Convolution Norm Convergence Theorem
Theorem
Let the function $g \in L^{1}$ be bounded and satisfy $\int_{\mathbb{R}}g(y)dy=1$. Assuming $f\in L^{2}$ and that the convolution $f \ast g$ of $f$ and $g$ is well-defined for all $x\in \mathbb{R}$, then $f \ast g_{\epsilon}$ converges in norm to $f$.
$$ \begin{equation} \lim \limits_{\epsilon \to 0} \left\| f \ast g_{\epsilon} -f \right\| = 0 \label{eq1} \end{equation} $$
In this case, $g_{\epsilon}(y)=\frac{1}{\epsilon}g \left( \frac{y}{\epsilon} \right)$.
The name ‘Convolution Norm Convergence Theorem’ is arbitrarily given as this theorem does not have an official name. The Convolution Convergence Theorem demonstrates that $f \ast g_{\epsilon}(x)$ converges pointwise to $f(x)$, while this theorem shows that the function $f \ast g_{\epsilon}$ itself converges to $f$.
Proof
To demonstrate $\eqref{eq1}$, organize the equation as follows.
$$ \begin{align*} f \ast g_{\epsilon}(x)-f(x) &=\int f(x-y)g_{\epsilon}(y)dy-f(x)\int g_{\epsilon}(y)dy \\ &=\int \big[ f(x-y)-f(x)\big]g_{\epsilon}(y)dy \\ &=\int \big[ f(x-y)-f(x) \big] \frac{1}{\epsilon}g\left(\frac{y}{\epsilon} \right)dy \end{align*} $$
Substituting with $y=\epsilon z$ gives:
$$ \begin{align*} f \ast g_{\epsilon}(x)-f(x) &=\int \big[ f(x-y)-f(x) \big] \frac{1}{\epsilon}g\left(\frac{y}{\epsilon} \right)dy \\ &=\int \big[ f(x-\epsilon z)-f(x) \big] g\left(z\right)dz \\ &=\int \big[ T_{\epsilon z}f(x)-f(x) \big] g\left(z\right)dz \end{align*} $$
Minkowski’s Inequality for Integrals
For $1\le p < \infty$, given $f\in L^{p}$ and $g \in L^{1}$, the following inequality holds:
$$ \left\| \int f(\cdot,y)g(y)dy \right\|_{p} \le \int \left\| f(\cdot,y) \right\|_{p} |g(y)|dy $$
Hence, by Minkowski’s inequality:
$$ \begin{align*} \left\| f \ast g_{\epsilon}-f \right\|_{2} &= \left\| \int \big[ T_{\epsilon z}f-f \big] g\left(z\right)dz \right\|_{2} \\ &\le \int \left\| T_{\epsilon z }f-f \right\|_{2} \left| g(z) \right|dz \end{align*} $$
Since $g\in L^{1}$ and $\left\| T_{\epsilon z }f-f \right\|_{2}\le 2\left\| f \right\|_{2}$, $\left\| f \ast g_{\epsilon}-f \right\|_{2}$ is bounded.
For $1\le p <\infty$, given $f\in L^{p}$ and $z\in \mathbb{R}^{n}$, the following inequality holds:
$$ \lim \limits_{y\to 0} \left\| T_{y+z}f-T_{z}f \right\|_{p}=0 $$
Here, $T$ refers to translation.
With the above facts, the following holds:
$$ \lim \limits_{\epsilon \to 0} \left\| T_{\epsilon z }f-f \right\|_{2}=0 $$
Therefore, satisfying the conditions of the Dominated Convergence Theorem, we obtain the following equation, concluding the proof.
$$ \begin{align*} \lim \limits_{\epsilon \to 0} \left\| f \ast g_{\epsilon}-f \right\|_{2} &\le \lim \limits_{\epsilon \to 0} \int \left\| T_{\epsilon z }f-f \right\|_{2} \left| g(z) \right|dz \\ &\le \int\lim \limits_{\epsilon \to 0} \left\| T_{\epsilon z }f-f \right\|_{2} \left| g(z) \right|dz \\ &= \int 0 \cdot \left| g(z) \right|dz \\ &=0 \end{align*} $$
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