Let $T(x)$ be a differentiable function from $\mathbb{R}$ to $\mathbb{R}$ such that $T(x) = x \iff x \in \mathbb{Z}$. Prove that there is always a set of points such that $|T'(x)| > 1$
Proof: Lets consider the interval $[n,n+1]$ where $n\in \mathbb{Z}$. By the Mean Value Theorem we know that there must exist a point $x_0 \in (n,n+1)$ such that: $$ T'(x_0) = \frac{T(n+1) - T(n)}{(n+1) - n} = \frac{n+1 - n}{n+1 - n} = 1 $$
Given that $T(x)$ has fixed points only when $x \in \mathbb{Z}$ we know that $T(x)$ is not the identity function and therefore we will have two cases:
- $T(x) - x > 0$, $\forall x \in (n,n+1)$
- $T(x) - x < 0$, $\forall x \in (n,n+1)$
For case 1. we can use the MVT again to show that there exist a point $x_1 \in (n,x_0)$ such that $$ T'(x_1) = \frac{T(x_0) - T(n)}{x_0-n} = \frac{T(x_0) - n}{x_0 - n} > \frac{x_0 - n}{x_0 - n} = 1 $$
For case 2. we can also use the MVT to show that there exist a point $x_2 \in (x_0, n+1)$ such that $$ T'(x_2) = \frac{T(n+1) - T(x_0)}{n+1 - x_0} = \frac{(n+1) - T(x_0)}{(n+1) - x_0} > \frac{n+1 - x_0}{n+1 - x_0} = 1 $$
This would show that there always points for which $T'(x) > 1$ and therefore $|T'(x)| > 1$. However, I would like to also show that there necessarily exist points where $T'(x) < -1$.
You can't, because it's not necessarily true. Take, for example, $$T(x) = \frac{1}{\pi}\sin(\pi x) + x.$$ Then $$T(x) = x \iff \frac{1}{\pi}\sin(\pi x) = 0 \iff x \in \Bbb{Z}.$$ Further, $$T'(x) = 1 + \cos(\pi x) \in [0, 2],$$ hence $T'(x) < -1$ never occurs.