Thread:

1. Prove that f=0

Let be function differentiable on and satisfy conditions

Prove that .

2. Let $g(x)=e^{-Kx}f(x),$ thus the only thing we need to prove is that $g(x)=0.$

We have that $g'(x)=e^{-Kx}\big(f'(x)-Kf(x)\big),$ but given $|f'|\le K|f|$ we get that $g(x)g'(x)\le0$ which is the same as $\left( \frac{g^{2}(x)}{2} \right)'\le 0,$ thus $\big(g(x)\big)^2\le0$ then $g(x)=0$ for $x\ge0.$

We proved a stronger result, given $f:[0,\infty)\to\mathbb R$ differentiable.

3. Yes, this is it. Congratulations for nice solution. And little correction:
from
$<br /> \left( \frac{g^{2}(x)}{2} \right)'\le 0,<br />$
you can't conclude
$<br /> \big(g(x)\big)^2\le0<br />$

but just, by integration, that $g^2(x)=C=const$,
and than from $g(0)=0$ that $C=0$.

4. how can we simply assume?

Originally Posted by Krizalid
Let $g(x)=e^{-Kx}f(x),$ thus the only thing we need to prove is that $g(x)=0.$

We have that $g'(x)=e^{-Kx}\big(f'(x)-Kf(x)\big),$ but given $|f'|\le K|f|$ we get that $g(x)g'(x)\le0$ which is the same as $\left( \frac{g^{2}(x)}{2} \right)'\le 0,$ thus $\big(g(x)\big)^2\le0$ then $g(x)=0$ for $x\ge0.$

We proved a stronger result, given $f:[0,\infty)\to\mathbb R$ differentiable.
assuming for just g(x)? Doesnot it mean that the proof is only for g(x)? please clarify and comment.

5. Originally Posted by Pulock2009
assuming for just g(x)? Doesnot it mean that the proof is only for g(x)? please clarify and comment.
$e^{f(x)}\ne0$

6. Originally Posted by ns1954
Yes, this is it. Congratulations for nice solution. And little correction:
from
$<br /> \left( \frac{g^{2}(x)}{2} \right)'\le 0,<br />$
you can't conclude
$<br /> \big(g(x)\big)^2\le0<br />$

but just, by integration, that $g^2(x)=C=const$,
and than from $g(0)=0$ that $C=0$.
Actually by definite integration I conclude that, so there's no problem.
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