Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f$ is differentiable on $[a,b]$, $f'(x) \leq A|f(x)|$ where $A$ is a non-negative constant.

If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$

I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.

I know $|f(x)|=|f'(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\ |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.

share|improve this question
1  
Is this $|f'(x)|\leqslant A|f(x)|$? Then: Gronwall's lemma. –  Did Jan 11 '12 at 16:04
    
Iterating the final inequality seems like the way to go. But I doubt that the point is that you keep "getting a smaller c". On the other hand, notice that you get an extra $A(x-a)$ factor; if $x$ is enough close (how close?) to $a$, this factor is strictly smaller than $1$. On iterating this $n$ times, you will get an $(A(x-a))^n$, which approaches $0$ as $n \to \infty$. Does this help? [There are some details to fill, but the idea is hopefully clear.] –  Srivatsan Jan 11 '12 at 16:06
add comment

1 Answer 1

\begin{align} |f(x)| &= |f(x) - f(a)| = |f'(c)||x-a| \le A|f(c)||x-a| \\ & = A|f(c) - f(a)||x-a| = A|f'(d)||d-a||x-a| \le A|f'(d)||x-a|^2 \\ & \le |f(d)||A(x-a)|^2. \end{align} Now suppose there exists $c \in ]a,x[$ such that $|f(x)| \le |f(d)| |A(x-a)|^n$, by induction on $n$. Then \begin{align} |f(x)| & \le |f(c)||A(x-a)|^n = |f(c) - f(a)||A(x-a)|^n \\ & = |f'(d)||d-a||A(x-a)|^n \le A|f(d)||d-a||A(x-a)|^n \\ & \le |f(d)||A(x-a)|^{n+1}. \end{align} Hence for every integer $n$ there exists $c_n$ such that $|f(x)| \le |f(c_n)||A(x-a)|^{n+1}$. Since $[a,b]$ is compact and $f$ is differentiable, $f$ is continuous, hence $|f|$ is continuous and attains a maximum $M$ over $[a,b]$, so that $|f(x)| \le |f(c_n)| |A(x-a)|^n \le M|A(x-a)|^n$. This means $f(x) = 0$ as long as $|x-a| < \frac 1A$, since the bound only depends on $n$, hence we can let it go to infinity. By continuity of $f$ you also get $f(x) = 0$ if $|x-a| \le \frac 1A$.

Since you can split the interval $[a,b]$ as $[a,a+\frac 1A]$, $[a+\frac 1A, a + \frac2A]$, $\dots$, $[a+\frac mA, b]$, you can repeat this process finitely many times and show that $f$ is zero over $[a,b]$.

Hope that helps,

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.