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Let \( \mathcal{D} \) be an interval and let \(f \) be a differential function such that \(f'(x)=0 \;\;\;\;\forall x \in \mathcal{D} \) . Prove , without the use of MVT, that \( f\) is constant.

Given two points \(x,y \in \mathcal{D}\) we define \[\displaystyle{D(x,y) = \frac{f(x)-f(y)}{x-y}}.\] Note that for \(x < y < z\) we have \[ D(x,y) = \alpha D(x,z) + (1-\alpha)D(z,y)\] where \(\alpha = \alpha(x,y,z) = \displaystyle{\frac{z-x}{x-y} \in (0,1)}.\) It follows that either \(D(x,z) \leqslant D(x,y) \leqslant D(y,z)\) or \(D(x,z) \geqslant D(x,y) \geqslant D(y,x).\)

If \(f\) was not constant we could find \(x_1 < y_1\) with \(f(x_1) \neq f(y_1)\). Say without loss of generality that \(f(x_1) < f(y_1)\). Then \(D(x_1,y_1) > 0\). Furthermore, for \(z = (x_1+y_1)/2\) we have that \(D(x_1,z) \geqslant D(x_1,y_1)\) or \(D(z,y_1) \geqslant D(x_1,y_1)\). In the first case we set \(x_2=x_1,y_2=z\) and in the second case we set \(x_2=z,y_2=y_1\).

Proceeding in the same manner we obtain two sequences \((x_n)\) and \((y_n)\) with the first being increasing and bounded above by \(y_1\) and the second being decreasing and bounded below by \(x_1\). Therefore both sequences converge and since the sequence \(z_n = y_n-x_n\) satisfies \(z_{n+1} = z_n/2 \) they actually converge to the same point, say \(c\).

Furthermore, the sequence \(d_n = D(x_n,y_n)\) is increasing and bounded below by \(d_1 > 0\). Since \(f'(c) = 0\), given this \(d_1\) we can pick \(\delta > 0\) such that if \(0<|x-c|<\delta\) then \[\displaystyle{\left|\frac{f(x) - f(c)}{x-c}\right| < d_1.}\] In particular, since \((a_n)\) and \((b_n)\) converge to \(c\) from above and below respectively, if \(n\) is large enough we will have \(f(b_n) < f(c) + (b_n-c)d_1\) and \(f(a_n) > f(c) - (c-a_n)d_1\). However from those we obtained \(D(b_n,a_n) < d_1\), a contradiction.

Good morning Demetres. Thank you for the solution.
What do you think of this one?

Let \(a, b \) such that \(a< b \) and \(\epsilon >0 \). For any \( x \in [a, b] \) let \(U_x \) be an interval such that \( \forall y , y \in U_x \) holds:
$$|f(x)-f(y)|\leq\epsilon |x-y|$$
Because \( \displaystyle \bigcup_{x \in [a, b]}U_x\) covers \( [a, b]\), we can take a finite subcover. It is easy to see that \( \left | f(a)-f(b) \right |\leq \epsilon \left | a-b \right | \) and since \( \epsilon \) is arbirtary small that implies \(f(a)=f(b) \) and we're done.

I found it in an old textbook but perhaps it can be found elsewhere too!!

Hi Apostole,

It is a nice proof! I have not seen it anywhere.