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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?

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    $\begingroup$ Use the squeeze theorem: What happens to $M_\epsilon$ and $m_\epsilon$ as $\epsilon\to0$ when $f$ is continuous at 0? $\endgroup$ – Jose27 Nov 18 '17 at 4:53
  • $\begingroup$ @Jose27 Both becomes $f(0)$ to the limit, right?? $\endgroup$ – Abishanka Saha Nov 18 '17 at 4:56
  • $\begingroup$ Yep, that's correct. $\endgroup$ – Jose27 Nov 18 '17 at 5:00
  • $\begingroup$ But $\rho\left(\frac{x}{\epsilon}\right)=0$ for $\left|\frac{x}{\epsilon}\right|\geq1$ $\endgroup$ – Abishanka Saha Nov 18 '17 at 5:10
  • $\begingroup$ @AbishankaSaha that means $\rho$'s maximum is either 0 or something positing inside of the interval $[-\epsilon, \epsilon]$. Perhaps you should write explicitly upper and lower bounds to determine what happens by squeeze theorem. Also, this has been asked before here(duplicate?): math.stackexchange.com/questions/2033892/… $\endgroup$ – DaveNine Nov 18 '17 at 5:27
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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.

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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.

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  • $\begingroup$ I have done it in the same way following @Jose27 's hint. Thanks anyway. $\endgroup$ – Abishanka Saha Nov 18 '17 at 5:49

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