6
$\begingroup$

Let $\eta \in \mathcal{C}^{\infty}_{0}(\mathbb{R})$, where $\mathcal{C}^{\infty}_{0}(\mathbb{R})$ is the set of compactly supported infinitely differentiable function, be a function which is

  • non-negative, non-vanishing and constant in a neighborhood of $x=0$,
  • symmetric about $x=0$,
  • supported in $[-1,1]$,
  • bounded between $0$ and $\eta(0)$, and
  • normalized so that $\int \eta d\mu =1$.

Let $\varphi$ be a compactly supported absolutely continuous function with a piecewise continuous derivative $\varphi'$. I have been told (by a user of MSE whom I thank very much again) the following, which I think to be interesting enough to be the object of a specifical question:

Then one defines$$\eta_{n}(x) = n\eta(x/n)$$so that $\int_{\mathbb{R}}\eta_{n}\,d\mu =1$ for all $n$. For any compactly supported absolutely continuous function $\varphi$, define$$\varphi_{n}=\int_{\mathbb{R}}\eta_{n}(x-y)\varphi(y)\,d\mu(y).$$The function $\varphi_{n}$ is in $\mathcal{C}^{\infty}_{0}(\mathbb{R})$. Because $\eta_{n}$ is supported in $[-1/n,1/n]$, and $\varphi$ is continuous, then $\varphi_{n}$ converges uniformly to $\varphi$ as $n\rightarrow\infty$. And, because $\varphi$ is absolutely continuous, then$$\varphi_{n}'=\int_{\mathbb{R}}\eta_{n}'(x-y)\varphi(y)\,d\mu(y)=-\int_{\mathbb{R}}\eta_{n}(x-y)\varphi'(y)\,d\mu(y).$$ For my case $\varphi'$ is piecewise continuous, and the right side then converges pointwise everywhere to mean of the left- and right-hand limits of $\varphi'$, and it remains uniformly bounded by any bound for $\varphi'$.

I must say that I have no knowledge of the theory of mollification until now.

I would like to understand why

  1. the fact that $\eta_{n}$ is supported in $[-1/n,1/n]$, and $\varphi$ is continuous, implies that $\varphi_{n}$ converges uniformly to $\varphi$ as $n\rightarrow\infty$;
  2. the absolute continuity of $\varphi$ implies $\varphi_{n}'=\int_{\mathbb{R}}\eta_{n}'(x-y)\varphi(y)\,d\mu(y)=-\int_{\mathbb{R}}\eta_{n}(x-y)\varphi'(y)\,d\mu(y)$;
  3. the piecewise continuity of $\varphi'$ implies the facts that $-\int_{\mathbb{R}}\eta_{n}(x-y)\varphi'(y)\,d\mu(y)$ converges to $(\lim_{t\to x^+}\varphi'(t)+\lim_{t\to x^-}\varphi'(t))/2$ and $|\int_{\mathbb{R}}\eta_{n}(x-y)\varphi'(y)\,d\mu(y)|\le |\varphi'(x)|$.

I heartily thank both who told me these interesting facts and whoever will help me to understand the reason of the quoted facts.

$\endgroup$
  • 2
    $\begingroup$ Your definition of $\eta_n$ is incorrect. You mean $\eta (nx)$ instead of $\eta(x/n)$. $\endgroup$ – PhoemueX Jan 1 '15 at 21:35
3
$\begingroup$

In the following, I will assume that you use the version

$$ \eta_n (x) = n \cdot \eta(nx). $$

This ensures that ${\rm supp}(\eta_n) \subset [-1/n, 1/n]$ and that $\int \eta_n \, dx = 1$ for all $n$.

I will also assume that $\mu$ denotes the usual Lebesgue measure on $\Bbb{R}$.

