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Let $f:[a,b]\rightarrow \mathbb{R}$ s.t $f(a)=0$. If $f$ is differentiable on $[a,b]$ and there exist $C \in \mathbb{R}$ s.t $|f'(x)| \leq C|f(x)|$ for all x $ \in [a,b]$, then $f(x)=0$ for all $x \in [a,b]$

My attempt:

|$f'(a)| \leq C|f(a)|=0\Rightarrow f'(a) = 0$

Let $\epsilon=\frac{1}{C}$, then there exist $\delta > 0$ s.t for all $x \in (a,b)$ if $x\in (a,a+\delta)$, then $\frac{|f(x)|}{x-a}<\frac{1}{C}$ thus $$|f'(x)|\leq C|f(x)|<|x-a|<\delta$$

I do not know what to do from here.

Edit: I could try to think about $f'(b)$: If $f'(b)>0$ or $f'(b)<0$, then find some contradiction.

Nocherto
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1 Answers1

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Note that $$ |f(x)|=|f(x)-f(a)|=\bigg|\int_a^xf'(t)dt\bigg|\le\int_a^x|f'(t)|dt\le C\int_a^x|f(t)|dt. $$ Let $F(x)=\int_a^x|f(t)dt$ and then $F(x)\ge 0, F(a)=0$. Then one has $$ F'(x)\le CF(x)$$ and hence $$ (e^{-Cx}F(x))'\le 0.$$ So $$ e^{-Cx}F(x)-e^{-Ca}F(a)=\int_a^x(e^{-Ct}F(t))'dt\le0$$ or $$ e^{-Cx}F(x)\le 0$$ or $$ F(x)\le0$$ So $F(x)\equiv0$ or $f(x)=0$ in $[a,b]$.

xpaul
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  • very nice.${}{}$ – Andres Mejia Nov 26 '17 at 01:38
  • The derivative of a function is not necessarily integrable, so this approach fails from the start. –  Nov 26 '17 at 01:39
  • @mathguy But in the context of this problem, the derivative is bounded by $C \max_{x\in[a,b]} f(x)$. – Xander Henderson Nov 26 '17 at 02:39
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    @XanderHenderson - even then, the derivative may not be Riemann integrable (and I don't think the question is at Lebesgue integration level - it only requires the mean value theorem for solution). –  Nov 26 '17 at 02:42