Doubt about convolution in the definition of a bump function

Question

Let be $\Omega\subset \mathbb{R}^n $ an open and let be $\Omega'\subset \subset \Omega$ open.

I want to find a function $\varphi\in C^\infty (\Omega)$ with ${\rm supp}\subset \Omega$ such that:

• $0\leq\varphi\leq1$
• $\varphi\equiv 1$ in $\Omega'$

I know that one such function is $\phi=\mathbb{1}_{\bar\Omega}*\rho _\epsilon$, with $\Omega \subset \subset \bar \Omega \subset \subset \Omega$ and $\rho_\epsilon$ the standard mollifier.

My question Why is not right the function $\phi=\mathbb{1}_{\Omega'}*\rho _\epsilon$? Why this function is not $1$ on $\Omega'$?

Thank you!

Answer 1

I will try to explain with an example. Let $\Omega=(-2,2)$ and $\Omega'=(-1,1)$.

As a mollifier $\rho$, we take the "concrete example" of this wikipedia page.

$$\rho(x) = \begin{cases} e^{-1/(1-|x|^2)}/I_n& \text{ if } |x| < 1\\ 0& \text{ if } |x|\geq 1 \end{cases}$$

Where $I_n$ is a coefficient of normalization so the integral is $1$. As we can see, this mollifier takes its support in $(-1,1)$. When we do the convolution product, there is only one point where all the support is contained in the support of $\mathbb{1}_{\Omega'}$. Then there will be only one point where the value is 1.

Because an image worths thousands words, here is the image :

In red you have the indicator function of $\Omega'$, in blue the mollifier moving around, in green the convolution product and the point shows the value of the convolution product depending on the position of the blue curve.

As you can see, the problem is "the thickness" of the mollifier. But you will always have some thickness, so the key is to take an indicator function of a set slightly bigger than $\Omega'$. You can also try to decrease the "thichkness" of your mollifier.

For this example, we can take the indicator function of $(-1.5,1.5)$ and as a new mollifier $\tilde\rho (x)=2\rho(2x)$. And now we have what we want :

Answer 2

The real problem is when $x$ tends to boundary $\partial \Omega '$, the bump function $\phi(x) <1$, i.e, take $\Omega'=(0,1)$, note that

$$ \phi(x)=\int 1_{\Omega'}(x-y)\rho_{\varepsilon}(y)dy = \int_{x-1}^x \rho_{\varepsilon}(y)dy = \frac{1}{\varepsilon}\int_{x-1}^x \rho(y/\varepsilon)dy $$

Taking, for instance $x=1$, and doing a changes of variable, we have that

$$ \phi(1)=\frac{1}{\varepsilon}\int_{0}^1 \rho(y/\varepsilon)dy = \int_{0}^{1/\varepsilon} \rho(w)dw < 1, \quad \forall \varepsilon >0. $$

Recall that $\rho$ is a function $C_c^{\infty}$ with $supp \rho = B_1(0)$ and

$$\int_{-1}^{1} \rho(w)dw = 1$$.