Proposition 3.1 (Properties of Convolutions)
Proposition 3.1. Suppose that \(f\), \(g\), and \(h\) are \(2\pi\)-periodic (Riemann) integrable functions. Then:
- Distributivity: \(f*(g+h)=(f*g)+(f*h)\)
- Scalar multiplication: \((cf)*g=c(f*g)=f*(cg)\) for any \(c\in\mathbb{C}\).
- Commutativity: \(f*g=g*f\).
- Associativity: \((f*g)*h=f*(g*h) \).
- Continuity: \(f*g\) is continuous.
- Fourier coefficients of a convolution
\(\widehat{f*g}(n)=\widehat{f}(n)\,\widehat{g}(n).\)
Proof (III)
Start with the definition:
$$
\begin{align*}
(f*g)(x) = \frac{1}{2\pi}\int_{0}^{2\pi}f(y)g(x-y)\,dy.
\end{align*}
$$
Our goal is to transform this expression into
$$
\begin{align*}
(g*f)(x) = \frac{1}{2\pi}\int_{0}^{2\pi}g(z)f(x-z)\,dz.
\end{align*}
$$
The first expression has \(f(y)g(x-y)\). the desired expression has \(g(z)f(x-z)\). This suggests using the substitution
$$
\begin{align*}
z=x-y
\end{align*}
$$
differentiate both sides, we get \(dz=-dy\) so \(dy=-dz\). When \(y=0\), \(z=x-0=x\) and when \(y=2\pi\), \(z=x-2\pi\).
Therefore, the bounds change as follows:
$$
\begin{align*}
y:0\longrightarrow 2\pi
\qquad\Longrightarrow\qquad
z:x\longrightarrow x-2\pi.
\end{align*}
$$
Now substitute \(y=x-z\), \(x-y=z\) and \(dy=-dz\) into
into the convolution integral:
$$
\begin{align*}
(f*g)(x)
=
-\frac{1}{2\pi}
\int_{x}^{x-2\pi}
f(x-z)g(z)\,dz \\
&=
\frac{1}{2\pi}
\int_{x-2\pi}^{x}
f(x-z)g(z)\,dz.
\end{align*}
$$
For fixed \(x\), define
$$
\begin{align*}
F(z)=f(x-z)g(z).
\end{align*}
$$
This function is \(2\pi\)-periodic because
$$
\begin{align*}
F(z+2\pi)
&=f\bigl(x-(z+2\pi)\bigr)g(z+2\pi)\\
&=f(x-z-2\pi)g(z)\\
&=f(x-z)g(z)\\
&=F(z).
\end{align*}
$$
Here we used the \(2\pi\)-periodicity of both \(f\) and \(g\). Because \(F(z)=f(x-z)g(z)\) is \(2\pi\)-periodic, its integral is the same over any interval of length \(2\pi\).
Both \([x-2\pi,x]\) and \([0,2\pi]\) have length \(2\pi\), so
$$
\begin{align*}
\int_{x-2\pi}^{x}f(x-z)g(z)\,dz
=
\int_{0}^{2\pi}f(x-z)g(z)\,dz.
\end{align*}
$$
Therefore,
$$
\begin{align*}
(f*g)(x) =
\frac{1}{2\pi}
\int_{0}^{2\pi}f(x-z)g(z)\,dz.
\end{align*}
$$
Proof (v)
We want to show \(f*g\) is continuous. Fix any point \(x_0\). By the definition of continuity, we need to show
$$
\begin{align*}
\lim_{x\to x_0}(f*g)(x)=(f*g)(x_0).
\end{align*}
$$
Equivalently, we need to show
$$
\begin{align*}
\left|(f*g)(x)-(f*g)(x_0)\right|\longrightarrow 0
\qquad\text{as }x\to x_0.
\end{align*}
$$
Because \(g\) is continuous and \(2\pi\)-periodic, it is uniformly continuous. Therefore, for every \(\varepsilon>0\), there exists a \(\delta>0\) such that
$$
\begin{align*}
|u-v|<\delta
\quad\Longrightarrow\quad
|g(u)-g(v)|<\varepsilon.
\end{align*}
$$
We will apply this with
$$
\begin{align*}
u=x-y,
\qquad
v=x_0-y.
\end{align*}
$$
Notice that
$$
\begin{align*}
|u-v|
=|(x-y)-(x_0-y)|
=|x-x_0|.
\end{align*}
$$
Thus, whenever \(|x-x_0|<\delta\), we have
$$
\begin{align*}
|g(x-y)-g(x_0-y)|<\varepsilon
\end{align*}
$$
for every \(y\). Next, define
$$
\begin{align*}
B=\int_{0}^{2\pi}|f(y)|\,dy
\end{align*}
$$
Because \(f\) is integrable, \(B\) is finite. This number measures the total size of \(f\) over one period. We will use it to bound the convolution difference. Now, observe that by definition of convolution, we have
$$
\begin{align*}
\left|(f*g)(x)-(f*g)(x_0)\right|
&= \left| \frac{1}{2\pi}\int_{0}^{2\pi}f(y)g(x-y)\,dy - \frac{1}{2\pi}\int_{0}^{2\pi}f(y)g(x_0-y)\,dy
\right| \\
&=
\frac{1}{2\pi}
\left|
\int_{0}^{2\pi}
f(y)\bigl[g(x-y)-g(x_0-y)\bigr]\,dy
\right| \\
&\leq
\frac{1}{2\pi}
\int_{0}^{2\pi}
\left|
f(y)\bigl[g(x-y)-g(x_0-y)\bigr]
\right|\,dy \\
&\leq
\frac{1}{2\pi}
\int_{0}^{2\pi}
|f(y)|\,|g(x-y)-g(x_0-y)|\,dy \\
&\leq
\frac{1}{2\pi}
\int_{0}^{2\pi}
|f(y)|\varepsilon\,dy \\
\end{align*}
$$
Since \(\varepsilon\) does not depend on \(y\), pull it outside so
$$
\begin{align*}
\left|(f*g)(x)-(f*g)(x_0)\right|
&<
\frac{\varepsilon}{2\pi}
\int_{0}^{2\pi}|f(y)|\,dy \\
&=
\frac{\varepsilon}{2\pi}B\,dy.
