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 $a<b.$

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 $f(a) = f(b), a<b.$ 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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