  1. Here, we only need the fact that $\varphi$ is uniformly continuous (because it is continuous and of compact support). Let $\varepsilon > 0$ be arbitrary. By uniform continuity of $\varphi$, there is some $\delta > 0$ with $|\varphi(x) - \varphi(y)| < \varepsilon$ for $|x-y|<\delta$. For $n > 1/\delta$, the fact that $\int \eta_n (x-y) \, dy = \int \eta_n \, dy =1 $ yields \begin{eqnarray*} |\varphi(x) - \varphi_n (x)| &=& \bigg| \int \eta_n (x-y) \cdot [\varphi(x) - \varphi(y)] \, dy\bigg| \\ &\leq & \int \eta_n (x-y) \cdot |\varphi(x) - \varphi(y)| \, dy \\ &\leq& \int \eta_n (x-y) \cdot \varepsilon \, dy = \varepsilon. \end{eqnarray*} Here, the second line used the triangle inequality as well as $\eta_n \geq 0$. The last line used the fact that $\eta_n (x-y) \neq 0$ implies $x-y \in {\rm supp}(\eta_n) \subset [-1/n, 1/n]$ and hence $|x-y| \leq 1/n < \delta$, which yields $|\varphi(x) - \varphi(y)| < \varepsilon$ by choice of $\delta$.

  2. By a standard argument using differentiation under the integral sign (see e.g. Differentiating an integral using dominated convergence), we see that $\varphi_n$ is (infinitely often) differentiable with derivative $$ \varphi_n ' (x) = \int \eta_n ' (x-y) \cdot \varphi(y) \, dy. $$ Now we observe that $$ \frac{d}{dy} \eta_n (x-y) = - \eta_n '(x-y). $$ Furthermore, if $f,g : \Bbb{R} \to \Bbb{R}$ are absolutely continuous and integrable, we have the following rule of partial integration (if this is not clear, I can elaborate. This is essentially a consequence of the fact that $f\cdot g$ is also absolutely continuous with derivative $(f\cdot g)' = f' \cdot g + f \cdot g'$): $$ \int f\cdot g' \, dx = - \int f' \cdot g \, dx . $$ Hence, we get \begin{eqnarray*} \varphi_n ' (x) &=& \int \eta_n ' (x-y) \cdot \varphi(y) \, dy \\ & = & - \int \varphi(y) \cdot \frac{d}{dy} \eta_n (x-y)\, dy \\ & = & \int \varphi'(y) \cdot \eta_n (x-y) \,dy. \end{eqnarray*}

