We assume that we have a $\alpha$-Hölder continuous function $f$ on an interval $[0,1].$


I am wondering if there exists an explicit construction of a sequence $f\_{n} \in C\_c^{\infty}(\mathbb R)$ such that


$$\lVert f-f\_n\rVert\_{C^{\beta}([0,1])} \le \frac{1}{n}$$


for fixed $\beta<\alpha$ and $\lvert f\_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition. In an earlier post, I mistakingly took $\beta=\alpha.$



Smooth approximation of Hölder functions "from below"
$\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}\newcommand{\de}{\delta}\newcommand{\J}{\mathcal J}
\newcommand{\be}{\beta}\newcommand{\ep}{\varepsilon}$Yes, such a construction exists.


Indeed, take any real $\ep>0$. For a real $C\ge0$ and $\al\in(0,1]$, let us say that a function $g$ is $(C,\al)$-Hölder on a set $S\subseteq[0,1]$ if $|g(y)-g(x)|\le C|y-x|^\al$ for all $x,y$ in $S$.


Without loss of generality (wlog), the function $f$ is $(1,\al)$-Hölder on $[0,1]$.


Let $Z:=\{z\in I:=[0,1]\colon f(z)=0\}$ and $N:=I\setminus Z$. Wlog, $\{0,1\}\in Z$ (otherwise, extend $f$ appropriately to an interval $[A,B]\supset I$ so that $f(A)=0=f(B)$ and then shrink the interval $[A,B]$ to $I$). So, $N=\bigcup\_{J\in\J}J$ for some (countable) set $\J$ of pairwise disjoint nonempty open subintervals of $I$.


It is enough to construct a smooth function $f\_\ep$ such that for each $J\in\J$
\begin{equation\*}
 \|f-f\_\ep\|\_{C^\be(J)}\le100\ep \tag{10}\label{10}
\end{equation\*}
and
\begin{equation\*}
 |f\_\ep|\le|f|\text{ on } J. \tag{20}\label{20}
\end{equation\*}


To begin such a construction, take any real $\de>0$. Let $\J\_\de$ denote the set of all intervals $J\in\J$ of length $>\de$. Of course, the set $\J\_\de$ is finite. Let
\begin{equation\*}
 f\_\ep(x):=0\text{ for }x\in I\setminus\bigcup\_{J\in\J\_\de}J. 
\end{equation\*}


If $\de$ is small (which will be henceforth assumed), then on $I\setminus\bigcup\_{J\in\J\_\de}J$ the function $f-f\_\ep=f$ is small and $(1,\al)$-Hölder. So, $f-f\_\ep=f$ is $(\de\_1,\be)$-Hölder on $I\setminus\bigcup\_{J\in\J\_\de}J$ for a small $\de\_1$. This follows because $|y-x|^\al<<|y-x|^\be$ if $\be\in(0,\al)$ (as given) and $|y-x|<<1$, whereas $|y-x|^\al\asymp|y-x|^\be\asymp1$ if $|y-x|\asymp1$. (We write $E<<F$ if $E=o(F)$, $E\ll F$ if $E=O(F)$, and $E\asymp F$ if $E\ll F\ll E$.)


We shall build the function $f\_\ep$ separately on each interval $J=(a,b)\in\J\_\de$, in such a manner that
\begin{equation\*}
\text{$f\_\ep=0$ near the endpoints of $J$. } \tag{30}\label{30} 
\end{equation\*}
Then the smoothness of $f\_\ep$ on each interval $J\in\J$ will be enough for the smoothness of $f\_\ep$ on the entire interval $I$.


