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All function spaces are over $\mathbb{R}^3$.

Let $u_n \in C^\infty_0$, $u_n\rightarrow u$ in $L^1$. Let $v\in L^1_\text{loc}$ be such that $uv \in L^1$. Does $u_n v \rightarrow uv$ in $L^1$?

What if we additionally require $u_n\geq 0$ and $u\geq 0$? And/or $u_n\rightarrow u$ in $L^p$ for all $1\leq p \leq $ some max value?

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2 Answers 2

up vote 2 down vote accepted

Let $u_n = \chi_{B_{1/n}(n)} * \phi_{1/n^2}$, the characteristic function of $B_{1/n}(n)$ (the ball of radius $1/n$ around $n$, mollified with a standard mollifier. Then $\|u_n\|_1 = \operatorname{vol}(B_{1/n}) \to 0$, so $u_n \to 0=: u$ in $L^1$ (and, as $\|u_n\|_p \le \|\chi_{B_{1/n}}\|_p = \operatorname{vol}(B_{1/n})^{1/p} \to 0$ in every $L^p$ for $1 \le p < \infty$). Then let $v = \sum_{n=1}^\infty n^3\chi_{B_{1/n+1/n^2}(n)} \in L^1_{\text{loc}}$, we have $uv = 0 \in L^1$, but (for large $n$) $\|u_n v\|_1 = \|n^3u_n\|_1 = \operatorname{vol}(B_1) \not\to 0$.

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Great counterexample. – Nemis L. Oct 24 '12 at 9:22

It is easier to start looking at $\mathbb{R}$ and then generalize.

When considering the case in $\mathbb{R}$, let us first look at something easier than $C_0^\infty$. Put $u_n=\frac{1}{n^2}\chi_{[-n,n]}$, then $$\int u_n dx=\int_{-n}^n\frac{1}{n^2}dx=\frac{2}{n}\to0$$ Now choose $v$ that blows up at $\infty$ ("faster than $n$") e.g. $$v=\sum_1^\infty n^2 u_n$$ since compact sets are bounded they live in an some bounded interval so $v\in L^1_\text{loc}$ and $$\int u_n v =\int_{-n}^n\frac{1}{n^2}v=\frac{1}{n^2}\int_{-n}^n\sum_0^nk^2 u_kdx=\\\frac{1}{n^2}\sum_1^n\int_{-k}^kk^2 \frac{1}{k^2}dx=\frac{n(n+1)}{n^2}\to1\not=0.$$ Surely, we may choose smooth $u_n$ using e.g. through a convolution or redefining them at the boundaries.

When this is done, we can consider the case in $\mathbb{R^3}$ using balls instead of intervals or just put $u_n(x,y,z)=u_n(x)u_n(y)u_n(z)$.

Edit: The above is for $p=1$, for $p>1$ may look at consider $U_n=u_n^{1/p}$ and $V=v^{1/p}$.

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