If $f\in C([0,\infty))$, does $\delta>0$ and $g\in C^1((0,\delta))\cap C([0,\delta])$ s.t. $g\geq f$ on $[0,\delta]$ and $g(0)=f(0)$ exist?

Question

The question is the following:

Suppose $f : [0,\infty) \rightarrow \mathbb{R}$ is a continuous function. Can I find $\delta \in (0,\infty)$ and a function $g : [0,\delta] \rightarrow \mathbb{R}$ such that $g$ is continuous on $[0,\delta]$, differentiable in $(0,\delta)$, and satisfies $g \geq f$ on $[0,\delta]$ and $g(0) = f(0)$?

I am not sure if this is true, as continuous functions can indeed be non-differentiable everywhere (for example, paths of Brownian motion or the Weierstrass function). It seems the question ultimately hinges on how problematic differentiability becomes near zero. However, I haven’t been able to come up with a proof or a counterexample.

Answer 1

Sure you can find such a $g$ and without much difficulty. Without loss of generality we assume that $f(0) = 0$ (otherwise just subtract it). Construct the sequence $\delta_n>0$ inductively in such a way that $\delta_{n+1}\le \frac{\delta_n}{2}$ and $f(x) \le 2^{-n}$ for $0\le x \le \delta_n$ (for $\delta_1$ the first condition is void). This sequence exists because we first satisfy the second condition by the definition of continuity at $0$ (we don't even need to assume that $f$ is continuous anywhere else) and then if the $\delta_{n+1}$ we got is bigger than $\frac{\delta_n}{2}$ we artificially decrease it and make it equal to $\frac{\delta_n}{2}$.

Now, we will construct $g$ on each interval $I_n = [\delta_{n+1}, \delta_n]$ separately. Chop $I_n$ into three intervals of equal length $J_n, L_n, K_n$. On $J_n$ we put $g(x) = 2^{-n+1}$, on $K_n$ we put $g(x) = 2^{-n+2}$ and on $L_n$ we connect them by your favourite $C^\infty$ increasing function for which all the derivatives vanish at the end points, so that the resulting function $g$ is $C^\infty$ and on $I_n$ it is at least $2^{-n+1}$ and at most $2^{-n+2}$. Since by the definition of $\delta_n$ on $I_n$ we have $f(x) \le 2^{-n+1}$ this implies that on $(0, \delta_1]$ we have $f(x) \le g(x)$. It remains to define $g(0) = 0$, this extension is continuous and we still have $f(x) \le g(x)$ on $[0, \delta_1]$. In fact it is not hard to make $g$ be defined on the whole $[0, +\infty)$.

This problem can also be killed by dropping a nuke and using strong $C^\infty$ topology on $(0, +\infty)$, but this is clearly excessive.

Answer 2

$\newcommand\R{\Bbb R}$Another construction:

For real $x\ge0$, let $$F(x):=\max_{y\in[0,x]}f(y).$$ Then $F$ is a continuous nondecreasing real-valued function on $[0,\infty)$ such that $F\ge f$ and $F(0)=0$. Let now $$g(x):=\int_0^1 dt\,h(t) F(x+tx)=\frac1x\int_x^{2x}du\,F(u)h\Big(\frac ux-1\Big) \\ =\frac1x\int_0^\infty du\,F(u)h\Big(\frac ux-1\Big)$$ for real $x>0$, with $g(0)=0$, where $h(t):=c\exp(-\frac1{(1-t)t})$ for $t\in(0,1)$, $h(t):=0$ for real $t\notin(0,1)$, and $c:=1/\int_0^1 dt\,\exp(-\frac1{(1-t)t})$, so that $h$ is smooth on $\R$ and $\int_0^1 dt\,h(t)=\int_\R dt\,h(t)=1$.

Then $g\ge F\ge f$, $g(0)=0$, and $g$ is continuous on $[0,\infty)$ and differentiable $(0,\infty)$. $\quad\Box$