Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\epsilon$.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
Yeah, thanks for the in-depth response! – user1736 Dec 2 '10 at 6:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.