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Let $I \subset R$ be an interval. A differentiable function $F: I \rightarrow R$ is called a primitive for the function $f : I \rightarrow \mathbb{R}$ if $$F' (x) = f(x)$$ for all $x \in I$.

Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant $C \in \mathbb{R}$ such that $F_2 \equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x \in I$.

What we have covered so far is the formal definition of the derivative in terms of the limit: $$ \lim_{h \rightarrow 0}\frac{F_1(x+h)-F_1(x)}{h}= f(x)$$ $$ \lim_{h \rightarrow 0}\frac{F_2(x+h)-F_2(x)}{h}= f(x)$$

I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.

We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.

3 Answers3

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Hint: What is the derivative of $F_1-F_2$?

Arthur
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  • Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$ –  Dec 11 '18 at 10:49
  • And then apply the MVT to the interval. –  Dec 11 '18 at 10:49
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    @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$? – Arthur Dec 11 '18 at 10:52
  • It means the function $F_1 - F_2$ is constant, see my proof below using the MVT. –  Dec 11 '18 at 12:20
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Hint:

If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,

$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$

Remains to prove that the antiderivative of $0$ is a constant function.

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Let $G(x)= F_1 - F_2.$

We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.

Suppose on the contrary the function is non-constant this means there must exist $a, b \in I$ and where $a<b$ such that $G(a) \neq G(b)$.

We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that: Which is a contradiction. $$G'(c)= \frac{G(b)-G(a)}{b-a} \land G'(c)=0,$$ $$ \frac{G(b)-G(a)}{b-a}=0 \implies G(a)=G(b). $$ We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $\square$.