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Let $f$ and $g$ be functions defined in a neighborhood of $0$ in $\Bbb R$, such that $g(x)\neq 0$ in this neighborhood.

Prove that for all $L\in[-\infty,+\infty]$ and for all non-negative functions $h$ with compact support such that $\int_{\Bbb R} h(y) \,\mathrm dy=1$:

$$\lim_{x\to 0}\frac{f(x)}{g(x)}=L\quad\text{implies that}\quad \lim_{r\to 0}\frac {\int_{\Bbb R} f(ry) h(y)\,\mathrm dy}{\int_{\Bbb R} g(ry)h(y)\,\mathrm dy}=L$$

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  • $\begingroup$ if f,g are continous in 0 this is trivial: $lim_{r\to 0}\int_R f(ry)h(y)dy=\lim_{r\to 0}F(r)=F(0)=\int f(0)h(y)dy=f(0)\int h(y)dy=f(0)$. the same is true for g. Can anyone help me to avoid using the continuity of f and g? $\endgroup$ – user62138 Aug 15 '13 at 9:59
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    $\begingroup$ With the (rather strong) hypothesis on $h$ that $h$ has compact support, the usual epsilon-delta approach works fine, once the integrals on $\mathbb R$ are rewritten as integrals on $[-K,K]$. (But the hypothesis on $g$ is probably incorrect, since it does not prevent the denominator to be zero.) $\endgroup$ – Did Aug 15 '13 at 10:08
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    $\begingroup$ @user62138: how does either integral exist if $f$ or $g$ is not integrable? $\endgroup$ – robjohn Aug 15 '13 at 10:46
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    $\begingroup$ To add to @robjohn: just let $f(x) = g(x) = 1/x^2$ if $x \ne 0$ and $1$ otherwise. For every $x\in \mathbb{R}$ we have $f(x)/g(x) = 1$ so the hypothesis holds for $L = 1$. But the ratio in the conclusion is undefined for for any $r > 0$. $\endgroup$ – Willie Wong Aug 15 '13 at 12:10
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    $\begingroup$ Please do not deface questions with good answers. Others have devoted effort to answer your question; defacing the question is disrespectful of their effort and prevents others from benefiting from your question and its answers. $\endgroup$ – robjohn Aug 21 '13 at 17:49
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Let $[-K,K]$ denote some bounded interval containing the support of $h$. Assume first that $L$ is finite, then for every $u\lt L\lt v$, there exists some $z$ such that $u\lt f(t)/g(t)\lt v$ for every $|t|\lt z$. Add to the hypotheses in the post the hypothesis (H):

(H) $g(t)\gt0$ for every $|t|$ small enough.

Then, for every $|t|\lt z$, $ug(t)\lt f(t)\lt vg(t)$. Thus, for every $|r|\lt z/K$ and for every real number $t$, $ug(rt)h(t)\leqslant f(rt)h(t)\leqslant vg(rt)h(t)$ (this uses the hypothesis that $h\geqslant0$ everywhere). Integrating this and using once again the nonnegativity of $g$ and $h$, one sees that the ratio $$ A(r)=\frac{\int_\mathbb Rf(rt)h(t)\mathrm dt}{\int_\mathbb Rg(rt)h(t)\mathrm dt}, $$ is such that $u\leqslant A(r)\leqslant v$ for every $|r|\leqslant z/K$. Since $u$ and $v$ can be as close to $L$ as one wants, this proves the claim that $A(r)\to L$ when $r\to0$. If $L=+\infty$ or $L=-\infty$, copy this proof, replacing $(u,v)$ by $w$ large enough and of the sign of $L$.

For a counterexample when (H) fails, consider $h=\mathbf 1_{[-1/2,1/2]}$ and $f(t)=t+t^2$ and $g(t)=t+t^4$ if $t\ne0$. Then $L=1$ and $$ \int_\mathbb Rf(rt)h(t)\mathrm dt=\frac{r^2}{12},\qquad \int_\mathbb Rg(rt)h(t)\mathrm dt=\frac{r^4}{80}, $$ hence $A(r)\to+\infty$ when $r\to0$. To get the limit $A(r)\to0$ when $r\to0$, exchange the functions $f$ and $g$.

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  • $\begingroup$ This is the epsilon-delta approach. // Re continuity: if $g$ is continuous at $0$ and $g(0)\ne0$ then $g$ or $-g$ is positive in a neighbourhood of $0$ hence, basically, hypothesis (H) is very much satisfied. $\endgroup$ – Did Aug 18 '13 at 16:57
  • $\begingroup$ Listen, I begin to be a bit lost by your queries... If $g$ is continuous and $g(0)\ne0$, this is the setting of hypothesis (H). If $g$ is continuous and $g(0)=0$, see the (counter) example at the end of my post. $\endgroup$ – Did Aug 18 '13 at 17:11
  • $\begingroup$ ((At least two comments by the OP are now deleted.)) $\endgroup$ – Did Aug 21 '13 at 10:15

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