Convergence of Mollification
Theorem
Let’s assume the following holds for a Mollifier $\eta_{\epsilon}$.
- $\displaystyle \alpha= \int_{-\infty}^{0} \eta_{\epsilon}(x) dx$
- $\displaystyle \beta=\int_{0}^{\infty} \eta_{\epsilon} (x) dx$
- Let’s say $\alpha + \beta = 1$.
And say $f$ is piecewise continuous and bounded. Then, the mollification of $f$ converges as follows.
$$ \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x) = \alpha f(x+) + \beta f(x-) $$
If $\eta_{\epsilon}(x)$ is an even function,
$$ \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x) = \dfrac{1}{2} \big[ f(x+) + f(x-) \big] $$
Furthermore, if $f$ is a continuous function,
$$ \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x) = f(x) $$
Explanation
Even if $f$ is not sufficiently smooth, the mollification of $f$ becomes a smooth function that converges to $f$.
Auxiliary theorem
$$ \int_{a}^b \eta_{\epsilon}(x)dx=\int_{a/\epsilon}^{b/\epsilon}\eta (x)dx $$
Proof
As a corollary
It holds by the Convolution Convergence Theorem.
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By direct calculation
Given the assumptions, the following holds.
$$ \begin{align*} f \ast\eta_{\epsilon}(x) -\alpha f(x+)-\beta f(x-) &= \int_{-\infty}^{\infty} f(x-y)\eta_{\epsilon}(y)dy - \int_{-\infty}^{0} f(x+)\eta_{\epsilon}(y)dy - \int_{0}^{\infty} f(x-)\eta_{\epsilon}(y)dy \\ &= \int_{-\infty}^{0} \big[ f(x-y) -f(x+) \big] \eta_{\epsilon}(y)dy - \int_{0}^{\infty} \big[ f(x-y) - f(x-)\big] \eta_{\epsilon}(y)dy \end{align*} $$
First, let’s demonstrate that the second term’s integral converges to $0$. Given $\delta >0$ and let’s say $N=\int |\eta_{\epsilon}(y) | dy$. Then, for every $0<y<c$, one can choose sufficiently small $c$ to satisfy the following equation.
$$ |f(x-y)-f(x-)| < \dfrac{\delta}{2N} \quad \mathrm{when}\ 0<y<c $$
Then, think of splitting the integral into $\displaystyle \int_{0}^{\infty}=\int_{0}^{c} +\int_{c}^{\infty}$. Looking at $\int_{0}^{c}$ first, by the above inequality, the following holds.
$$ \begin{align*} \left| \int_{0} ^{c} \big[ f(x-y)-f(x-) \big] \eta_{\epsilon}(y)dy \right| &\le \dfrac{\delta}{2N}\int_{0}^{c}|\eta_{\epsilon}(y)|dy \\ &= \dfrac{\delta}{2N}N \\ &=\dfrac{\delta}{2} \end{align*} $$
Now, since $f$ is bounded, let’s say $ f(x) \le M$. Since $\int | \eta (y)|dy=1<\infty$, there exists a positive number $R>0$ that satisfies the following equation.
$$ \int_{R^{\prime}}^{\infty}|\eta (y)|dy < \dfrac{\delta}{4M} \quad \mathrm{for}\ R^{\prime}>R $$
Let’s say $\epsilon < c/R$. Then, since $R< c/\epsilon$, $\int_{c/\epsilon}^{\infty} |\eta (y)|dy <\frac{\delta}{4M}$. Therefore, the following holds.
$$ \begin{align*} \left| \int_{c}^{\infty} \big[ f(x-y)-f(x-) \big] \eta_{\epsilon}(y)dy \right| &\le 2M \int _{c}^{\infty} | \eta_{\epsilon}(y) | dy \\ &= 2M \int _{c/\epsilon}^{\infty} | \eta (y) | dy \\ &\le 2M\dfrac{\delta}{4M} \\ &= \dfrac{\delta}{2} \end{align*} $$
Combining the two results, we obtain the following.
$$ \left| \int_{0}^{\infty} \big[ f(x-y)-f(x-)\big]\eta_{\epsilon}(y)dy \right|<\delta $$
Since $\delta$ is any positive number, the value of the above equation is $0$. The case of $\displaystyle \int_{-\infty}^{0}$ can also be proven in the same way. Therefore, we obtain the following.
$$ \begin{align*} && \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x) -\alpha f(x+)-\beta f(x-) &= 0 \\ \implies && \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x) &=\alpha f(x+)+\beta f(x-) \end{align*} $$
If the mollifier $\eta (x)$ is an even function, obviously since $\alpha=\beta=\frac{1}{2}$, the following holds.
$$ \lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x) = \dfrac{1}{2} \big[ f(x+) + f(x-) \big] $$
Furthermore, when $f(x)$ is continuous, in the same way,
$$ f \ast\eta_{\epsilon}(x)-f(x) <\delta $$
can be demonstrated, ultimately showing that $\lim \limits_{\epsilon \rightarrow 0} f \ast \eta_{\epsilon}(x)=f(x)$.
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