Does the sandwich (squeeze) theorem + continuity imply the derivatives are squeezed, too?

Question

Say I have three functions $f_-(x)$, $f_+(x)$, and $f(x)$ that obey: \begin{align} f_-(x) &\le f(x) \le f_+(x), \\ f_+'(x) &\le f_-'(x), \\ f_0 & = \lim_{x \rightarrow x_0} f_-(x) = \lim_{x \rightarrow x_0} f_+(x),\ \mathrm{and} \\ f'_0 & = \lim_{x \rightarrow x_0} f_-'(x) = \lim_{x \rightarrow x_0} f_+'(x), \end{align} for all $x$ in some (possibly infinite) open interval $[a,b]$ ($x_0$ in the same interval). Both the upper and lower bounds are at least $C_1$ continuous. Naturally, this is sufficient to show that $\lim_{x\rightarrow x_0} f(x) = f_0$ by the sandwich (squeeze) theorem.

I want to also show that $f'(x)$ is similarly sandwiched. Are the above properties sufficient to show that $\lim_{x\rightarrow x_0} f'(x) = f'_0$? If not, is it sufficient to require that $f(x)$ also be $C_1$ continuous to get the desired result? If not, what additional limitations/properties need to be imposed on $f(x)$ to ensure the desired result?

Answer 1

Suppose $f \le g \le h$, with $f(x_0) = g(x_0) = h(x_0)$, and $f$ and $h$ are differentiable at $x_0$ with $f'(x_0) = h'(x_0)$. Then, $g$ is also differentiable at $x_0$. Note that $$\frac{f(x_0 + t) - f(x_0)}{t} \le \frac{g(x_0 + t) - g(x_0)}{t} \le \frac{h(x_0 + t) - h(x_0)}{t},$$ If we extend these three functions to $t = 0$ by setting them to be $f'(x_0) = h'(x_0)$, then the regular squeeze theorem holds, so, $$g'(x_0) = \lim_{t\rightarrow 0} \frac{g(x_0 + t) - g(x_0)}{t} = f'(x_0).$$

In fact, the assumption that $f'(x_0) = h'(x_0)$ can be deduced. We have,

$$\frac{f(x_0 + t) - f(x_0)}{t} \le \frac{h(x_0 + t) - h(x_0)}{t} \implies f'(x_0) \le h'(x_0),$$ but on the other hand, $$\frac{f(x_0 - t) - f(x_0)}{-t} \ge \frac{h(x_0 - t) - h(x_0)}{-t} \implies f'(x_0) \ge h'(x_0).$$