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I'm trying to prove that if f'=0 then is $f$ is constant WITHOUT using the Mean Value Theorem.

My attempt [sketch of proof]: Assume that $f$ is not constant. Identify interval $I_1$ such that $f$ is not constant. Identify $I_2$ within $I_1$ such that $f$ is not constant. Repeat this and by the Nested Intervals Principle, there is a point $c$ within $I_n$ for any $n$ such that $f(c)$ is not constant... This is where I realized that my approach might be wrong. Even if it isn't I don't know how to proceed.

Thanks for reading and any help/suggestions/corrections would be appreciated.

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    Well one could boil it down to Rolle's theorem, but that wouldn't really be a different solution ;)2011-11-06

3 Answers 3

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So we have to prove that f'(x)\equiv0 $\ (a\leq x\leq b)$ implies $f(b)=f(a)$, without using the MVT or the fundamental theorem of calculus.

Assume that an $\epsilon>0$ is given once and for all. As f'(x)\equiv0, for each fixed $x\in I:=[a,b]$ there is a neighborhood $U_\delta(x)$ such that $\Biggl|{f(y)-f(x)\over y-x}\Biggr|\leq\epsilon\qquad\bigl(y\in\dot U_\delta(x)\bigr)$ ($\delta$ depends on $x$). For each $x\in I\ $ put U'(x):=U_{\delta/3}(x). Then the collection \bigl(U'(x)\bigr)_{x\in I} is an open covering of $I$. Since $I$ is compact there exists a finite subcovering, and we may assume there is a finite sequence $(x_n)_{0\leq n\leq N}$ with $a=x_0 such that I\subset\bigcup_{n=0}^N\ U'(x_n). The $\delta/3$-trick guarantees that $|f(x_n)-f(x_{n-1})|\leq \epsilon(x_n-x_{n-1}).$ By summing up we therefore obtain the estimate $|f(b)-f(a)|\leq \epsilon(b-a)$, and as $\epsilon>0$ was arbitrary it follows that $f(b)=f(a)$.

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    Ah, this is basically the proof I was trying to write down, but with no luck. +12011-11-06
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Does the real line have gaps? That's the issue. Suppose you can partition the line into two sets $A$ and $B$, so that

  • Every real number belongs to either $A$ or $B$;
  • No number belongs to both;
  • Every member of $A$ is less than every member of $B$;
  • For every member of $A$, there is a larger number that is still a member of $A$;
  • For every member of $B$, there is a smaller number that is still a member of $B$.

In that case, there would be no boundary point, such that every number less than that point is in $A$ and every number greater than that is in $B$. That would be a gap.

Now suppose $f(x) = 0$ if $x\in A$ and $f(x)=1$ if $x\in B$. Then f\;'(x)=0 for every value of $x$, but $f$ is not constant.

You can't prove every function whose derivative is everywhere $0$ is constant unless you rule out gaps. The proof of the mean value theorem conventionally relies on Rolle's theorem, which in turn relies on the fact that a continuous function on a closed interval has a maximum and a minimum in that interval. That theorem is not true unless the real line is gapless. A continuous function could increase on the set $A$ described above and decrease on $B$, and it would have no maximum.

The mean value theorem is how the gaplessness of the line gets involved in the proof that if f\;'=0 everywhere then $f$ is constant.

Probably you could find other ways of proving that, but they'd have to invoke gaplessness somehow.

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    I had in mind things more like $(-\infty,0)\cup(0,\infty)$. The set $[0,1]\cup[2,3]$ does not have a "gap" in the sense defined in the answer I posted.2011-11-08
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Depending on how much technology you want to use, you could perhaps use the fact that $H^0_{\text{dR}}(\mathbb{R}^n) \cong \mathbb{R}$. (This follows from $\mathbb{R}^n$ being homotopy equivalent to a point) Hence any closed $0$-form (so any function smooth function $f:\mathbb{R} \rightarrow \mathbb{R}$ with $df=0$) is constant.

I think that all of this doesn't use the Mean Value Theorem, but I guess it's a bit of an overkill...