Is there a sandwich method for uniform convergence? 

Is there a sandwich method that allows you to claim that If given
  sequences $(f_n), (g_n), (h_n)$
We know that $g_n, h_n$ are uniformly convergent and $\forall x,
> \forall n$
$$g_n \leq f_n \leq h_n$$ (maybe not exact same format)
Then $f_n \rightrightarrows f$?

By the way, is anyone else hungry?

 A: I'm going to assume that we also have that $(g_n)$, and $(h_n)$ converge (not necessarily uniformly) to the same function $f$, otherwise there is no reason to believe that $f_n$ converges at all (take for a counterexample, $g_n$ to always be the constant $0$ function, and $h_n$ to always be the constant $1$ function. Then letting the $f_n = g_n$ for even $n$ and $f_n = h_n$ for odd $n$ gives a sequence $(f_n)$ that does not converge at all, much less uniformly.
Now we prove the claim.
Suppose $(g_n)$ and $(h_n)$ converge uniformly to some function $f$.
We seek to show that for every $\epsilon > 0$, there exists some integer $N$ such that for all integers $n > N$, we have $|f_n(x) - f(x)| < \epsilon$ for all $x\in \mathbb{R}$.
So, we let $\epsilon > 0$ be arbitrary.
Since $(g_n)$ converges uniformly to $f$, there exists an integer $N_1$ such that for all integers $n > N_1$, we have
$$
|g_n(x) - f(x)| < \frac{\epsilon}{3}
$$
for all $x\in \mathbb{R}$.
Similarly, there exists an integer $N_2$ such that for all integers $n > N_2$, we have
$$
|h_n(x) - f(x)| < \frac{\epsilon}{3}
$$
for all $x\in \mathbb{R}$.
Now define $N = \max(N_1, N_2)$.
Notice that for $n > N$, we have
$$
|g_n(x) - h_n(x)| \leq |g_n(x) - f(x)| + |h_n(x) - f(x)| < \frac{2\epsilon}{3}
$$
for all $x\in \mathbb{R}$ by the triangle inequality.
By the relation that $g_n(x) \leq f_n(x) \leq h_n(x)$ for all $x\in \mathbb{R}$, we have that
$$
|f_n(x) - g_n(x)| < \frac{2\epsilon}{3}
$$
for all $n > N$.
Therefore, putting everything together, we have
$$
|f_n(x) - f(x)| \leq |f_n(x) - g_n(x)| + |g_n(x) - f(x)| < \frac{\epsilon}{3} + \frac{2\epsilon}{3} = \epsilon
$$
for all integers $n > N$ and all $x\in \mathbb{R}$.
Therefore, $f_n$ converges uniformly to $f$, as desired.
