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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.

  • 0
    Something is missing here: how do you know that the integral $\int_{\mathbb R} f(rx)w(x)\,dx$ exists? Are $f$ and $g$ assumed to be integrable in a neighborhood of $0$?2013-06-15

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