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)}\le L(\al,\be)\ep \tag{10}\label{10}
\end{equation*}\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*}
where $L(\al,\be)$ is a real number depending only on $\al,\be$.\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 $>4\de$$>\de$. Of course, the set $\J_\de$ is finite. Let
\begin{equation*}
f_\ep(x):=0\text{ for }x\in I\setminus\bigcup_{J\in\J_\de}J.
\end{equation*}
If $\de$ is small (which will be henceforth assumed), then on $I\setminus\bigcup_{J\in\J_\de}J$ the function $f-f_\ep=f$ is small and $(1,\al)$-Hölder. So, $f-f_\ep=f$ is $(\de_1,\be)$-Hölder on $I\setminus\bigcup_{J\in\J_\de}J$ for a small $\de_1$. This follows because $|y-x|^\al<<|y-x|^\be$ if $\be\in(0,\al)$ (as given) and $|y-x|<<1$, whereas $|y-x|^\al\asymp|y-x|^\be\asymp1$ if $|y-x|\asymp1$. (We write $E<<F$ if $E=o(F)$, $E\ll F$ if $E=O(F)$, and $E\asymp F$ if $E\ll F\ll E$.)
We shall build the function $f_\ep$ separately on each interval $(a,b)\in\J_\de$ we have$J=(a,b)\in\J_\de$, in such a manner that
\begin{equation}
\text{$f_\ep:=0$ on $(a,a+\de)\cup(b-\de,b)$ }\tag{30}\label{30}
\end{equation}\begin{equation*}
\text{$f_\ep=0$ near the endpoints of $J$. } \tag{30}\label{30}
\end{equation*}
andThen the smoothness of $f_\ep$ is smooth on $(a,b)$, theneach interval $f_\ep$$J\in\J$ will be smoothenough for the smoothness of $f_\ep$ on the entire interval $I$.
Take nowindeed any interval $(a,b)\in\J_\de$ indeed. It is enough to construct a function $f_\ep$ which is smooth on $J$ and such that \eqref{10}, \eqref{20}, and \eqref{30} hold$J\in\J_\de$.
Note
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,
\begin{equation*}
f\ge2c\text{ on }[a+\de,b-\de] \tag{40}\label{40}
\end{equation*}
for $M:=f(x_*)\ge f(x)$ for some real $c>0$$x_*\in J$ and all $x\in J$.
For a very smallany $\eta\in(0,\de]$$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*}
Then
\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 aany smooth nonnegative function supported on the interval $[-\eta,\eta]$ such that $\int K_\eta=1$. Let then
\begin{equation*}
g_\eta:=g*K_\eta,
\end{equation*}
where
\begin{equation*}
g:=(f-c)1_{[a+2\de,b-2\de]}.
\end{equation*}
ThenMoreover, by \eqref{40}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 $g\ge0$ and hence$\eta^\al\le c$, in view of conditionswhich will be indeed assumed henceforth.
Further, for $\eta\in(0,\de]$ and \eqref{40}$x\in[x_+(c),x_+(c)+\eta]$,
\begin{align}
g_\eta(x)&=\int_{-\de}^\de du\,K_\eta(u)(f(x-u)-c)1(x-u\in[a+2\de,b-2\de]) \tag{45}\label{45} \\
&\le\eta^\al+\int_{-\de}^\de du\,K_\eta(u)(f(x)-c)1(x-u\in[a+2\de,b-2\de]) \notag \\
&\le\eta^\al+(f(x)-c)1(x\in[a+\de,b-\de]) \notag \\
&\le\eta^\al+(f(x)-c)\le f(x) \notag
\end{align}\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}
forOn the other hand, again for $x\in(a,b)$
if$x\in[x_+(c),x_+(c)+\eta]$, 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*}
\eta\in(0,\min(\de,c^{1/\al})), \tag{50}\label{50}
\end{equation*}\begin{equation*}
2\eta^\al\le c, \tag{50}\label{50}
\end{equation*}
and \eqref{50}which will be henceforthindeed assumed henceforth.
AlsoSimilarly, the integrand in$g_\eta(x)\le\eta^\al$ given \eqref{4550} is $\ge0$ for all $x\in(a,b)$ and $=0$ for $x\in(a,a+\de)\cup(b-\de,b)$. So
Also, $g_\eta\ge0$ on $(a,b)$ and $g_\eta=0$ on$g_\eta(x)=0\le f(x)$ for $(a,a+\de)\cup(b-\de,b)$$x\in J\setminus J_c$.
SoWe 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),
\end{equation*}\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 condition$f_\ep$ is uniformly close to $f$ on $J$, also uniformly in $J\in\J_\de$.
By \eqref{1045} for our interval, on $J=(a,b)$ holds if$J_c$ the function $\de$ and$g$ is uniformly close to $f$ if $c$ are chosen to beis small enough. This is rather straightforward and, which will be detailed laterhenceforth assumed. Also, for $x\in J\setminus J_c$ we have $0\le f\le\de^\al$, which is small (Itsince $\de$ was assumed to be small) and hence close to $g$ (which is rather late here now.$0$ 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 $J\in\J_\de$. $\quad\Box$
