Does this integral go to zero? I have $f_1,f_2$, $C^\infty$ functions with compact support and $f_3$ a smooth and bounded function; let $a\in\mathbb{R}^3$. I have to evaluate this limit
$$\lim_{|a|\to\infty}\int_{\mathbb{R}^6}f_1(y_1)f_2(y_2-a)f_3(y_1+y_2)d^3y_1d^3y_2$$
I've followed this reasoning: the integrand goes to zero if $|a|\to\infty$ because $f_2(y_2-a)\to 0$ (the argument of the function goes away from the support); so applying the theorem of dominated convergence the thesis follows. Is there anything wrong?
 A: At first, one would think this function of $a$ is compactly supported, because it looks very much like the convolution of two compactly supported functions. But this is completely wrong. We integrate over $\mathbb{R}^3\times\mathbb{R}^3$ and not $\mathbb{R}^3$: $y_1$ and $y_2$ are independent. There is indeed an underlying convolution, but over $\mathbb{R}^6$, e.g. of functions $g_1(y_1,y_2)=f_1(y_1)f_3(y_1+y_2)$ and $g_2(y_1,y_2)=f_2(y_2)$. And these need no longer be compactly supported.
Without further assumptions on $f_3$, there does not necessarily exist a limit. So assume $f_3=1$ first. The integrand is compactly supported, so it is integrable. By the change of variable $(y_1,y_2)\longmapsto (y_1,y_2+a)$ and Fubini, we get
$$
I(a)=\int_{ \mathbb{R}^6}f_1(y_1) f_2(y_2-a) dy_1dy_2=\int_{y_1\in\mathbb{R}^3}f_1(y_1) dy_1\cdot\int_{y_2\in \mathbb{R}^3}f_2(y_2) dy_2=J.
$$
This already proves that $I(a)$ does not necessarily tend to $0$. Now assume that $\lim_{\|y\|\rightarrow +\infty}f_3(y)=L$ exists. We first note that $I(a)$ is the integral of an integrable function, as the integrand is dominated by $|f_1(y_1)||f_2(y_2-a)|\sup|f_3|$, which is compactly supported. Let $K$ be a compact containing both supports of $f_1$ and $f_2$. Then by change of variable $(y_1,y_2)\longmapsto (y_1,y_2+a)$, we see that the integral is equal to
$$
I(a)=\int_{\mathbb{R}^6}f_1(y_1)f_2(y_2)f_3(y_1+y_2+a) dy_1dy_2 
$$
$$=\int_{K^2}f_1(y_1)f_2(y_2)f_3(y_1+y_2+a)dy_1dy_2.
$$
Now, with $J$ as above,
$$
|I(a)-LJ|\leq\int_{K^2}|f_1(y_1)||f_2(y_2)||f_3(y_1+y_2+a)-L|dy_1dy_2
$$
Since $y_1+y_2$ is bounded on $K^2$, we see that $f_3(y_1+y_2+a)$ converges uniformly to $L$ on $K^2$ as $\|a\|$ tends to $+\infty$. Therefore

$$
\lim_{\|a\|\rightarrow+\infty}I(a)=\int_{\mathbb{R}^3}f_1(y_1) dy_1\cdot\int_{\mathbb{R}^3}f_2(y_2) dy_2\cdot \lim_{\|y\|\rightarrow+\infty}f_3(y).
$$

In the general case, there needs not be a limit. Assume to simplify the matter that functions are over $\mathbb{R}$, and take $f_1=f_2=1_{[0,1]}$. Denoting $H$ an antiderivative of an antiderivative of $f_3$, we get in this case
$$
I(a)=H(a+2)-2H(a+1)+H(a).
$$
This does not necessarily have a limit. Think of trigonometric functions, for instance. The argument can be adapted to smooth functions over $\mathbb{R}^3$.
A: @Julien I don't succed in posting as comment. I've read again my previous comment and I note that it is wrong; in fact
$$\int_{\mathbb{R}^3}f_2(y_2-a)$$ is different from zero iff $y_2-a\in K=supp f_2$ and so it becomes
$$\int_{y_2-a\in K}f_2(y_2-a)=\int_{K}f_2$$
Moreover I forgot to write that I know the behaviour of $f_3$ at $\infty$; in particular it goes to zero at $\infty$. So following your reasoning  in the answer I have the thesis. I apologize to you but I was quite confused and I made a terrible error.
Thanks for your answer
