# Evaluate the limit: $\lim_{\epsilon \to 0} {{1}/{\epsilon}}\int_{-\infty}^{\infty} \rho(x/ \epsilon)f(x)dx$

If $\rho: \mathbb R \to \mathbb R$ continuous function with $\rho(x) \ge 0$ and $\rho(x)=0$ if $|x| \ge 1$, $f: \mathbb R \to \mathbb R$ continuous function and $\int_{-\infty}^{\infty}\rho(t)dt=1$

Evaluate the limit: $\lim_{\epsilon \to 0} {{1}/{\epsilon}}\int_{-\infty}^{\infty} \rho(x/ \epsilon)f(x)dx$

• Where are you stuck? And shouldn't it be $\epsilon\to0$? Nov 28 '16 at 5:45
• the limt is $\lim\int_{-1}^{1} \rho(y)f(y \epsilon)dy$? Then what to do? Nov 28 '16 at 5:57

$1/\epsilon\int_{-\infty}^{\infty}\rho(x/\epsilon)f(x)dx=1/\epsilon\int_{-\epsilon}^{\epsilon}\rho(x/\epsilon)f(x)dx$ (by assumption of $\rho$)
Then we can see like this : $$\left|\frac{1}{\epsilon}\int_{-\epsilon}^{\epsilon}\rho(x/\epsilon)[f(x)-f(0)]dx\right|\leq \frac{1}{\epsilon}\int_{-\epsilon}^{\epsilon}\rho(x/\epsilon)|f(x)-f(0)|dx$$ Since $f$ is continuous on real line, there exists $\delta$ such that $$|x-0|<\epsilon \implies |f(x)-f(0)|<\delta$$ so, $$\frac{1}{\epsilon}\int_{-\epsilon}^{\epsilon}\rho(x/\epsilon)|f(x)-f(0)|dx <\delta$$ so, $$\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_{-\epsilon}^{\epsilon}\rho(x/\epsilon)f(x)dx=f(0)$$
• 'if i am not mistaken then " there exists $\delta$ such that... " should be replaced by " for every $\delta >0$ " in your post . Nov 28 '16 at 8:10