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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 $\beta=\alpha.$

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    $\begingroup$ Looks like you can get away with first taking the composition $G\circ f$ where $G(x)=0$ of $[-\delta,\delta]$, $x-\delta$ for $x>\delta$ and $x+\delta$ for $x<-\delta$ and then convolving this composition with a smooth narrow bump, cannot you? $\endgroup$
    – fedja
    Commented Apr 12, 2023 at 2:19
  • $\begingroup$ As @fedja suggested or also: first approximate $f$ with smooth $f_n$ in $C^\gamma$ with $\beta<\gamma<\alpha$ and then take $f \wedge f_n$ and so on. I am using the fact that a Lipschitz map $F$ defines a continuous map $u \mapsto F(u)$ from $C^\gamma \to C^\beta$ (if you want $\beta=\gamma$ you need $F \in C^1$). You need also to write $f \wedge g=(f-g)^-+f$. $\endgroup$ Commented Apr 12, 2023 at 10:44
  • $\begingroup$ @fedja : I am afraid that where $f$ was $0$ (or very close to $0$) (and remained so after the composition with $G$) it might be not necessary that the (near) vanishing property is preserved after the convolution. $\endgroup$ Commented Apr 12, 2023 at 11:04
  • $\begingroup$ @GiorgioMetafune : Can you explain what you mean by "so on"? $\endgroup$ Commented Apr 12, 2023 at 11:05
  • $\begingroup$ @IosifPinelis Sorry for being unclear. Having $f_n$, use $(f \wedge f_n) \vee (-f)$. $\endgroup$ Commented Apr 12, 2023 at 11:10

1 Answer 1

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$\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}\newcommand{\de}{\delta}\newcommand{\J}{\mathcal J} \newcommand{\be}{\beta}\newcommand{\ep}{\varepsilon}$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 $J\in\J_\de$. $\quad\Box$

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  • $\begingroup$ Does your proof give that for any continuous and positive $f$, there exists a smooth $g$ such that $0 \leq g \leq f$ poitwise? Also that does not look obvious to me (unless $f$ is strictly positive). $\endgroup$ Commented Apr 12, 2023 at 17:49
  • $\begingroup$ @GiorgioMetafune : I don't think that this proof implies the result you mention, for continuous (rather than Hölder-continuous) functions. However, I think the just-continuous case should be somewhat simpler. $\endgroup$ Commented Apr 12, 2023 at 18:15
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    $\begingroup$ For continuous $f \geq 0$ this simple construction works. If $M$ is the maximum of $f$, $K=\{f \geq M/2\}$, $V=\{f >M/4\}$ let $g$ be smooth with compact support in $V$ with $0 \leq g \leq M/4$ and $g=M/4$ in $K$. Then $0 \leq g \leq f$ and $f-g \leq (3/4)M$. Iterating this argument one finds $0 \leq f-(g_1+\dots g_n) \leq (3/4)^n M$ (in the previous comment I forgot to say that $f-g$ should be small). $\endgroup$ Commented Apr 13, 2023 at 7:03
  • $\begingroup$ @GiorgioMetafune : Nice argument! $\endgroup$ Commented Apr 13, 2023 at 13:57

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