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Let $f(x)$ a function defined at $I\subseteq \mathbb{R}$ and assume that $F(x)$ and $G(x)$ are the antiderivatives of $f(x)$ in $I$, so there is a $c$ such that for all $x\in I$, $F(x)=G(x)+c$

Let us define $H(X)=F(x)-G(X)$ therefore $H'(X)=F'(x)-G'(X)=f(x)-f(x)=0$. Which theorem should be used to finalize the proof?

quid
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gbox
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2 Answers2

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First, you need that $I$ is an interval, I assume this is intended. If not it is just not true. (Abstractly, you want the domain to be connected.)

Then use the Mean Value Theorem.

If $H$ is non-constant there are $a,b$ such that $H(a) \neq H(b)$, so by MVT you get a contradiction to $H'$ identically $0$.

quid
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  • Just to be sure I fully grasped the contradiction, if we have $H(a)\neq H(b)$ by MVT there is $c$ such that $H'(c)\neq 0$ in contradiction ? – gbox Jul 17 '16 at 15:04
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    Yes. Because there is a $c$ such that $H'(c) = (H(b)-H(a))/(b-a)$ and the latter is not $0$ – quid Jul 17 '16 at 15:05
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This is only true if $I$ is an interval.

Let $x_1,x_2\in I$, with $x_1<x_2$; then there is $\xi\in(x_1,x_2)$ with $$ D(x_1)-D(x_2)=(x_2-x_1)D'(\xi) $$ where $D(x)=F(x)-G(x)$. This is the mean value theorem applied to $[x_1,x_2]\subseteq I$.

Since $D'(\xi)=F'(\xi)-G'(\xi)=0$, …

More generally, if a differentiable function defined over an interval has zero derivative, then the function is constant.

The standard counterexample is $$ f(x)=\arctan x+\arctan\frac{1}{x} $$ defined on $(-\infty,0)\cup(0,\infty)$. It has everywhere zero derivative, but it is not constant.

egreg
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