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#1 h Mar 14, 2025, 6:32 AM

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In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?

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#2 h Mar 14, 2025, 8:19 AM

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Suppose that there exists a point $t_0 \in [0, 1]$ , where $f(t_0) \neq 0$ . Since $f$ is continuous, there exists a neighborhood $(a, b) \ni t_0$ , where $f(t) \neq 0$ for all $t \in (a, b)$ . In this neighborhood, the condition implies $f'(t) = 1$ . Integrating, we obtain $f(t) = t + C$ for some constant $C$ . Consider the maximal interval $(a, b) \subset [0, 1]$ , where $f(t) \neq 0$ . On this interval $f(t) = t + C$ . If $a > 0$ , then at the point $a$ we must have $f(a) = 0$ (by the definition of the boundary of the interval). However, the derivative $f'(a)$ to the left of $a$ (where $f \equiv 0$ ) is $0$ , and to the right (where $f'(t) = 1$ ) is $1$ . This contradicts the continuity of the derivative $f'$ . Similarly for $b < 1$ . Consequently, the interval $(a, b)$ must coincide with $[0, 1]$ , that is, $f(t) = t + C$ on the entire segment. If $f(t) = t + C$ , then

1. $C = 0$ , then $f(t) = t$ , which satisfies the condition: at $t = 0$ the value $f(0) = 0$ , but the derivative $f'(0) = 1$ , and for $t > 0$ $f(t) \neq 0$ , therefore $f'(t) = 1$

2. $C \neq 0$ , then $f(t) = t + C$ never vanishes on $[0, 1]$ , and the derivative is equal to $1$ everywhere

In both cases $f'(t) \equiv 1$ . If $f(t) \equiv 0$ , the condition is trivially satisfied. If $f(t) \not\equiv 0$ , then then from the points above it follows that $f'(t) \equiv 1$

This post has been edited 2 times. Last edited by Tip_pay, Mar 14, 2025, 8:24 AM

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#3 h Mar 14, 2025, 8:47 AM

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Let $A:=\{f'=1\}$ and $B:=\{f=0\}$ then $A,B$ are closed and $A\cup B=[0,1]$ . We will show that $A$ is also open in $[0,1]$ , hence $A=\emptyset$ or $A=[0,1]$ and we are done.
Suppose $x_n\not \in A, x_n\rightarrow x$ and $x\in A$ . Then $x_n\not=x$ and $f(x_n)=0$ for all $n$ .
Hence $f(x)=0$ . Since $x$ is an accumulation point of $B$ we have $f'(x)=0$ , contradiction.

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#4 h Mar 14, 2025, 11:02 AM

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@Alex! What happens when B is finite? Then it has no limit point. So The entire argument fails.

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#5 h Mar 14, 2025, 4:38 PM

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@rljmano: let me repeat myself. The hypothesis was $x_n\not\in A, x_n\rightarrow x, x\in A$ , which is possible if $A^c$ is not closed. Since all $x_n\in B$ and all $x_n\not=x$ , it follows that $x$ is an accumulation point of $B$ .
Hence the case that $B$ is finite is not compatible with the hypothesis.
The rest of the proof remains unchanged.

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