Suppose $f,g$ are two uniformly continuous functions on $\mathbb R$, and $h$ is a continuous function on $\mathbb R$ that satisfies:$$f(x)\le h(x) \le g(x)$$Does that mean $h$ is a uniformly continuous function? How to prove that? Or what other requirements should be added to make this be true?
I came across this idea when proving a question, but I can't find an answer.
Take the counterexample$$ -1 \le \sin(x^2) \le 1 \quad \forall x \in \mathbb{R}, $$ but the function $h(x) = \sin(x^2)$ is not uniformly continuous.
The required condition would be;
Let $f, g, h: \mathbb{R} \to \mathbb{R}$ be continuous functions. Suppose:
Then $h$ is uniformly continuous on $\mathbb{R}$.
Proof: Let $\epsilon > 0$ be given.
Since $\lim_{|x| \to \infty} (g(x) - f(x)) = 0$, there exists an $M > 0$ such that for all $|x| \ge M$: $$|g(x) - f(x)| < \frac{\epsilon}{3}$$ Using the sandwich condition, we know $|h(x) - f(x)| \le |g(x) - f(x)|$. Thus, for $|x| \ge M$: $$|h(x) - f(x)| < \frac{\epsilon}{3}$$
Since $f$ is uniformly continuous on $\mathbb{R}$, there exists $\delta_1 > 0$ such that for any $x, y \in \mathbb{R}$: $$|x - y| < \delta_1 \implies |f(x) - f(y)| < \frac{\epsilon}{3}$$
Consider the compact interval $K = [-(M+1), M+1]$. Since $h$ is continuous on $K$, it is uniformly continuous on $K$ (by the Heine-Cantor theorem). Thus, there exists $\delta_2 > 0$ such that for any $x, y \in K$: $$|x - y| < \delta_2 \implies |h(x) - h(y)| < \epsilon$$
Let $\delta = \min(1, \delta_1, \delta_2)$. Consider any $x, y \in \mathbb{R}$ with $|x - y| < \delta$.
Case 1: $x \in [-M, M]$
Since $|x - y| < 1$, $y$ must be in $[-(M+1), M+1]$. Therefore, both points are in $K$. $$|h(x) - h(y)| < \epsilon$$
Case 2: $|x| \ge M$, $|y| \ge M$.
We use the triangle inequality, using $f$ as a bridge between $x$ and $y$: $$|h(x) - h(y)| \le |h(x) - f(x)| + |f(x) - f(y)| + |f(y) - h(y)|$$ Substituting the bounds: $$|h(x) - h(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$
Since $|h(x) - h(y)| < \epsilon$ in all cases, $h$ is uniformly continuous.
This is not true. Take $f(x)=|x|$ and $g(x)=-|x|$ they are both uniformly continuous on $\mathbb{R}$. Let $h(x)=x\sin(x)$. This will not be uniformly continuous since the values move apart faster and faster as $x$ grows.
