# Proof of $n \int_0^1 f(x+t) g(nt)\, dt \rightarrow f(x)$

Suppose $g\in L^1([0,\infty))$ and $\int_0^\infty g(x)\, dx=1$.

How to prove that if $f:[0,\infty) \rightarrow \mathbb{R}$ is a continuous function then $n \int_0^1 f(x+t) g(nt)\, dt \rightarrow f(x)$ as $n\rightarrow \infty$ for $x\in \mathbb{R}$?

Say $nt=u$ in the integral then we get $\int_0^n f(x+u/n) g(u) \,du$. Estimating this integral from above and letting $f_n(x)= f(x+u/n) \chi_{[0,n]}$ gives nothing since we don't know whether $f$, being pointwise limit of this sequence, is integrable on $\mathbb{R}$. How can I use the continuity of $f$? Somehow Lebesgue Dominated Convergence Theorem is gonna play a role but I cannot get it, could you help?

Obrigado.

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Hint: Consider $|\int_0^{\infty}f(x+u/n) g(u) \,du-\int_0^{\infty}f(x) g(u) \,du|\leq|\int_0^{A}M\epsilon\,du|+|\int_A^{\infty}\epsilon g(u)\,du|$ – 89085731 Apr 17 '12 at 0:51
Why is $g \leq \epsilon$ on the interval $[0,A]$? Do you mean this? – беркай Apr 17 '12 at 1:22
Sorry I need to make a correction.See below$|\int_0^{\infty}f(x+u/n) g(u) \,du-\int_0^{\infty}f(x) g(u) \,du|\leq|\int_0^{A}M_1\epsilon|+|\int_A^{\infty}M_2 g(u)|$ – 89085731 Apr 17 '12 at 1:26
$\epsilon$ is for f(x+u/n)-f(x) in a finite interval – 89085731 Apr 17 '12 at 1:27

## 3 Answers

I can't help but see these questions in terms of probability.

Let $T$ be a non-negative random variable with density $g(t)$; then $ng(nt)$ is the density of $T/n$, and so $$n\int_0^1f(x+t)g(nt)\,dt=\int f(x+T/n)\,1_{[0,1]}(T/n)\,dP.$$

The random variables $f(x+T/n)\,1_{[0,1]}(T/n)$ are uniformly bounded by $M:=\sup_{x\leq y\leq x+1}|f(y)|$ and converge to the constant $f(x)$ so by dominated convergence we have $$\int f(x+T/n)\,1_{[0,1]}(T/n)\,dP\to \int f(x)\,dP=f(x).$$

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This is just Byron's answer again:

Let $g_n(t) = f(x+t/n)\chi_{[0,n]}g(t)$. Note that for $t\in[0,n]$ we have $x+t/n\in[x,x+1]$. Let $M$ be an upper bound for $|f|$ on $[x,x+1]$. We then have $|g_n(t)|\le M |g(t)|$ for all $t\ge0$. Also $\lim\limits_{n\rightarrow\infty} g_n(t)= f(x)g(t)$ for all $t\ge0$. Thus, by the Dominated Convergence Theorem: $$\int_0^n f(x+t/n) g(t)\,dt=\int_0^\infty g_n(t)\,dt\ \ \buildrel{n\rightarrow\infty} \over\longrightarrow\ \ \int_0^\infty f(x)g(t)\,dt=f(x).$$

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You can easily prove the statement for linear combinations of indicator functions. Take $g=\frac{1}{b-a}\chi_{[b-a]}$: $$n\int_0^1 f(x+t)g(nt)dt=\frac{n}{b-a}\int_{a/n}^{b/n}f(x+t)dt=f(x+c),$$ for a $c\in[a/n,b/n]$ (this is the meanvalue theorem). For $n\rightarrow\infty$ $c$ converges to $0$ and so $f(x+c)\rightarrow f(x)$. Now approximate $g$ by such functions to get the result.

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