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Let $u(x,y),v(x,y)\in L^2$. What can we say about $\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{d}{dy}(uv) \, dy \, dx$? Does it equal zero? If so, why?

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    $\begingroup$ It's possible that the limit $\lim_{y\to +\infty}u(x,y)$ doesn't exist. $\endgroup$ Jun 20, 2012 at 12:35
  • $\begingroup$ Plus: your notation is confusing, I do not understand it. $\endgroup$
    – GEdgar
    Jun 20, 2012 at 13:41
  • $\begingroup$ I've changed the notation. $\endgroup$ Jun 20, 2012 at 17:07

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The question as stated is not well-defined. This is true even in the case where $u = v$, when you are differentiating $(u(x,y))^2$ with respect to to $y$. There are plenty of examples of nonnegative $L^1$ functions $f(y)$ which are nowhere differentiable, so if you take $u(x,y)$ to be of the form $\chi_{[0,1]}(x)\sqrt{f(y)}$ for example, you won't be able to take the $y$ derivative.

If you are taking distributional derivatives, there is a way of making sense out of the question, but I'm not sure that's what you mean here.

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I think your integral might not make sense. First of all, I would write

$$\lim_{a\to\infty}\left[u(x,y)v(x,y)\right]_{y=-a}^{y=a}$$

rather than

$$\left(u\left(x,y\right)v\left(x,y\right)\right)|_{-y=\infty}^{y=\infty}$$ but that's just style. More importantly, this limit might not exist. The fact that a function is in $L^1$ (like $uv$ in your case) does not imply that it has a limit (which, if any, has to be $0$) at infinity. Indeed, consider the function

$$f(x) = \begin{cases}n\qquad \text{ if }n-\dfrac{1}{n^3}\leq x\leq n\\0\qquad\text{else}\end{cases}$$ for $n\in \mathbb{N}\setminus\{0\}$. This function is in $L^1(\mathbb{R})$, since

$$\int_{\infty}^{+\infty}f(x)dx=\int_0^{+\infty}f(x)dx = \lim_{m\to\infty}\int_0^m f(x)dx =$$ $$=\lim_{m\to\infty}\sum_{n=1}^{m}\int_{n-\frac{1}{n^3}}^n ndx = \lim_{m\to\infty}\sum_{n=1}^{m}\dfrac{1}{n^2}=\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$ However, $f$ does not have a limit as $x$ goes to infinity. All you can say for an integrable (over $\mathbb{R}$) function is that

$$\liminf_{|x|\to\infty}f(x)=0$$ which is not enough to compute the limit that appears in your question.

If your functions have more regularity, namely they are not just $L^2$ but also $H^1$, then your integral exists and it's equal to zero, since the derivative cannot grow at infinity, so there can't be spikes whose amplitude does not go to zero. As a consequence, also the $\limsup$ is zero, which tells you that the limit exists and it is zero.

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