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We all know that a primitive $F(x)$ of a function $f(x)$ is the function which derivative is $f(x)$. For example:

$$\int 2x dx = x^2 + C$$

where $x^2 + C$ is the set of all primitives of the function. However, how can I be sure that $x^2 + C$ is the only possible solution? How can I be sure that a function $f(x) \ne x^2 + C\ ,\ f'(x) = 2x$ does not exist?

How could I create a mathematical proof of this fact (in this specific case and in the general case)? Does it make sense to seek such a proof? If not, why?

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3 Answers 3

up vote 6 down vote accepted

What you said is a consequence of the following:

If $f^{\prime}=g^{\prime}$ then $f=g+c$ for some constant $c\in \mathbb{R}$.

To prove this let $h=f-g$. Then $h$ is differentiable and $h^{\prime}=f^{\prime}-g^{\prime}=0$. $h$ is therefore constant (*), that is $\exists c\in \mathbb{R}:h(x)=c$. Then, $f(x)=g(x)+c$

Therefore, a continuous function always has a primitive, unique up to additive constant.

(*) This is an important consequence of the Mean Value Theorem. Here is a proof of that

Let $f$ be continuous in $[a,b]$ and differentiable in $(a,b)$ so that $f^{\prime}\equiv 0$. Let $x,y\in [a,b]$ and WLOG $x<y$. By the MVT in $[x,y]$, \begin{equation}\exists \xi \in (x,y):f^{\prime}(\xi)=\frac{f(y)-f(x)}{y-x}\implies\frac{f(y)-f(x)}{y-x}=0\implies f(y)=f(x)\end{equation} and so $f$ is constant

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The "$h$ is therefore constant" claim needs proof. –  Ted Dec 20 '12 at 17:18
    
As I said to Pete, I can't see the answer in your text. I've updated my question to better explain my problem. –  user16538 Dec 20 '12 at 17:26
    
@user16538 What exactly do you not understand in my answer? –  Nameless Dec 20 '12 at 17:27
1  
@user16538 he is saying: given two anti-derivatives, $f$ and $g$, they may only differ by a constant. –  Jonathan Dec 20 '12 at 17:29
    
Your proof tells me that $C$ must be constant (this is fine), but not that $f$ is unique (and that's my problem). –  user16538 Dec 20 '12 at 17:30

Your question is eminently sensible and indeed important.

The keyword here is Mean Value Theorem, or more specifically its special case that a function defined and differentiable on an interval with identically zero derivative must be constant. For more details, see (e.g.) $\S 5.1$ of these notes.

Here are some further remarks:

$\bullet$ Of course it is understood here that all functions are defined on an interval $I$. If we look at functions with more complicated -- and in particular disconnected -- domains, then uniqueness of primitives up to a constant need not hold. One sees the ugly head of this in freshman calculus when students are taught (not by me!) that the antiderivatives of $\frac{1}{x}$ are $\log |x| +C$.

$\bullet$ The fact that a function which is differentiable on an interval with identically zero derivative is constant -- I call this the Zero Velocity Theorem -- is actually quite deep. As mentioned above, it is a consequence of the Mean Value Theorem. In turn it holds in an ordered field iff the field is Dedekind complete, as is shown in Jim Propp's nice article Real Analysis in Reverse. It is easy to construct counterexamples over $\mathbb{Q}$: just take a piecewise constant function with discontinuities at irrational numbers.

$\bullet$ A proof of the Zero Velocity Theorem directly from an equivalent of the least upper bound axiom -- specifically, using Real Induction -- is given here.

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Thanks for your answer. I do not think it addresses my question (or, better, I cannot see how). I've updated my question to better explain my problem. –  user16538 Dec 20 '12 at 17:25
    
@user16538: Unless I am very much mistaken, I am answering your question. I am explaining the proof that if $F$ is one primitive of a function $f: I \rightarrow \mathbb{R}$, then any other primitive $G$ is of the form $F + C$ for some constant $C$. –  Pete L. Clark Dec 20 '12 at 17:35
    
This real induction is real nice. –  Baby Dragon Dec 20 '12 at 18:20

Let $F(x)$ be a function such that $F'(x)=2x$ for all $x$. Let $G(x)=F(x)-x^2$. Then $G'(x)=0$ for all $x$.

Let $G(0)=C$. By the Mean Value Theorem, for any $x\ne 0$ there is a $t_x$ between $0$ and $x$ such that $$\frac{G(x)-G(0)}{x-0}=G'(t_x).$$ Since $G'(x)$ is identically $0$, it follows that $G(x)-G(0)=0$, that is, that $G(x)=C$. Thus $F(x)=2x+C$ for all $x$.

Remark: Let us find all the antiderivatives of $\dfrac{1}{x^2}$. Here the correct answer is not quite the conventional one. By the argument above, one can show that if $F'(x)=\dfrac{1}{x^2}$ then there is a constant $C$ such that for all positive $x$, we have $F(x)=-\dfrac{1}{x}+C$.

Similarly, there is a constant $D$ such that for all negative $x$, we have $F(x)=-\dfrac{1}{x}+D$.

But the constants $C$ and $D$ need not be equal. The singularity at $x=0$ prevents the derivative from transferring information from the positives to te negatives.

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Sometimes you can go around the singularity. –  Baby Dragon Dec 20 '12 at 18:35

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