  3. Let us first note that the estimate $|\varphi_n '(x)| = |\int \eta_n (x-y) \varphi'(y) \, dy| \leq |\varphi' (x)|$ is in general false. To see this, consider e.g. (you can probably think of a better (less trivial) example once you understand the idea) $$ \varphi'(x)=\begin{cases} 1, & x>0,\\ 0, & x\leq0. \end{cases} $$ Then but what we will show below, $$ \varphi_n '(0) \to \lim_{x\downarrow 0} \varphi'(x) + \lim_{x \uparrow 0}\varphi'(x) = \frac{1}{2}, $$ which would not be possible if $|\varphi_n '(0)| \leq |\varphi'(0)| = 0$ would hold.
    But what is correct is that if $|\varphi'(y)| \leq M$ holds for all $|y-x| < \delta$ for some $\delta > 0$, then $|\varphi_n '(x)| \leq M$ holds for $n > 1/\delta$, because of $$ |\varphi_n '(x)| \leq \int \eta_n (x-y) \cdot |\varphi'(y)| \, dy \leq M \cdot \in \eta_n (x-y) \, dy = M, $$ where we used that $\varphi_n (x-y) \neq 0$ implies (as above) $|x-y| \leq 1/n < \delta$.
    In particular, if $|\varphi'(y)| \leq M$ holds for all $y$, then $|\varphi_n '(x)| \leq M$ holds for all $x$. In this sense, any bound for $\varphi'$ is also valid for $\varphi_n'$.
    Finally, let us show the claimed pointwise convergence. Let $x_0 \in \Bbb{R}$ and set $y_1 := \lim_{x \downarrow x_0} \varphi'(x)$ and $y_2 := \lim_{x \uparrow x_0} \varphi'(x)$. Note that these limits exist by piecewise continuity of $\varphi'$.
    Let $\varepsilon > 0$. Then there is $\delta > 0$ with $|\varphi(x) - y_1| < \varepsilon$ for $x \in (x_0, x_0 + \delta)$ and $|\varphi(x) - y_2| < \varepsilon$ for $x \in (x_0 - \delta, x_0)$. We now use the fact that $\eta$ is symmetric around $0$. This implies that $$ 1 = \int_{\Bbb{R}} \eta(x) \, dx = 2 \int_{\Bbb{R}_+} \eta(x) \, dx = 2 \int_{\Bbb{R}_-} \eta(x) \, dx. \qquad (\dagger) $$ This is actually the only thing we will use now, symmetry is not really relevant as long as the above equation $(\dagger)$ is true. We have \begin{eqnarray*} \varphi_{n}'\left(x_0\right) & = & \int\eta_{n}\left(x-y\right)\cdot\varphi'\left(y\right)\,{\rm d}y\\ & \overset{z=x_0-y}{=} & \int\eta_{n}\left(z\right)\cdot\varphi'\left(x_0-z\right)\,{\rm d}z \end{eqnarray*} Together with $(\dagger)$, we derive \begin{eqnarray*} \bigg|\frac{y_1}{2} + \frac{y_2}{2} - \varphi_n '(x_0)\bigg| &=& \bigg|\int_{\Bbb{R}_+} \eta_n (z) \cdot [y_2 - \varphi '(x_0-z)] \, dz + \int_{\Bbb{R}_-} \eta_n (z) \cdot [y_1 - \varphi' (x_0-z) \, dz\bigg| \\ & \leq & \int_{\Bbb{R}_+} \eta_n (z) \cdot |y_2 - \varphi '(x_0-z)| \, dz + \int_{\Bbb{R}_-} \eta_n (z) \cdot | y_1 - \varphi' (x_0-z)| \, dz \\ & \leq & \int_{\Bbb{R}_+} \eta_n (z) \cdot \varepsilon \, dz + \int_{\Bbb{R}_-} \eta_n (z) \cdot \varepsilon \, dz = \varepsilon. \end{eqnarray*} Here, the first and very last step used $(\dagger)$. The estimate in the last line holds for $n > 1/\delta$, because (as always) if $\eta_n (z) \neq 0$ and $z \in \Bbb{R}_+ = (0,\infty)$, then $|z| \leq 1/n < \delta$ and hence $x_0 - z \in (x_0 - \delta, x_0)$, so that $|\varphi'(x_0 -z) - y_2| < \varepsilon$ by choice of $\delta$. This takes care of the first integral and the second one is handled analogously.

$\endgroup$
  • $\begingroup$ I heartily thank you for your wonderfully clear answer and your kindness and willingness to be even clearer by explaining why $(fg)'=f'g+fg'$ holdes a.e. for absolutely continuous $f,g$ (which is a thing that I susprisingly understand because my textbook, Kolmogorov-Fomin's, gives me the tools to understand it). There is only one thing that isn't clear to me: I understand that $\int_\mathbb{R}\eta_n(x)dx=1$ but I don't understand equation $(\dagger)$... Thank you so much again and have a happy $5\cdot13\cdot31$! $\endgroup$ – Self-teaching worker Jan 2 '15 at 16:38
  • 1
    $\begingroup$ @Self-teachingDavide: There was actually an error in $(\dagger)$: I wrote $\varphi$ everywhere, but should have written $\eta$. Now note that $\int_{\Bbb{R}} \eta(x) \, dx = \int_{\Bbb{R}_+} \eta(x) \, dx + \int_{\Bbb{R}_-} \eta(x) \, dx = \int_{\Bbb{R}_+} \eta(x) \, dx + \int_{\Bbb{R}_+} \eta(-y) \, dy = 2\int_{\Bbb{R}_+} \eta(x) \, dx$. Here, we used the substitution $y=-x$ and in the last step we made use of the fact that $\eta(-y) = \eta(y)$ by assumption on $\eta$. $\endgroup$ – PhoemueX Jan 4 '15 at 12:15
  • $\begingroup$ $\infty$ thanks! $\endgroup$ – Self-teaching worker Jan 5 '15 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.