What is $\eta (x) = \int_{\mathbb{R}^n} \chi_{G_{2 \varepsilon}} (y) \rho_{\varepsilon} (x - y) dy$? Is it about "smoothing" non-smooth functions?

Question

Let $G_{\varepsilon} = \bigcup_{x \in G} K(x, \varepsilon)$ and $G \subset \mathbb{R}^n$

We've got $$\eta (x) = \int_{\mathbb{R}^n} \chi_{G_{2 \varepsilon}} (y) \rho_{\varepsilon} (x - y)\ \mathrm dy$$ where $$\rho_{\varepsilon} (x) = 0 \text{ for } |x| \ge \varepsilon$$ $$\rho_{\varepsilon} (x) = \frac{k}{\varepsilon^n} \exp\left(\frac{- \varepsilon}{\varepsilon^2 - |x|^2}\right) \text{ for } |x| < \varepsilon$$ where $k = \left(\displaystyle\int_{\overline{K}(0,1)} \exp\left(\frac{1}{|t|^2 - 1}\right)\ \mathrm dt\right)^{-1}$

Our Prof told us that $\rho_{\varepsilon}$ has something to do with "smotthing" non-smooth functions, so I guess $\eta$'s purpose should be similar?

Does anyone recognize what our Prof was talking about? And could maybe someone demonstrate the use of $\rho_{\varepsilon}$ and $\eta$ on an example?

Answer 1

This is a convolution, $(f*g)(x)=\int f(y) \, g(x-y) \, dy.$ Such can be used to smooth functions.

Let $f$ be the function to smooth and $g$ be a smooth nascent delta function, e.g. a smooth function with compact support such that $\int \rho(x) \, dx = 1.$ Then $f*\rho$ will be smooth and almost equal to $f.$ In your post, $f=\chi_{G_{2 \varepsilon}}$ and $\rho=\rho_{\varepsilon}.$

I was looking for pictures and the best I found was in this question on Math.SE.

Answer 2

I think what you eventually want to get is $\eta_\epsilon(x)\rightarrow \chi_G$ in $L^p(\mathbb R^n)$ (if $G$ is a bounded set) so this is a smooth approximation of the indicator function. The following theorem states exactly that in a more general setting:

Theorem: Let $\rho\in \mathcal D(\mathbb R^n)$ with $\rho\ge0$, $\int_{\mathbb R^n}\rho(x)dx=1$ and define $\rho_\epsilon(x)=\frac{1}{\epsilon^n}\rho(\frac{1}{\epsilon})$. Then for all $g\in L^p(\mathbb R^n)$, $\eta_\epsilon:=\rho_\epsilon\star g\rightarrow g$ in $L^p(\mathbb R^n)$ and $\eta_\epsilon $ is smooth.

Proof (of convergence):

$$\begin{aligned} |\rho_\epsilon\star g(x)-g(x)|&=\bigg|\int_{\mathbb R^n}\rho_\epsilon(y)(g(x-y)-g(x))dy\,\bigg|\\ &=\int_{\mathbb R^n}\rho(z)|g(x-\epsilon z)-g(x)|dz \end{aligned}$$

by setting $y=\epsilon z$. From Minkowski's inequality,

$$\begin{aligned} \Vert \rho_\epsilon\star g-g\Vert_{L^p}&\le \bigg[\int_{\mathbb R^n}\bigg|\int_{\mathbb R^n}\rho(z)|g(x-\epsilon z)-g(x)|dz\bigg|^pdx\bigg]^{\frac{1}{p}}\\ &\le\int_{\mathbb R^n}\bigg[\int_{\mathbb R^n}\rho(z)^p|g(x-\epsilon z)-g(x)|^pdx\bigg]^{\frac{1}{p}}dz\\ &=\int_{\mathbb R^n}\rho(z)\Vert g(\cdot-\epsilon z)-g\Vert_{L^p}dz\rightarrow 0\end{aligned}$$ as $z\rightarrow 0$ by dominated convergence theorem.

If $G$ is not bounded, you could instead use $L^p_{\text{loc}}(\mathbb R^n)$ space.