Prove that f=0

**ns1954** · January 21st 2010, 08:25 AM

Let

be function differentiable on

and satisfy conditions

Prove that

.

**Krizalid** · January 21st 2010, 07:05 PM

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.

**ns1954** · January 21st 2010, 07:34 PM

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

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

**Pulock2009** · January 21st 2010, 07:37 PM

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.

**Drexel28** · January 21st 2010, 07:42 PM

Originally Posted by Pulock2009

assuming for just g(x)? Doesnot it mean that the proof is only for g(x)? please clarify and comment.

$e^{f(x)}\ne0$

**Krizalid** · January 21st 2010, 08:10 PM

Originally Posted by ns1954

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

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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