Assume that $f$ is a real continuous function defined on $[a,b]$, which is also differentiable on $(a,b)$, where $a,b \in R$. Then the statement " $f'$ is bounded " is false according to this If a differentiable function has bounded derivative, Must it be that its derivative continuous?. However I thought at first that the statement was true and so I wrote a "proof":
Assume that $f'$ is unbounded on $(a,b)$. Then there exists $x_1 \in (a,b)$ such that $f'(x_1) > 1$. Similarly for every $n > 1$, there exists $x_n \in (a,b)$ such that $f'(x_n) > max(f'(x_{n-1}), n)$. Then for $\epsilon = f'(x_n) - n \;(n\in N)$, there exists $\delta_n $ such that if $t \in (x_n, x_n + \delta_n) \subset (a,b)$ then $|\frac {f(t) - f(x_n)}{t-x_n} - f'(x_n)| < \epsilon$, which means that $ n < f(x_n) + (t-x_n)n < f(t)$. So $f$ is unbounded.
Please identify the mistake in my proof. Thanks.
The problem lies in the jump from$$\left|\frac{f(t)-f(x_n)}{t-x_n}-f'(x_n)\right|<f'(x_n)-n$$to$$n<f(x_n)+(t-x_n)n<f(t),$$for which you provided no justification.