# Let $\rho$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a continuous function such that

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 $$\rho(x) =\left\{\begin{array}{ll} x+1 & \text{for } x \in [-1,0],\\ -x+1 & \text{for } x \in [0,1],\\ 0 & \text{otherwise.} \end{array}\right.$$

• 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?
– Jeff
Sep 4 '16 at 19:27

$\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int_{-\infty}^{\infty}\rho(\frac{x}{\varepsilon})f(x))=\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\int_{-\varepsilon}^{\varepsilon}\rho(\frac{x}{\varepsilon})f(x))$

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 $f(0)$.

• i can't get it on the first line.
– baam
Sep 4 '16 at 19:56
• $\rho(x)=0$ except on $[-1;1]$. $\frac{x}{\varepsilon}$ belongs to $[-1;1]$ when $x$ belongs to $[-\varepsilon;\varepsilon]$ Sep 4 '16 at 19:59
• You're welcome, mate. Sep 4 '16 at 20:07

Hint: The expression equals $\int_{-1}^1 \rho (y)f(\epsilon y)\, dy.$

• can u suggest me some book on real analytic diffeomorphism.
– baam
Sep 4 '16 at 20:08

Another hint: Use that $f$ is continuous and look closely into: $$\int_{\Bbb R} \frac{1}{\epsilon} \rho\left( \frac{x}{\epsilon} \right) \left( f(x) -f (0) \right) \; dx$$