Kernels

A sequence of kernels $K_n(x)$ on $[-\pi ,\pi ]$ is called *good* if

1. $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{K}_n(x)dx=1$,
2. $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }|K_n(x)|dx$ is bounded, and
3. $\frac{1}{2\pi }{\int }_{|x|\ge c}|K_n(x)|dx\to 0$ as $n\to \mathrm{\infty }$ for each $c>0$.

Theorem. Suppose $|f|\le B$ 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 ${\int }_{-\pi }^{\pi }|K_n(y)|dy\le C$ uniform in all $n$ by the second property.  Pick $ϵ>0$. By continuity, there is some $\delta >0$ such that $|y–x|<\delta$ implies $|f(y–x)–f(x)|<ϵ$.  We have

$$\begin{array}{rcl}2\pi |(K_n\star f)(x_0)–f(x_0)|& =& |{\int }_{-\pi }^{\pi }{K}_n(y)f(x_0–y)dy–{\int }_{-\pi }^{\pi }{K}_n(y)f(x_0)dy|\\ & =& |{\int }_{-\pi }^{\pi }{K}_n(y)(f(x_0–y)–f(y))dy|\\ & \le & {\int }_{-\pi }^{\pi }|K_n(y)||f(x_0–y)–f(y)|dy\\ & \le & {\int }_{y<\delta }|K_n(y)||f(x_0–y)–f(y)|dy+{\int }_{y\ge \delta }|K_n(y)||f(x_0–y)–f(y)|dy\\ & \le & ϵ{\int }_{y<\delta }|K_n(x)|dy+2B{\int }_{y\ge \delta }|K_n(y)|dy\\ & \le & ϵC+2B{\int }_{y\ge \delta }|K_n(y)|dy.\end{array}$$

By the third property, the second term converges to zero. So the original quantity can be made arbitarily small. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

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 $x_0$. $\mathrm{<img\; draggable="false"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/11/svg/25fc.svg">}$

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:

$$\begin{array}{rcl}2\pi |(D_n\star f)(x)–f(x)|& =& |{\int }_{-\pi }^{\pi }{D}_n(t)f(x-t)dt-{\int }_{-\pi }^{\pi }{D}_n(t)f(x)dt|\\ & =& |{\int }_{-\pi }^{\pi }{D}_n(t)(f(x-t)–f(x))dt|\\ & =& |{\int }_{-\pi }^{\pi }\frac{\sin ((n+1/2)t)}{\sin (t/2)}(f(x-t)–f(x))dt|\\ & =& |{\int }_{-\pi }^{\pi }\frac{f(x-t)–f(x)}{\sin (t/2)}\sin ((n+1/2)t)dt|.\end{array}$$

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 $n\to \mathrm{\infty }$.

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 $\underset{t\to 0}{lim sup}|h(t)|\le \underset{t\to 0}{lim sup}|Kt/\sin (t/2)|=2K$ 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.