I am reading a proof that uses the following fact which I cannot show:
Let $b_1, b_2 : \mathbb{R} \rightarrow \mathbb{R}$ be two continuous functions such that $b_1 (x) < b_2 (x),$ for all $x \in \mathbb{R}$. Then, for each integer $m \geq 3$, there exists a Lipschitz-continuous function $b_m: \mathbb{R} \rightarrow \mathbb{R}$ such that $$ b_1 (x) \leq b_m (x) \leq b_2 (x), \text{ for } |x| \leq m.$$
Any ideas on how to show this?
For given $m$, on the bounded and closed interval $[-m,m]$ the difference of the functions is bounded from below by a constant $\eta>0$. In other words, there is a 'tube' of width $\eta $ between the graphs of these functions.
Moreover, on that interval, they are both uniformly continuous. So you may assume that there is $\delta>0$ such that for both functions $b_i$, the implication $|x-y|\le \delta \Rightarrow |b_i(x)-b_i(y) |\le \frac{1}{2}\eta$, say, is true.
Now with this information you can create a piecewise linear function by looking at a partition of width $\delta$ of the interval $[-m,m]$ by adding linear pieces from left to right which remain in the tube between the graphs of the $b_i$. (This is a bit of a nasty task which I will not carry out in detail, but it's not too difficult). That piecewise linear function will be Lipshitz, the Lipshitz constant given by the subinterval of lenght $\delta$ with the steepest slope of that piecewise linear function.
Alternatively you may use the fact (if you know it) that the polynomials (or the smooth functions) are dense in the space of continuous functions. With $\eta$ as above, choose a polynomial which is $\frac{\eta}{4}$ close to the upper function $b_2$ and move it down by $\frac{\eta}{2}$