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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.

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  • $\begingroup$ If a function goes to infinity, its slope will also increase to infinity to get to infinity within finite distance. See a graph. $\endgroup$
    – evil999man
    May 22, 2014 at 14:53

1 Answer 1

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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: $|f(x_0)|\le |f(x_0)|+M(b-a).$

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  • $\begingroup$ why do you need to make assumptions whether $x\in (x_0,b)$ ? $\endgroup$ May 22, 2014 at 14:58
  • $\begingroup$ @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 $c\in(x,x_0)$. $\endgroup$ May 22, 2014 at 15:05
  • $\begingroup$ Maybe it is also worth mentioning why the bound holds for $x = x_0$. You do not seem to cover this case? :) $\endgroup$
    – Nikolaj
    Feb 7, 2021 at 16:23
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    $\begingroup$ @Nikolaj Case $x=x_0$ covered. Thanks for noticing it. $\endgroup$
    – mfl
    Feb 7, 2021 at 17:19

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