-\pi}^{\pi} D_n(t)f(x) dt \right| \\
&=& \left| \int_{-\pi}^{\pi} D_n(t)(f(x-t) – f(x)) dt \right| \\
&=& \left| \int_{-\pi}^{\pi} \frac{\sin((n+1/2)t)}{\sin(t/2)} (f(x-t) – f(x)) dt \right| \\
&=& \left| \int_{-\pi}^{\pi} \frac{f(x-t) – f(x)}{\sin(t/2)} \sin((n+1/2)t) dt \right|.
\end{eqnarray}
$$
By $f$’s differentiability at $x$ and l’Hopital’s rule, the left factor of the integrand is continuous at $t=0$, hence integrable on $[-\pi, \pi]$. Thus the Riemann-Lebesgue lemma implies the last integral tends to zero as $n \to \infty$.
In fact, this still holds if $f$ is merely Lipschitz in some neighborhood of $x$. For then if $h(t) = \frac{f(x-t)-f(x)}{\sin(t/2)}$, we have $\limsup_{t\to 0} |h(t)| \leq \limsup_{t\to 0}|Kt/\sin(t/2)| = 2K$ and so $h$ is integrable, hence Riemann-Lebesgue still applies.
The Fejer kernel $F_n(t) = \frac{1}{n}\sum_{k=0}^{n-1} D_n(t)$ is a good kernel. Hence $F_n \star f$ converges uniformly to $f$ on $[-\pi, \pi]$; but this convolution can be written as a linear combination of the form $a_n e^{inx} \int_{-\pi}^{\pi} e^{-int}f(t) dt$, it follows that all integrable functions can be uniformly approximated by trigonometric polynomials.