Existence of increasing, smooth modulus of continuity First, recall the definition: Given a function $f:M\to N$, where $M$ and $N$ are metric spaces, a modulus of continuity for $f$ is a function $\omega:[0,\infty)\to[0,\infty)$ such that


*

*$\omega(0)=\lim_{t\to 0^+}\omega(t)=0$;

*For every $x,x'\in M$,  $d(f(x),f(x'))\leq\omega(d(x,x'))$.


According to Wikipedia, the existence of a modulus of continuity $\omega$ for a function $f$ implies the existence of a increasing and smooth (in $(0,\infty)$) modulus of continuity for $f$. I don't see how we can obtain this.
The way suggested is the following: Given a modulus of identity $\omega$, let $\omega_1(t)=\sup_{s\leq t}\omega(s)$. Then $\omega_1$ is also a modulus of identity and is increasing (hence measurable, and locally integrable). Then, let $\omega_2(t)=\frac{1}{t}\int_t^{2t}\omega_1(s)ds$. $\omega_2$ is also a modulus of identity and it is continuous. I haven't checked, but I believe that with Lebesgue's Differentiation Theorem or something of the kind, we can show that $\omega_2$ is non-decreasing (alternatively, simply use $\omega_3(t)=\sup_{s\leq t}\omega_2(t)$).
Then, it is stated that a suitable adaptation in the definition of $\omega_2$ will give us a smooth function. By this "suitable adaptation" I believe we should use some kind of molification: Notice that the definition of $\omega_2$ is pretty much molification by the pulse function $p(x)=\begin{cases}1&\text{, if }|x|\leq 1/2\\0&\text{, otherwise}\end{cases}$. This is what I was thinking:
Let $\rho:\mathbb{R}\to\mathbb{R}$ be smooth, non-negative, $\operatorname{supp}\rho\subseteq[-1,1]$, and $\int_{-1}^1\rho(x)dx=1$. Making an abuse of notation, let's define, for $t>0$, $\rho(x,t)=\frac{2}{t}\rho(\frac{2x}{t}-3)$, so that $\operatorname{supp}\rho(\cdot,t)\subseteq[t,2t]$ and $\int_t^{2t}\rho(x,t)dx=1$. Finally, we define $\widetilde{w}(t)=\int_t^{2t}\rho(x,t)w(x)dx$. Using the usual arguments about molifications/Dominated Convergence, it is easy to show that $\widetilde{w}$ is smooth in $(0,\infty)$, and Dominated Convergence also implies that $\lim_{t\to 0}\widetilde{w}(t)=0$. The problem is showing that $\widetilde{w}$ is increasing.
Alternative approachs are appreciated.
 A: While this is an old question, since this assertion is mentioned on wikipedia without a proof nor reference I think it would be useful to have a solution written somewhere. The approach suggested in the question works and I am merely completing the details; the main addition is a change of variables, which also makes it clear why $\omega_2$ is increasing.

We assume $\omega \colon [0,\infty) \to [0,\infty)$ is non-decreasing and is continuous at zero, where it vanishes. Fix a mollifier $\rho \in C^{\infty}_c(\Bbb R)$ which is non-negative with $\operatorname{supp} \rho \subset (0,2)$ such that $\int_1^2 \rho(t) \,\mathrm{d} t = 0,$ we define
$$ \widetilde\omega(t) = \frac1t \int_{t}^{2t} \rho(s/t)\omega(s)\,\mathrm{d}s = \frac1t \int_{\Bbb R} \rho(s/t)\omega(s) \,\mathrm{d} t. \tag{$\dagger$}$$
We claim this works; differentiating the under the integral sign we see $\widetilde\omega$ is indeed smooth on $(0,\infty).$ For the remaining properties we use the change of variables $s \mapsto s/t$ to get
$$ \widetilde\omega(t) = \int_1^2 \rho(s) \omega(ts) \,\mathrm{d}s.\tag{$\ddagger$}$$
By monotonicity of $\omega$ we infer that
$$ \omega(t) \leq \widetilde\omega(t) \leq \omega(2t), $$
so in particular $\widetilde\omega(t) \to 0$ as $t \to 0^+.$
For monotonicity if $t_1<t_2$ we have $\omega(t_1s) \leq \omega(t_2s)$ for all $0\leq s \leq 1,$ so it follows by ($\ddagger$) that $\widetilde\omega(t_1) \leq \widetilde\omega(t_2).$ Note a similar argument shows that if $\omega$ is concave, the same holds for $\widetilde\omega.$
