# real analysis help $f(x)\ f'(x)$

Let $f(x)$ be a differentiable function such that $f ′ (x)$ is continuous. If $f(0)=0$ and $f(x)\ f ′ (x)\le 0$ for all $x\gt 0$ , prove that $f(x)=0$ for all $x\ge 0$ .

attempt:

I know you can multiply $f(x)f'(x)$ by $2$ to make it greater than $0$ but where does the proof go from there?

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Observe that $2f(x) f'(x)=\frac{d}{dx} \left( f(x)^2 \right)$. –  user1337 Dec 15 '13 at 23:03
but then where would i go from there –  sarah Dec 15 '13 at 23:10

$2f(x)f'(x)\leq 2\cdot 0=0$. This means that $(f^2(x))'\leq 0$. Thus $f^2$ is decreasing(this doen't mean that $f^2$ is not constant-the constants are also decreasing-not strcitly).Now $f^2(x)\geq 0$ for every $x\geq 0$. Also $f(0)=0=>f^2(0)=0$. This means that for every $x>0$, $0\leq f^2(x)\leq f^2(0)=0$.Thus $f^2(x)=0$, for $x\geq 0$. This means that $f(x)=0$ for $x\geq 0$.