Building a function $f$ such that $\| f - f_n \|_{L^p(B(x,r) \cap \Omega)} \to 0$ as $n \to \infty$ and $f \in L^p_{\text{loc}}(\Omega)$.

Question

Consider an arbitrary open set $\Omega \subset \mathbb R^n$ and an arbitrary element $1 \leqslant p < \infty$. Moreover, let $(f_n)_{n \in \mathbb N} \subset L^p(B(x,r) \cap \Omega)$ denote a convergent sequence in $L^p(B(x,r) \cap \Omega)$, for every $x \in \Omega$ and $r > 0$. In other words, for each $x \in \Omega$ and $r > 0$ there exists a function $f_{x,r} \in L^p(B(x,r) \cap \Omega)$ such that

$$ \lim_{n \to \infty}\| f_{x,r} - f_n \|_{L^p(B(x,r) \cap \Omega)} = 0. $$

My goal is, if possible, to build a function $f$ such that

$$ \tag{1} f \in L^p_{\text{loc}}(\Omega) \quad \text{ and }\quad \lim_{n \to \infty}\| f - f_n \|_{L^p(B(x,r) \cap \Omega)} = 0, $$

for every $x \in \Omega$ and $r > 0$.

My attempt. Clearly, to prove the R.H.S condition of $(1)$, it is sufficient to establish that

$$ \lim_{n \to \infty} \| f - f_n \|_{L^p(B(x,r) \cap \Omega)} \leqslant \lim_{n \to \infty} \| f_{x,r} - f_n \|_{L^p(B(x,r) \cap \Omega)}, $$

for every $x \in \Omega$ and $r > 0$. Furthermore, simple calculations yield

$$ \| f - f_n \|_{L^p(B(x,r) \cap \Omega)} \leqslant \| f - f_{x,r} \|_{L^p(B(x,r) \cap \Omega)} + \| f_{x,r}- f_n \|_{L^p(B(x,r) \cap \Omega)}, $$

for every $n \in \mathbb N$ and $x \in \Omega, r > 0$. This inequality turns out to not provide a viable choice since the functions $f_{x,r}$ are not necessarily equal for different values of $x \in \Omega$ and $r > 0$. Hence, I am looking for an alternative approach to prove this result.

Thanks for any help in advance.

Answer 1

This question is much simpler than you thought. The key is that the following caution in your OP

This inequality turns out to not provide a viable choice since the functions $f_{x,r}$ are not necessarily equal for different values of $x\in \Omega$ and $r>0$.

is almost unnecessary.

The reason is that $L^p(X)$ for a measure space $X$ is a normed space, so limit in $L^p$ is unique. That is, two limit functions are equal almost everywhere.

Therefore, if $B(x_1, r_1)\cap B(x_2,r_2)\neq \emptyset$, then the limits $$ f_{x_1,r_1}=f_{x_2,r_2} \text{ a.e. in }B(x_1, r_1)\cap B(x_2,r_2), $$ since $f_n\to f_{x_1,r_1}$ and $f_n\to f_{x_2,r_2}$ in $L^p(B(x_1, r_1)\cap B(x_2,r_2))$. Therefore we have a function $$ F_2(y):=\begin{cases} f_{x_1,r_1}(y) & y\in B(x_1,r_1)\\ f_{x_2,r_2}(y) & y\in B(x_2,r_2)\backslash B(x_1,r_1) \end{cases} $$ defined on $B(x_1, r_1)\cup B(x_2,r_2)$. Furthermore, $F_2=f_{x_2,r_2}$ a.e. on $B(x_2,r_2)$.

Now since ${\mathbb R}^n$ is second countable, there exist a countable family of $\{B(x_i,r_i)\}$ such that $$ \Omega = \bigcup_{i=1}^\infty B(x_i, r_i). $$ Let $E_k=\bigcup_{i=1}^k B(x_i, r_i)$. By induction, we can build a sequence of functions $F_k$ on $E_k$, using $F_{k-1}$ on $E_{k-1}$ and $f_{x_k,r_k}$ on $B(x_k,r_k)\backslash E_{k-1}$, just as how we build $F_2$ above.

Eventually the sequence $F_k$ totally extend each other, and defines a function $F$ on $\Omega$ by $$ F(y) := F_k(y) \quad\text{if }y\in E_k. $$

Furthermore, by its construction and the uniqueness of $L^p$ limit, $$ F=f_{x_i,r_i} \text{ a.e. on }B(x_i,r_i). $$ Then $F$ is the limit $f$ you were looking for.

Now we show that $$ F\in L^p_{\text{loc}}(\Omega). $$ Let $K\subset \Omega$ be a compact subset. Then by the definition of compactness, there exists finitely many $B_j=B(x_{i_j}, r_{i_j}), j=1,\dots, N$ such that $K\subset \bigcup B_j$. Then $$ \int_K |F|^p \leq \sum_{j=1}^N \int_{B_j} |F|^j = \sum_{j=1}^N \int_{B(x_{i_j}, r_{i_j})} |f_{x_{i_j},r_{i_j}}|^p <\infty. $$

The last thing to do is to consider a general $B(x, r)$. But since $B(x, r)\cap \Omega\subset \bigcup_{i=1}^\infty B(x_i,r_i)$, by the uniqueness of $L^p$-limit, we see that the limit here $$ f_{x, r} = F \text{ a. e. on }B(x, r)\cap \Omega. $$ Therefore $$ \lim_{n\to \infty} \|f_n-F\|_{B(x,r)\cap \Omega}=\lim_{n\to \infty} \|f_n-f_{x,r}\|_{B(x,r)\cap \Omega} = 0,\quad \forall x\in \Omega, r>0. $$