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Let $\rho$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a continious function such that $\rho(x) \geqslant 0 $ for all $x \in \mathbb{R}$, $\rho =0 $ if $\mid x \mid \geqslant 1$ and $$\int_{-\infty}^{\infty} \rho(t) dt = 1.$$ Let $f(x): \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Evaluate $$ \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{-\infty}^{\infty} \rho (\frac{x}{\epsilon})f(x)dx .$$

I have constructed such a function $\rho(x)$ which satisfies the the condition given above, but I am unable to compute the limit for my constructed function. Kindly help me. My constructed function $\rho(x)$ is $$\rho(x) =\left\{\begin{array}{ll} x+1 & \text{for } x \in [-1,0],\\ -x+1 & \text{for } x \in [0,1],\\ 0 & \text{otherwise.} \end{array}\right. $$

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  • $\begingroup$ A couple of hints: Write $\rho_\varepsilon(x) = \frac{1}{\varepsilon}\rho(x/\varepsilon)$ for simplicity. What is $\int_{-\infty}^\infty \rho_\varepsilon(x) dx$? Where does $\rho_\varepsilon$ vanish? $\endgroup$ – Jeff Sep 4 '16 at 19:27
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$\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int_{-\infty}^{\infty}\rho(\frac{x}{\varepsilon})f(x))=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int_{-\varepsilon}^{\varepsilon}\rho(\frac{x}{\varepsilon})f(x))$

Denote the integral inside the limit by $I_\varepsilon$

$m_\varepsilon\int_{-\varepsilon}^\varepsilon\rho(\frac{x}{\varepsilon})dx\leq I_\varepsilon \leq M_\varepsilon\int_{-\varepsilon}^\varepsilon\rho(\frac{x}{\varepsilon})dx$

where $m_\varepsilon$ and $M_\varepsilon$ are the bounds on $f$ on the interval $[-\varepsilon;+\varepsilon]$

The integral we have is obviously $\varepsilon$ and thus

$m_\varepsilon\varepsilon\leq I_\varepsilon \leq M_\varepsilon\varepsilon$

Divide both sides by $\varepsilon$, note that $m_\varepsilon$ and $M_\varepsilon$ go to $f(0)$ by continuity of $f$, and we are done. The limit is $f(0)$.

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  • $\begingroup$ i can't get it on the first line. $\endgroup$ – baam Sep 4 '16 at 19:56
  • $\begingroup$ $\rho(x)=0$ except on $[-1;1]$. $\frac{x}{\varepsilon}$ belongs to $[-1;1]$ when $x$ belongs to $[-\varepsilon;\varepsilon]$ $\endgroup$ – Hasan Saad Sep 4 '16 at 19:59
  • $\begingroup$ You're welcome, mate. $\endgroup$ – Hasan Saad Sep 4 '16 at 20:07
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Hint: The expression equals $\int_{-1}^1 \rho (y)f(\epsilon y)\, dy.$

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  • $\begingroup$ can u suggest me some book on real analytic diffeomorphism. $\endgroup$ – baam Sep 4 '16 at 20:08
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Another hint: Use that $f$ is continuous and look closely into: $$ \int_{\Bbb R} \frac{1}{\epsilon} \rho\left( \frac{x}{\epsilon} \right) \left( f(x) -f (0) \right) \; dx$$

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