Convergence of $u * \eta_\epsilon$

Question

Let $\eta \in C_c^\infty(B(0,1)), \eta \ge 0, \eta$ radially symmetric and $\int_{\mathbb{R}^n} \eta d\mathcal{L}^n = 1$. $\eta_r := r^{-n} \eta(\frac{x}{r}) \in C_c^\infty(B(0,r))$. Integral of $\eta_r$ is still 1.

Now we are using a few convergences which I don't get in lecture:

$L^1$-convergence

Let $\Omega$ be a open subset of $\mathbb{R}^n$, $u \in L^1(\Omega), u_ \epsilon (x) := u*\eta_\epsilon (x)$ (the convolution). Then $u_\epsilon \to u$ in $L^1$ for $\epsilon \to 0$

Uniform convergence

Now let $u \in C_c^0(\mathbb{R}^n)$, then $u_\epsilon \to u$ uniformly (and thus also pointwise)

Also, out of interest: if you drop the compact support of $u$, would it still converge pointwise?

Thanks for helping!

Answer 1

Note that

\begin{equation} \begin{split} |u_\epsilon (x) - u(x)| &= \left|\int_B u(y) \eta_\epsilon (y-x) dy - u(x) \right|\\ &= \left|\int_B (u(y) - u(x)) \eta_\epsilon(y-x) dy \right|\\ \end{split} \end{equation}

But the support of $\eta_\epsilon$ is so small, so the integration is over where $|y-x|$ is small. (Then try to use the uniform continuity of $u$).