# Smooth approximation of Hölder functions "from below"

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 $$\beta=\alpha.$$

• 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? Commented Apr 12, 2023 at 2:19
• 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 $f \wedge g=(f-g)^-+f$. Commented Apr 12, 2023 at 10:44
• @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. Commented Apr 12, 2023 at 11:04
• @GiorgioMetafune : Can you explain what you mean by "so on"? Commented Apr 12, 2023 at 11:05
• @IosifPinelis Sorry for being unclear. Having $f_n$, use $(f \wedge f_n) \vee (-f)$. Commented Apr 12, 2023 at 11:10

$$\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}\newcommand{\de}{\delta}\newcommand{\J}{\mathcal J} \newcommand{\be}{\beta}\newcommand{\ep}{\varepsilon}$$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 $$E< if $$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 $$$$g_\eta\le f\text{ on }J.$$$$

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 $$J\in\J_\de$$. $$\quad\Box$$

• 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). Commented Apr 12, 2023 at 17:49
• @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. Commented Apr 12, 2023 at 18:15
• 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). Commented Apr 13, 2023 at 7:03
• @GiorgioMetafune : Nice argument! Commented Apr 13, 2023 at 13:57