Following Evan's PDE book, Appendix C4, PP. 629, let's define the function:

$\eta(x):=C\exp\left(\frac{1}{\|x\|^2 - 1} \right) \forall x \in \mathbb{R}^d$ when $\|x\|\leq 1$ and $0$ otherwise, where $C$ is a positive constant so that the integral of $\eta$ is $1$.

Define the mollifier $\eta_\varepsilon$ by :$\eta_\varepsilon(x): \frac{1}{\varepsilon^d}\eta(\frac{x}{\varepsilon})$.

Next, let's define the measure $P_\varepsilon$ by $dP_\varepsilon(x):=\eta_\varepsilon(x) \, dx$, where $dx$ denotes the Lebesgue measure on $\mathbb{R}^d$.

I'm trying to prove (if true?!), that:

$P_\varepsilon \to \delta_0$ weakly as $\varepsilon \to 0$, where $\delta_0$ denotes the Dirac measure at $0$. How do we prove this, if possible?

So to start, I want to prove that $\forall f$ continuous, bounded on $\mathbb{R}^d$, we must have:

\begin{align} & \int_{\mathbb{R}^d} f(x)\, dP_\varepsilon(x) = \int_{\mathbb{R}^d} f(x)\eta_\varepsilon(x) \, dx \\ \to {} & f(0)=\int_{\mathbb{R}^d} f(x) \, d\delta_0(x) \end{align} as $\varepsilon \to 0$. To achive this, I've done as follows:

\begin{align} & \int_{\mathbb{R}^d}f(x) \eta_{\varepsilon}(x)\,dx - f(0) \\ = {} & \int_{\mathbb{R}^d}(f(x)-f(0)) \eta_\varepsilon(x) \, dx \\ = {} & \int_{B(0;\varepsilon)}(f(x)-f(0)) \eta_\varepsilon(x) \, dx, \end{align}

as the support of $\eta_\varepsilon = \bar{B(0;\varepsilon)}$.

Fix any $\eta > 0$. Using the continuity of $f$ at $0$, we must have a $\varepsilon > 0$ so that for all $x$ with $\|x\|\leq \varepsilon$, $|f(x)-f(0)|\leq \eta$. Next, after we've taken the absolute values:

\begin{align} & \left|\int_{B(0;\varepsilon)}(f(x)-f(0)) \eta_\varepsilon(x) \, dx \right| \\ \leq {} & \int_{B(0;\varepsilon)}|f(x)-f(0)| |\eta_\varepsilon(x)| \, dx \\ = {} & \int_{B(0;\varepsilon)}|f(x)-f(0)| \eta_\varepsilon(x) \, dx \\ \leq {} & \eta \int_{B(0;\varepsilon)} \eta_\varepsilon(x)\,dx \\ \leq {} & \eta \int_{\mathbb{R}^d} \eta_\varepsilon(x) \, dx \\ = {} & \eta \end{align}

I think this finishes the proof, but if you could verify if it's correct, then it'd be greatly appreciated :)

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    $\begingroup$ You can't learn all of MathJax or LaTeX in a day, but see my edits for better usage. In particular, putting absolute-value delimiters outside of MathJax while the expression they enclosed was inside MathJax is contraindicated. $\endgroup$ Jan 27, 2020 at 16:05

1 Answer 1


Your proof looks good!

For the future, you may want to learn to use the align* environment to format multi-line equations; it looks better than what you have.

  • $\begingroup$ Thank you for checking, and fixed the typo! Yes, I'll definitely try to use the align* environment next time... $\endgroup$ Jan 27, 2020 at 15:06

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