real analysis - Finding a differentiable function from (0,1) to (0,1) dominating (point wise) a given continuous function from (0,1) to (0,1)

Suppose we have a continuous function $f: (0,1) \to (0,1)$. Does there exist a differentiable function $\phi: (0,1) \to (0,1)$ such that $f(x) \leq \phi(x)$?

There does exist such a differentiable function if the function $f$ has finitely many non-differentiable points. Also this can be done if there exists an $\epsilon >0$, no matter how small, such that the function $f$ is differentiable in $(0,\epsilon)$.

Note that the real challenge lies in finding the differentiable function $\phi$ to be strictly less than one at the same time. This seems very much possible, at least pictorially. Because the graph of the function $f$ lies strictly below the $Y=1$ line. And hence one can draw a smooth curve between the graph of $f$ and the $Y=1$ line. But I can not prove it mathematically (rigorously). I need your help. Thanks in advance!

Answer

Pick monotonic sequences $a_n\to 0$ and $b_n\to 1$.
Since $f$ is bounded away from $1$ on $[a_{n},b_{n}]$, you can readily find smooth (in fact constant) $\psi_n\colon[a_n,b_n]\to (0,1)$ that is differentiable and $\ge f$ on this interval.
Now you can recursively glue together $\phi_n$ on $[a_{n},b_{n}]$ such that $\phi_n(x)=\phi_{n-1}(x)$ for $x\in[a_{n-2},b_{n-2}]$, $\psi_n(x)\ge\phi_n(x)\ge\phi_{n-1}(x)\setminus[a_{n-2},b_{n-2}]$ for $x\in[a_{n-1},b_{n-1}]$, $\phi_n(x)=\psi_n(x)$ for $x\in[a_{n1},b_{n1}]\setminus [a_{n-1},b_{n-1}]$.
Finally let $\phi(x)=\phi_{n+1}(x)$ where $n$ is such that $x\in[a_{n},b_{n}]$.