Convolution Convergence Theorem
Theorem
Let’s assume that a function $g \in L^{1}$ satisfies the following condition.
$$ \begin{align*} \int_{\mathbb{R}}g(y)dy &= 1 \\ \int_{-\infty}^{0}g(y)dy &= \alpha \\ \int_{0}^{\infty}g(y)dy &=\beta \\ \alpha+\beta &= 1 \end{align*} $$
Furthermore, let’s say $f$ is piecewise continuous on $\mathbb{R}$. And either $f$ is bounded, or $g$ outside any interval $[-a,a]$ is $g=0$. That is, the convolution $f \ast g(x)$ is well-defined for all $x\in \mathbb{R}$. Now, let’s assume for $\epsilon >0$, we have $g_{\epsilon}(y)=\frac{1}{\epsilon}g(\frac{y}{\epsilon})$. Then, the below equation holds.
$$ \lim \limits_{\epsilon \to 0}f \ast g_{\epsilon}(x)=\alpha f(x+)+\beta f(x-) ,\quad x\in\mathbb{R} $$
Here, $f(x+)$ and $f(x-)$ are respectively the upper and lower limits of $f$ at $x$. Especially if $f$ is continuous at $x$, the following equation holds.
$$ \lim \limits_{\epsilon \to 0} f \ast g_{\epsilon}(x)=f(x) $$
Moreover, if $f$ is continuous on some closed interval, then the mentioned convergence is uniform.
The name ‘Convolution Convergence Theorem’ is tentatively assigned due to the absence of a specific name for the theorem. It serves as a useful auxiliary theorem in Fourier analysis and distribution theory.
Proof
The equation we need to demonstrate is as follows.
$$ \lim \limits_{\epsilon \to 0} \left| f \ast g_{\epsilon}(x)-\alpha f(x+)-\beta f(x-) \right|=0 $$
Arranging the inside of the absolute value gives the following.
$$ \begin{align} &f \ast g_{\epsilon}(x)-\alpha f(x+)-\beta f(x-) \nonumber \\ =&\ \int_{-\infty}^{\infty}f(x-y)g_{\epsilon}(y)dy- \int_{-\infty}^{0}g(y)dyf(x+)- \int_{0}^{\infty}g(y)dyf(x-) \nonumber \\ =&\ \int_{-\infty}^{\infty}f(x-y)g_{\epsilon}(y)dy- \int_{-\infty}^{0}g_{\epsilon}(y)dyf(x+)- \int_{0}^{\infty}g_{\epsilon}(y)dyf(x-) \nonumber \\ =&\ \int_{-\infty}^{0}\big[ f(x-y)-f(x+) \big]g_{\epsilon}(y)dy+\int_{0}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \label{eq1} \end{align} $$
Now, let’s suppose an arbitrary positive number $\delta >0$ is given. Then, due to the definition of lower and upper limits,
$$ \begin{equation} 0<y<c \implies \left| f(x-y)-f(x\pm) \right| <\delta \label{eq2} \end{equation} $$
there exists $c>0$ that satisfies the condition. Now, let’s solely arrange the second term of $\eqref{eq1}$. The integration interval can be divided as follows.
$$ \begin{align} &\int_{0}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \nonumber \\ =&\ \int_{0}^{c}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy+\int_{c}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \label{eq3} \end{align} $$
First, let’s investigate the first term of $\eqref{eq3}$. From the condition $\eqref{eq2}$, we can obtain the following equation.
$$ \begin{align*} \left| \int_{0}^{c}\big[ f(x-y)-f(x-)\big]g_{\epsilon}(y)dy \right| &< \int_{0}^{c}\delta \left| g_{\epsilon}(y) \right|dy \\ &=\delta\int_{0}^{c/\epsilon} \left| g(y) \right|dy \end{align*} $$
Therefore,
$$ \begin{equation} \lim\limits_{\epsilon \to 0}\left| \int_{0}^{c}\big[ f(x-y)-f(x-)\big]g_{\epsilon}(y)dy \right|=0 \label{eq4} \end{equation} $$
The process of dealing with the remaining integration interval can be divided into two cases.
Case 1. If $f$ is bounded
The second term of $\eqref{eq3}$ can be arranged as below due to the boundedness of $f$. Let’s say it is $\left| f \right| \le M$. Then, $$ \begin{align*} \left| \int_{c}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \right| & \le 2M \left|\int_{c}^{\infty} g_{\epsilon}(y)dy \right| \\ & \le 2M\int_{c/\epsilon}^{\infty}\left| g(y) \right|dy
\end{align*} $$Therefore,
$$ \begin{equation} \lim \limits_{\epsilon \to 0} \left| \int_{c}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \right| =0 \label{eq5} \end{equation} $$
Case 2. When $\left| x \right|>a$ for $g(x)=0$
Then, whenever $\left| x \right|>\epsilon a $, we have $g_{\epsilon}(x)=0$. Then for sufficiently small $\epsilon$,
$$ \left| x \right|>c \implies g_{\epsilon}(x)=0 $$
This holds. Therefore,
$$ \begin{equation} \lim \limits_{\epsilon \to 0} \left| \int_{c}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \right| =0 \label{eq6} \end{equation} $$
Thus, from $\eqref{eq3}$, $\eqref{eq4}$, $\eqref{eq5}$, and $\eqref{eq6}$, we get the following result.
$$ \lim \limits_{\epsilon \to 0}\left| \int_{0}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \right|=0 $$
Applying the same method to the first term of $\eqref{eq1}$ gives us the next.
$$ \lim \limits_{\epsilon \to 0}\left| \int_{-\infty}^{0}\big[ f(x-y)-f(x+) \big] g_{\epsilon}(y)dy \right|=0 $$
Therefore,
$$ \begin{align*} &\lim \limits_{\epsilon \to 0}\left| f \ast g_{\epsilon}(x)-\alpha f(x+)-\beta f(x-) \right| \\ \le& \lim \limits_{\epsilon \to 0}\left| \int_{-\infty}^{0}\big[ f(x-y)-f(x+) \big]g_{\epsilon}(y)dy \right| \\ &+\lim \limits_{\epsilon \to 0} \left| \int_{0}^{\infty}\big[ f(x-y)-f(x-) \big] g_{\epsilon}(y)dy \right| \\ &= 0 \end{align*} $$
Since this holds,
$$ \lim \limits_{\epsilon \to 0}f \ast g_{\epsilon}(x)=\alpha f(x+)+\beta f(x-) ,\quad x\in\mathbb{R} $$
Also, if $f$ is continuous at $x$, since $f(x+)=f(x-)$ and $\alpha+\beta =1$, then
$$ \lim \limits_{\epsilon \to 0} f \ast g_{\epsilon}(x)=f(x) $$
At this point, a bounded closed interval is compact, and if $f$ is continuous on some compact set, it is uniformly continuous. Thus, in selecting $c$ above, it becomes independent of $x$, which indicates that $f \ast g_{\epsilon}$ converges uniformly to $f$.
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