Take indeed any interval $J\in\J\_\de$.
Note that (i) $f(a)=0=f(b)$ and (ii) either $f>0$ on $(a,b)$ or $f<0$ on $(a,b)$. Wlog, $f>0$ (on $(a,b)$). Moreover, $f$ is continuous. So, $M:=f(x\_\*)\ge f(x)$ for some $x\_\*\in J$ and all $x\in J$.
For any $c\in(0,M]$, the points
\begin{equation\*}
 x\_+(c):=\min\{x\in[x\_\*,b)\colon f(x)\le c\},\quad x\_-(c):=\max\{x\in(a,x\_\*]\colon f(x)\le c\} 
\end{equation\*}
are well defined. Moreover,  

\begin{equation\*}
 f\ge c\text{ on }J\_c:=[x\_-(c),x\_+(c)],\quad f(x\_\pm(c))=c. \tag{40}\label{40}
\end{equation\*}


Let
\begin{equation\*}
 g:=(f-c)1\_{J\_c}. \tag{45}\label{45}
\end{equation\*}
Then $g$ is $(1,\al)$-Hölder on $\R$ and hence so is
\begin{equation\*}
 g\_\eta:=g\*K\_\eta, \tag{47}\label{47}
\end{equation\*}
where, for each $\eta\in(0,\de)$, we let $K\_\eta$ be any smooth nonnegative function supported on the interval $[-\eta,\eta]$ such that $\int K\_\eta=1$.


Moreover, for $x\in J\_c=[x\_-(c),x\_+(c)]$,  

\begin{align}
 g\_\eta(x)&=\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x-u)-c)1(x-u\in J\_c) \notag \\ 
 &\le\eta^\al+\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x)-c)1(x-u\in J\_c) \notag \\ 
 &\le\eta^\al+(f(x)-c)\le f(x) \notag 
\end{align}
assuming that $\eta^\al\le c$, which will be indeed assumed henceforth.


Further, for $x\in[x\_+(c),x\_+(c)+\eta]$,  

\begin{align}
 g\_\eta(x)&=\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x-u)-c)1(x-u\in J\_c) \notag \\ 
&=\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x-u)-f(x\_+(c))1(x-u\in J\_c) \notag \\ 
&\le\int\_{-\eta}^\eta du\,K\_\eta(u)(x\_+(c)-(x-u))^\al 1(x-u\in [x\_-(c),x\_+(c)]) \notag \\ 
&\le\int\_{-\eta}^\eta du\,K\_\eta(u)\eta^\al 1(x-u\in [x\_-(c),x\_+(c)]) 
\le\eta^\al. \notag 
\end{align}
On the other hand, again for $x\in[x\_+(c),x\_+(c)+\eta]$, we have $f(x)\ge f(x\_+(c))-(x-x\_+(c))^\al\ge c-\eta^\al\ge\eta^\al\ge g\_\eta(x)$ assuming that
\begin{equation\*}
 2\eta^\al\le c, \tag{50}\label{50}
\end{equation\*}
which will be indeed assumed henceforth.
Similarly, $g\_\eta(x)\le\eta^\al$ for $x\in[x\_-(c)-\eta,x\_-(c)]$ given \eqref{50}.
Also, $g\_\eta(x)=0\le f(x)$ for $x\in J\setminus J\_c$.
We conclude that
\begin{equation}
 g\_\eta\le f\text{ on }J. 
\end{equation}


So, letting
\begin{equation\*}
 f\_\ep:=g\_\eta\text{ on }J=(a,b), \tag{60}\label{60}
\end{equation\*}
we see that $f\_\ep$ is smooth on $J=(a,b)$, and conditions \eqref{20} and \eqref{30} hold.
Moreover, $g\_\eta$ is $(1,\al)$-Hölder on $\R$ and hence $f\_\ep$ is $(1,\al)$-Hölder on $J$.


So, it remains to check that $f\_\ep$ is uniformly close to $f$ on $J$, also uniformly in $J\in\J\_\de$.


By \eqref{45}, on $J\_c$ the function $g$ is uniformly close to $f$ if $c$ is small enough, which will be henceforth assumed.


Also, on $J\setminus J\_c$ we have $0\le f\le\max((b-x\_+(c))^\al,(x\_-(c)-a)^\al)$, which is small (since $c$ was assumed to be small), and hence $f$ is close to $g$ (which is $0$ on $J\setminus J\_c$). So, $g$ is uniformly close to $f$ on $J$, uniformly in $J\in\J\_\de$.


Finally, because $\eta$ is small, and in view of \eqref{60} and \eqref{47}, we conclude that indeed $f\_\ep$ is uniformly close to $f$ on $J$, uniformly in $J\in\J\_\de$. $\quad\Box$


