About mollifiers

Question

When we define mollifiers as in Evans's book, we use in one dimension say, $\phi_\epsilon=(1/\epsilon)\phi(x/\epsilon)$ where $\phi$ is standard mollifier. My question is what is the rate of convergence relative to $\epsilon$ for the norm $\|f_\epsilon-f\|_{L^1}$ where $f_\epsilon=\phi_\epsilon*f$?

Answer 1

The answer is not universal, it depends on the properties of the integrable function $f$. Suppose for example that $f$ is Lipschitz continuous with constant $K$. Then

$$||f_\epsilon-f||_1=\int_{-\infty}^{\infty}\frac{1}{\epsilon}\phi\left(\frac{t-t'}{\epsilon}\right)|f(t')-f(t)|dt'\leq K\epsilon\int_{-\infty}^{\infty}x\phi(x)dx$$

and the rate of convergence is at least $\mathcal{O}(\epsilon)$. If you choose $f$ to be a generic integrable function, then you can show that, in one dimension at least, the rate of convergence may be non-uniform and in fact arbitrarily slow as $\epsilon\to 0$ for certain points. A counterexample is given by the function $f(x)=|x|^{a}, 0<a<1$, which is integrable in any arbitrary interval containing the origin. Choose $t=0$ and then the norm in question reads

$$||f_\epsilon-f||_1=\epsilon^a\int_{-\infty}^{\infty}|x|^{a}\phi(x)dx$$

Also, the convergence rate may be sped up arbitrarily, if several derivatives of $f$ are zero at a certain point. In fact you can show that $||f_{\epsilon}(t_0)-f(t_0)||_1=\mathcal{O}(\epsilon^n)$ where $n$ is the order of the first non-zero derivative at $t=t_0$.