# Integrating mollifiers and their limits

Suppose I have a function $f \in L^2(R^n)$ with compact support, and a mollifier $\phi(x) \in C^\infty(R^n)$ with compact support s.t. $\int \phi(x) dx = \int \phi_\epsilon(x) = 1$ and $lim_{\epsilon \to 0} \phi_\epsilon(x) = \delta(x)$ where $\phi_{\epsilon}(x) = \epsilon^{-n} \phi(x/\epsilon)$ and $\delta(x)$ denotes the dirac delta.

Let $f_\epsilon = f * \phi_\epsilon$, where $*$ denotes convolution.

In general, is it possible to pass the limit of the integral of $f_\epsilon$ inside and say that $\lim_{\epsilon \to 0} \int f * \phi_\epsilon = \int \lim_{\epsilon \to 0} f *\phi_\epsilon$?

And if it is not, what type of stronger condition would I need to be able to do that?

I thought about using the Dominated Convergence Theorem, but I wasn't sure how to come up with a dominating function that works for every $\epsilon$.

## 1 Answer

Note that $$\int f \ast \phi_{\epsilon}(x)\,dx= \int(\int f(x - y)\phi_{\epsilon}(y)\,dy)\,dx$$ So by applying Fubini's theorem and integrating with respect to $y$ first, $\int f \ast \phi_{\epsilon}(x)\,dx = \int f \int \phi_{\epsilon} = \int f$. So to show your equality, it suffices to show that $\lim_{\epsilon \rightarrow 0} f \ast \phi_{\epsilon}(x) = f(x)$ for almost all $x$. It's actually true at all Lebesgue points of $f(x)$ which is a standard fact about these mollifiers. So this gives you what you want.

Another way: Since your functions are compactly supported you can use that $\int |f - f \ast \phi_{\epsilon}| \leq C (\int |f - f \ast \phi_{\epsilon}|^2)^{1 \over 2}$. (This follows by Schwarz's inequality for example). Then you can use Plancherel's theorem on the right hand side and get that $$\int |f - f \ast \phi_{\epsilon}|^2 = \int |\hat{f} - \hat{\phi_{\epsilon}}\hat{f}|^2$$ Since $\phi_{\epsilon}$ are mollifiers, $\hat{\phi_{\epsilon}}(x) = \hat{\phi}(\epsilon x)$ converges to 1 for all $x$ as $\epsilon$ goes to zero. So $\hat{f} - \hat{\phi_{\epsilon}}\hat{f}$ converges to zero. Since $|\hat{f} - \hat{\phi_{\epsilon}}\hat{f}|^2$ has the dominating function $|2\hat{f}|^2$, one can now apply the dominated convergence theorem to the right-hand side above and get that zero is the limit. As a result, the left-hand side must also converge to zero. So since $\int |f - f \ast \phi_{\epsilon}| \leq C (\int |f - f \ast \phi_{\epsilon}|^2)^{1 \over 2}$, one also has that $\int |f - f \ast \phi_{\epsilon}|$ goes to zero. So the same is true for $|\int f - \int f \ast \phi_{\epsilon}| = |\int (f - f \ast \phi_{\epsilon})| \leq \int |f - f \ast \phi_{\epsilon}|$, which is what you are looking for.

• Yeah, thanks for the in-depth response! – user1736 Dec 2 '10 at 6:19