# Why should this result be true? If $f: [0,1] to mathbb{R}$ is differentiable, $g$ is continuous, and $f'(t) = g(f(t)),$ then $f$ is monotonic.

Mathematics Asked on January 7, 2022

If $$f: [0,1] to mathbb{R}$$ is differentiable, $$g$$ is continuous, and
$$f'(t) = g(f(t)),$$ then $$f$$ is monotonic.

Example 1: Suppose $$g(x) = x Rightarrow f’ = f.$$ Then $$f = ce^x$$ which is indeed monotonic.

Example 2: Suppose $$g(x) = x^n Rightarrow f’ = f^n$$ for $$n ne 1.$$ Then $$df/f^n = dx Rightarrow x+C = f^{1-n}/(1-n) Rightarrow f = ((1-n)x + C)^{1/(1-n)},$$ monotonic again.

It seems bizarre that $$g$$ can be anything we want, and yet the result will be true. Since $$f$$ is continuous, it suffices to prove $$f$$ is injective. So suppose $$f(a) = f(b)$$ with $$a

Avenue 1: Then $$f'(a) = g(f(a)) = g(f(b)) = f'(b).$$ But what’s next? This seems like a dead end.

Avenue 2: There exists $$c in (a,b)$$ such that $$f'(c) = 0$$ by Rolle’s Theorem. Thus, $$g(f(c)) = 0.$$ But what’s next? Looks like another dead end again.

Is there a 3rd avenue? I’ve been looking for one. Until I find it, I cannot believe the result.

Update: The answer here solves my question. The question also appeared previously on MSE, but I do not like the method used there since it seems to assume $$f’$$ is continuous.

This is an explanation of an answer found here, which works but leaves out a few details.

Suppose $$f(a) = f(b), a It will suffice to show $$f$$ is constant on $$[a,b].$$ Let $$gamma$$ be the path parameterized by $$gamma(t) = f(t), t in [a,b].$$ Then $$int_a^b f'(t)^2 , dt = int_a^b g(f(t))f'(t) = int_{gamma} g(z) , dz = 0$$ since $$g$$ is continuous and $$gamma$$ is a closed loop due to the fact that $$gamma(a) = f(a) = f(b) = gamma(b).$$

Thus, $$f'(t) = 0$$ on $$[a,b],$$ which implies $$f$$ is constant on $$[a,b].$$

Answered by Display name on January 7, 2022

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