About $\lim\limits_{r \to 0}\frac{\int_{\mathbb{R}}f(ry)w(y)\mathrm dy}{\int_{\mathbb{R}}g(ry)w(y)\mathrm dy}$

Let $f,g$ be real functions defined in a neighbourhood of $x=0$ and suppose $g(x) \ne 0$ for every $x$. Let also be $w \colon\mathbb R \to \mathbb R$ a non-negative function with compact support and such that $\int_{\mathbb R}w(x)\mathrm{d}x=1$.

I am wondering if it is true that $$\lim_{x \to 0}\frac{f(x)}{g(x)}= L \Rightarrow \lim_{r \to 0}\frac{\int_{\mathbb{R}}f(ry)w(y)\mathrm dy}{\int_{\mathbb{R}}g(ry)w(y)\mathrm dy}=L$$ where $L \in \mathbb R \cup \{-\infty,+\infty\}$.

Well, I am pretty sure that this fact is true when $f$ and $g$ are continuous in $x=0$: indeed, in this case, the function $F(r):=\int_{\mathbb R} f(ry)w(y)\mathrm dy$ would be continuous in $r=0$, and so $$\lim_{r \to 0} F(r) = F(0) = \int_{\mathbb R} f(0)w(y)\mathrm dy = f(0)$$ and similar for $g$.

Am I right?

And what can we conclude without the request of continuity? Thank you very much.

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$g(x) \ne 0$ for every $x$. –  Romeo Apr 14 '12 at 8:12
Whoops, nevermind then. Your reasoning is correct, but in general one wants to be careful about things like $\lim F(r) = f(0)$, because that involves an interchange of two limits (which is not always valid). Here however it works easily because $w$ has compact support... –  anon Apr 14 '12 at 8:15
as you observe, you don't need the ratio. Write $\int f(rz)w(z)dz = \int f(z)\frac {w(\frac zr)}{r}dz. \frac {w(\frac zr)}{r}$is called a summability kernel and you can say a lot about the convergence. –  mike Apr 14 '12 at 12:26
@mike: thank you for your comment. Would you be so kind to give me some more details about the "summability kernels"? I have never heard about them. Of course, I had already tried with the substitution method, without success. –  Romeo Apr 14 '12 at 13:04
You can look them up in wikipedia or Katznelson book (Harmonic Analysis ...?) is a good source. –  mike Apr 24 '12 at 19:00