\end{align*}
$$
Since \(B/(2\pi)\) is a fixed finite constant and \(\varepsilon\) can be made arbitrarily small, we obtain
$$
\begin{align*}
\left|(f*g)(x)-(f*g)(x_0)\right|\longrightarrow 0
\qquad\text{as }x\to x_0.
\end{align*}
$$
Therefore, \(\lim_{x\to x_0}(f*g)(x)=(f*g)(x_0)\). Because \(x_0\) was arbitrary, \(f*g\) is continuous.
Proof (vi)
Note that (vi) is the property necessary to study the Fourier partial sum.
$$
\begin{align*}
\widehat{f*g}(n)
=\frac{1}{2\pi}\int_{-\pi}^{\pi}(f*g)(x)e^{-inx}\,dx.
\end{align*}
$$
Substitute the definition of convolution:
$$
\begin{align*}
(f*g)(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)g(x-y)\,dy.
\end{align*}
$$
in
$$
\begin{align*}
\widehat{f*g}(n)
=\frac{1}{2\pi}\int_{-\pi}^{\pi}
\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)g(x-y)\,dy\right)
e^{-inx}\,dx.
\end{align*}
$$
Here \(y\) is the inner integration variable. \(x\) is held fixed while that inner integral is evaluated. Next, since \(e^{inx}\) doesn’t depend on \(y\), we can insert it into the integral as follows
$$
\begin{align*}
\widehat{f*g}(n)
= \frac{1}{2\pi} \int_{-\pi}^{\pi}
\left(\frac{1}{2\pi}\int_{-\pi}^{\pi} f(y) g(x-y)e^{-inx} \,dy\right)
\,dx.
\end{align*}
$$
Next split the exponent so that \(e^{-inx}=e^{-iny}e^{-in(x-y)}\). Then
$$
\begin{align*}
\widehat{f*g}(n)
=\frac{1}{2\pi}\int_{-\pi}^{\pi}
\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)e^{-iny}\,g(x-y)e^{-in(x-y)}\,dy\right)\,dx.
\end{align*}
$$
What we want to do next is to split the order of the integration, so integrate with respect to \(x\) first and then with respect to \(y\). We can do this by Fubini’s theorem (Since \(f\) and \(g\) are continuous, \(f(y)e^{-iny}\,g(x-y)e^{-in(x-y)}\) remains continuous on the rectangle \(-\pi\le x\le\pi\) and \(-\pi\le y\le\pi\) TODO more on this?). Therefore
$$
\begin{align*}
\widehat{f*g}(n)
=\frac{1}{2\pi}\int_{-\pi}^{\pi}
\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)e^{-iny}\,g(x-y)e^{-in(x-y)}\,dx\right)\,dy.
\end{align*}
$$
Next, pull \(f(y)e^{-iny}\) outside the inner integral, since it does not depend on \(x\) so
$$
\begin{align*}
\widehat{f*g}(n)
=\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)e^{-iny}
\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}
g(x-y)e^{-in(x-y)}\,dx\right)\,dy.
\end{align*}
$$
Let’s now focus on the inner integral:
$$
\begin{align*}
\frac{1}{2\pi}\int_{-\pi}^{\pi}g(x-y)e^{-in(x-y)}\,dx.
\end{align*}
$$
Let \(u=x-y\). Since \(y\) is fixed here, then \(du=dx\), and the bounds become
$$
\begin{align*}
x &=-\pi\Rightarrow u=-\pi-y \\
x &= \pi\Rightarrow u=\pi-y.
\end{align*}
$$
So the inner integral becomes
$$
\begin{align*}
\frac{1}{2\pi}\int_{-\pi-y}^{\pi-y}g(u)e^{-inu}\,du.
\end{align*}
$$
But now recall that \(g(u)\) is \(2\pi\) periodic. So we can use periodicity to shift the bounds back:
$$
\begin{align*}
\frac{1}{2\pi}\int_{-\pi-y}^{\pi-y}g(u)e^{-inu}\,du
=
\frac{1}{2\pi}\int_{-\pi}^{\pi}g(u)e^{-inu}\,du.
\end{align*}
$$
This works because \(g(u)e^{-inu}\) A periodic function has the same integral over any full period, and both intervals have length \(2\pi\). But now this is just the \(n\)th Fourier coefficient of \(g\). Hence
$$
\begin{align*}
\widehat{f*g}(n)
=\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)e^{-iny}\,\widehat g(n)\,dy.
\end{align*}
$$
Since \(\widehat g(n)\) does not depend on \(y\), we can pull it outside the integral and hence we get
$$
\begin{align*}
\widehat{f*g}(n) &=\widehat g(n)\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)e^{-iny}\,dy\right) \\
&= \widehat g(n) \widehat f(n).
\end{align*}
$$
So convolving \(f\) with \(g\) has a simple effect on its Fourier coefficients: each coefficient of \(f\) gets multiplied by the corresponding coefficient of \(g\).
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi