Does this sequence converge uniformly?

Question

We consider some sequence $(f^{(k)})_{k\in \mathbb{N}}\subset L^p(\mathbb{R}^3)$ such that $f^{(k)}\to f$ in $L^p(\mathbb{R}^3)$. Further assume there is some $R>0$ so that $\text{supp}(f^{(k)}),\text{supp}(f)\subset B_R(0)$ for all $k\in \mathbb{N}$. Now consider $$f_n^{(k)}=f^{(k)}\ast \psi_{\varepsilon_n},$$ where $\psi_{\varepsilon_n}=\frac{1}{\varepsilon^n}\psi(\frac{x}{\varepsilon_n})$ and $\psi$ is the standard mollifier. Is it true that $$f_n^{(k)}\to f^{(k)}$$ uniformly (in $k$) in $L^p$?

Answer 1

Here is a different (and kind of high level) approach. Working through the standard proof of convergence of mollification seems to be hard. One would need a uniformity for the density argument.

Let me denote by $T_n$ the mollification associated with $\epsilon_n$. We have $T_ny \to y$ in $L^p$ for all $y\in L^p$ and $f_k \to f$ in $L^p$. Moreover, $\|T_n\|_{\mathcal L(L^p,L^p)}\le 1$ by standard estimates.

Now $$ T_nf_k - f_k = T_n(f_k-f) + (T_nf-f) + (f_k-f). $$ Due to $\|T_n\|\le 1$ the first and third term converge to zero for $k\to\infty$ uniformly in $n$. The middle one converges to zero for $n\to \infty$: Let $\epsilon>0$. Then there is $K$ such that $\|f_k-f\|<\epsilon/3$ for all $k>K$. In addition, there is $N$ such that $\|T_nf-f\|<\epsilon/3$ and $\|T_nf_k-f_k\|<\epsilon$ for all $n>N$ and all $k=1\dots K$. Hence $\|T_nf_k-f_k\|<\epsilon$ for all $n>N$ and all $k$.

Answer 2

Let $\delta>0$. Due to the density of $C_0(\mathbb{R}^3)$ in $L^p(\mathbb{R}^3)$ there is $(f_k)_{\delta}\in C_0(\mathbb{R}^3)$ so that $$ \|f_k-(f_k)_{\delta}\|_{L^p(\mathbb{R}^3)}<\frac{\delta}{3}.$$ Moreover we know that $\text{supp}(f_k),\text{supp}((f_k)_{\delta})\subset \overline{B_{\tilde{R}(0)}}$ for all $k\in \mathbb{N}$ and for some $\tilde{R}>0$ big enough. Further for any $x\in B_R(0)$ $$|(f_k)_{\delta}(x)-((f_k)_{\delta})_{\varepsilon}(x)|\leq \int_{B_R(0)}\psi_{\varepsilon}(x-y)|(f_k)_{\delta}(x)-(f_k)_{\delta}(y)|dy\\ \leq \sup_{x\in B_R(0),y\in B_{\varepsilon}(x)}|(f_k)_{\delta}(x)-(f_k)_{\delta}(y)|<\frac{\delta}{3}$$ for fixed $k$. And further $$\|(f_k)_{\varepsilon}-((f_k)_{\delta})_{\varepsilon}\|_{L^p}=\|(f_k)-((f_k)_{\delta})\|_{L^p}<\frac{\delta}{3}.$$ Then by triangle inequality we have $$\|f_k-(f_k)_{\varepsilon}\|_{L^p}\leq \|f_k-(f_k)_{\delta}\|_{L^p}+\|(f_k)_{\delta}-((f_k)_{\delta})_{\varepsilon}\|_{L^p}+\|((f_k)_{\delta})_{\varepsilon}-(f_k)_{\varepsilon}\|_{L^p}.$$ In the first step I first fix $f_k$ and then choose $\delta$ so it is dependent on $k$. In the second step the same again. And for the third estimate I use the first one so it also depends on $k$. How can I get the independence of $k$ now?