Evaluate $\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_{-\infty}^{\infty}\rho\left(\frac{x}{\epsilon}\right)f(x)dx$ Let $\rho:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $\rho(x)\geq0$ for all $x\in\mathbb{R}$, $\rho(x)=0$ for $|x|\geq1$ and $$\int_{-\infty}^{\infty}\rho(t)dt=1$$ I have to evaluate $$\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_{-\infty}^{\infty}\rho\left(\frac{x}{\epsilon}\right)f(x)dx$$ for any continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$.
One can easily show for any $\epsilon>0$ via the substitution $y=x\epsilon$, that $$\frac{1}{\epsilon}\int_{-\infty}^{\infty}\rho\left(\frac{y}{\epsilon}\right)dy=1~~\Rightarrow~~\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int_{-\infty}^{\infty}\rho\left(\frac{x}{\epsilon}\right)dx=1$$
Thus for the constant function $f(x)=1~~\forall x$, we have the limit to be $1$. 
Now for general continuous $f$, it has maximum and minimum over the compact interval $[-\epsilon,\epsilon]$, say $M_{\epsilon}$ and $m_{\epsilon}$, respectively. Then for any $\epsilon>0$, $$m_{\epsilon}\leq\frac{1}{\epsilon}\int_{-\infty}^{\infty}\rho\left(\frac{y}{\epsilon}\right)f(y)dy\leq M_{\epsilon}$$
I guess for any general continuous function the limit should be $f(0)$. Is it true? How to prove it?
 A: Use the hypothesis of continuity on $f$. Fix $\delta>0$. Then for $|x|<\epsilon$ ($\epsilon$ sufficiently small), we have 
$$
f(0)-\delta<f(x)<f(0)+\delta
$$
and indeed 
$$
(f(0)-\delta)\frac{1}{\epsilon}\int_{-\epsilon}^\epsilon \rho\left(\frac{x}{\epsilon}\right)\mathrm dx\leq\frac{1}{\epsilon}\int_{-\epsilon}^\epsilon \rho\left(\frac{x}{\epsilon}\right)f(x)\mathrm dx\leq
(f(0)+\delta)\frac{1}{\epsilon}\int_{-\epsilon}^\epsilon \rho\left(\frac{x}{\epsilon}\right)\mathrm dx
$$
but note that taking $x\mapsto x/\epsilon$ yields 
$$
\frac{1}{\epsilon}\int_{-\epsilon}^\epsilon \rho\left(\frac{x}{\epsilon}\right)\mathrm dx=\int_{-1}^1 \rho\left(x\right)\mathrm dx=
\int_{-\infty}^\infty \rho\left(x\right)\mathrm dx=1
$$
Since the estimate clearly holds in the limit as $\epsilon\downarrow 0$, and $\delta>0$ was arbitrary, you are done.
A: Let $\frac{x}{\epsilon}=y$, then $\forall \epsilon>0$
$$I_{\epsilon}=\int_{-\infty}^{\infty} f(x)\rho\left(\dfrac{x}{\epsilon}\right)\dfrac{dx}{\epsilon} = \int_{-\infty}^{\infty} f(y\epsilon)\rho(y)dy = \int_{-1}^{1} f(y\epsilon)\rho(y)dy$$So, as you've noted, $\inf_{[-\epsilon,\epsilon]}f(x)=m_{\epsilon}\leq I_{\epsilon}\leq M_{\epsilon}=\sup_{[-\epsilon,\epsilon]}f(x)$, so it suffices to show $m_{\epsilon}=M_{\epsilon} = f(0)$ as $\epsilon\to0^+$.
Clearly $m_{\epsilon}\leq f(0)\leq M_{\epsilon}$ for all $\epsilon>0$. Regarding $m,M$ as functions of $\epsilon$, and $m$ is decreasing while $M$ is increasing. It's elementary that monotonic functions have left and right handed limits, so these limits exist and we can compute the limit using a sequence. Let $\epsilon=\frac{1}{n}$ and write $m_n = m_{1/n}$ and similarly for $M$. Now, this is a little confusing, but $m_n$ is an increasing sequence and $M_n$ is a decreasing sequence (convince yourself this is true). $m_n$ is bounded above by $f(0)$, and it is the least upper bound (this can be shown using continuity of $f$), so $$\lim_{n\to\infty}m_n=\lim_{\epsilon\to0^+}=f(0)$$Similarly, $M_n$ is bounded below by $f(0)$ and it is the greatest lower bound, so the same holds for $M_n/M_{\epsilon}$. Thus, by the squeeze theorem $$\lim_{\epsilon\to0^+}m_{\epsilon}=f(0)\leq \lim_{\epsilon\to0^+}I_{\epsilon}\leq \lim_{\epsilon\to0^+}M_{\epsilon}=f(0)$$Going to a sequence is unnecessary, but it helped me from getting confused about the directions of the limit, monotonicity of the functions, and the bounds.
