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Problem Let $f:I\rightarrow \mathbb{R}$ be differentiable in an arbitrary interval I. If $f'(x)=0 $ for all $x\in I$, then $f$ is constant.

My idea If $x \in \operatorname{int}(I)$, we know that $\exists$ $n_0\ge0$ such that $x\in [x-\frac{1}{n_0},x+\frac{1}{n_0}]$ then by the Intermediate Value Theorem, there exists a $c$ in this interval satisfying $$f'(c)=\frac{f(x+\frac{1}{n_0})-f(x-\frac{1}{n})}{n_0}.$$

But we know $f'(c)=0$ by hypothesis, so $f(x+\frac{1}{n_0})=f(x-\frac{1}{n_0})$ and its true for all $n\ge n_0$, then by continuity of $f$ we have that $f$ is constant on $[x-\frac{1}{n_0},x+\frac{1}{n_0}]$.

Since we don't know if $I$ is a closed interval, we can use $\operatorname{Cl}(I)=[a,b]$ because it is closed and also connected. So I can make a partition $$P: a<x_1<\dotsb<x_n<b$$ with $x_i \in I$ for all $i$.

Thus, $f$ is constant on $\operatorname{int}(I)$ and by continuity of $f$, if $I$ is closed we have that $f$ has the same value on $a$ and $b$.

Question

Is that approach enough to prove the statement?

Is there an easier way to show that?

Xander Henderson
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Powder
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    There are a lot of problems in your try. It is obvious that $x\in [x-\frac{1}{n_0},x+\frac{1}{n_0}]$ for every $n_0 >0$, this has nothing to do with $I$. Also, you can have $f\left(x+ \frac 1 {n_0} \right) = f\left(x- \frac 1 {n_0} \right)$ for all $n_0>0$ and $f'(c) =0$ but $f$ is not constant, consider $x=0$ and $f(t)=t^2$. – nicomezi Nov 17 '19 at 15:37
  • " and its true for all n≥n0, then by continuity of f we have that f is constant " Well, that doesn't follow. ${\frac 1n}$ are not "close together". For any $\frac 1{n_0}$ we can find a $\delta > 0$ where $|x - \frac 1{n_0}|< \delta$ implies that $x$ is not any $\frac 1n$... why didn't you just use all $\epsilon$? But even then having all $f(x-\epsilon) = f(x+\epsilon)$ doesn't make $f$ constant. – fleablood Nov 17 '19 at 17:29
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    Don't "balance" and don't try to "solve" for specific $f(x-k) = f(x+k)$. The strength is in the general. For any $a,b$ there is $f'(c) = \frac {f(b)-f(a)}{b-a}$ and so.... – fleablood Nov 17 '19 at 17:35

2 Answers2

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Yes, there is an easier way. Take $a,b\in I$ (which I assume to be an interval), with $a<b$. Then$$f(b)-f(a)=f'(c)\times(b-a)=0,$$for some $c\in[a,b]$.

Concerning your approach, I don't understand it, starting from “So I can make a partition”.

José Carlos Santos
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  • The partition is used just to extend the ideia to the whole interval. But isn't a good ideia, now I see. – Powder Nov 17 '19 at 15:51
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In fact the conditions on $f$ can be weakened: we assume that $f\colon I \to \mathbb{R}$ ( or a normed space) is continuous, and for every $x \in [a,b)$, and $\epsilon> 0$ there exists $x'> x$ such that $\|f(x')-f(x)\|\le (M+\epsilon)(x'-x)$. Then we have

$$\|f(b)-f(a)\| \le M(b-a)$$

Proof:

Fix $\epsilon> 0$. Now consider the set

$$\{\ t\in [a,b] \ | \ \|f(t) - f(a)\| \le (M+\epsilon)(t-a)\ \}$$

The set contains $a$, so it's non-void. Since $f$ is continuous, the set is closed, so it contains its supremum. Now, using the condition given on $f$, we can show that the supremum is in fact $b$ (if it was some $x< b$, choose $x'> x$ such that $\|f(x')-f(x)\|\le (M+\epsilon) (x'-x)$. Now add two inequalities and get $x'$ is also in the set, contradiction).

Take now $\epsilon\to 0$ in $$\|f(b)-f(a)\|\le (M+\epsilon)(b-a)$$ and get the conclusion.

orangeskid
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