# prove that a function whose derivative is bounded also bounded

I got this problem:

Let $f$ be a differentiable function on an open interval $(a,b)$ such that $f'$ (the derivative of $f$) is bounded on $(a,b)$ (meaning there exist $0<M$ such that $\forall x\in(a,b), |f'(x)|\leq M$), Prove that $f$ is also bounded on $(a,b)$.

I tried to prove it but wasn't able to proceed. Thanks.

• If a function goes to infinity, its slope will also increase to infinity to get to infinity within finite distance. See a graph. – evil999man May 22 '14 at 14:53

Fix a point $$x_0\in (a,b).$$ Assume $$x\in(x_0,b).$$ By using the Lagrange's theorem there exists $$c\in(x_0,x)$$ such that $$f(x)-f(x_0)=f'(c)(x-x_0).$$ Thus
$$|f(x)|=|f(x_0)+f'(c)(x-x_0)|\leq |f(x_0)|+|f'(c)||(x-x_0)|\leq |f(x_0)|+M(b-a).$$ Proceeding in the same way you get the bound for $$x\in(a,x_0).$$ Thus we have shown that the function is bounded.
Edit: If $$x=x_0$$ then we have the bound: $$|f(x_0)|\le |f(x_0)|+M(b-a).$$
• why do you need to make assumptions whether $x\in (x_0,b)$ ? – Gabriel Romon May 22 '14 at 14:58
• @G.T.R Just to not worry about signs and relative order. In fact if $x\in(a,x_0)$, the only difference is that $c\in(x,x_0)$. – Hagen von Eitzen May 22 '14 at 15:05
• Maybe it is also worth mentioning why the bound holds for $x = x_0$. You do not seem to cover this case? :) – Nikolaj Feb 7 at 16:23
• @Nikolaj Case $x=x_0$ covered. Thanks for noticing it. – mfl Feb 7 at 17:19