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Suppose that $f:[0,1]\to\mathbb{R}$ is differentiable on (0,1) and continuous on [0,1]. Would like to assert that if f(0)=0, and $|f'(x)|\leq |f(x)|$ for each $x\in (0,1)$, then f is the zero function.

I have tried applying several different variations of the mean value theorem, but nothing useful has come of this. I also have that the derivative has a max and min (based on the compact domain and the inequality condition). What am I missing here to get that $f=0$?

reluctant mathematician
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    If $0<x<1$, there is a $c_1\in(0,x)$ with $|f(x)|\le |f(c_1)|x$ by MVT. Apply MVT again to obtain $|f(x)|\le|f(c_2)|c_1 x$ with $0<c_2<c_1$. Keep going... – David Mitra Dec 05 '14 at 09:47

2 Answers2

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$$ |f(x)|=\Bigl|\int_0^xf'(t)\,dt\Bigr|\le\int_0^x|f'(t)|\,dt\le\int_0^x|f(t)|\,dt. $$ Let $g(x)=\int_0^x|f(t)|\,dt$. Then $$ g(x)\ge0,\quad g(0)=0,\quad g'(x)-g(x)\le0. $$ Multiplynig the last inequality by $e^{-x}$ we get $$ \bigl(e^{-x}g(x)\bigr)'\le0\implies e^{-x}g(x)\le g(0)=0. $$

Actualy, this is a particular case of Gronwall's lemma.

Julián Aguirre
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Denote by $M$ the quantity $\max_{t\in [0,1]}|f(t)|$. Using the mean value theorem we get for $x\in (0,1)$, $$|f(x)|\leqslant x M.$$ We can prove by induction that $|f(x)|\leqslant x^nM$ for each $x\in (0,1)$. Indeed, it is true for $n=1$, and if it is true for $n$, then by the mean value theorem, $$f(x)=f'(c)x $$ for some $c\in (0,x)$, hence $|f(x)|\leqslant x|f'(c)|\leqslant x|f(c)|$. Using the induction hypothesis, we get $|f(x)|\leqslant xc^nM\leqslant x^{n+1}M$.

Since $x^n\to 0$ as $n$ goes to infinity, we get the wanted conclusion.

Davide Giraudo
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