Valid Proof on Boundedness for $f $and $f'$?

Question

Proposition: let $f:(a,b)\rightarrow R$ be a differentiable and unbounded function. Then $f'$ is unbounded.

Proof by contrapositive

Suppose $f'$ is bounded. So, there exists a $B$, such that $|f'(c)|\leq B$ for all $c$ in $(a,b)$. So, using definition of differentiation, and limits:

$|x-c|<\delta$ $\rightarrow$ $|\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon$.

So, $|\frac{f(x)-f(c)}{x-c}|<\epsilon+B$.

Therefore $|f(x)|<(\epsilon+B)\delta+B$ so $f(x)$ is bounded.

I feel like I am playing with the definitions of limits and derivatives too loosley or something. Any help or error pointing would be gratly appreciated. Thanks

edit: I know it is differentiable so delta and epsilon must exist. So I can choose delta to be $b-a$, in which case, the "$<\epsilon"$ side of the implication is true for all x in $(a,b)$.

Answer 1

The link between $f$ and $f'$ is given by mean value theorem. You can prove the contrapositive as follows.

Choose any point $c\in(a, b) $ and then we have via mean value theorem $$f(x) =f(c) +(x-c) f'(d) $$ where $d$ lies between $c$ and $x$. If $f'$ is bounded then by the above equation $f$ is also bounded and we are done.

Also note that the above proof as well as the result holds only when the interval $(a, b) $ is bounded. As an example $f(x) =\log x$ is unbounded on $[1,\infty)$ but its derivative is bounded on $[1,\infty) $.

Answer 2

Without loss of generality, assume $f$ is unbounded above. (If not, change $\max$ to $\min$ in what follows.)

Now, choose $x_n\in I_n=[a+1/n,b-1/n]$ such that $I_n\subset (a,b)$ and such that $f(x_n)=\max \{f(x);x\in I_n\}$. Then $\{f(x_n\}_n$ is unbounded and increasing.

On the other hand, $\{x_n\}_n$ contains a monotonic subsequence, which we may assume without loss of generality is increasing, and which we still write $\{x_n\}_n$ for convenience.

Then, by the mean value theorem, we can find numbers $c_n$ such that

$\frac{f(x_{n+1})-f(x_1)}{x_{n+1}-x_1}=f'(c_n);\ x_1\le c_n\le x_{n+1}$.

And since $\frac{f(x_{n+1})-f(x_1)}{x_{n+1}-x_1}\ge \frac{f(x_{n+1})-f(x_1)}{b-a},\ $

we have $f'(c_n)\ge \frac{f(x_{n+1})-f(x_1)}{b-a}$.

The result follows.