Let $f\in C^\infty(B_R(0))$, where $R>0$. For $0<\sigma<1$ require the following properties on $f$:
$$ 0\leq f\leq 1,\ f=1 \text{ on } B_{(1-\sigma)R},\ f=0 \text { on }\partial B_R,\ |\nabla f|\leq C(\sigma R)^{-1},\ \text{ and } |\Delta f|\leq C(\sigma R)^{-2}$$
The constant $C$ depends only on the Dimension and the order of derivatives. The $\Delta$ may be replaced by the full Hessian if necessary.
What I want to show, and what seems to be used in several articles is the following inequality:
$$f(x) |\nabla f(x)| \leq C (f^2(x)) (\sigma R)^{-2}$$ where the constant may denepnd on $R$ (if necessary). Of course, this looks a lot like an application of Taylor's formula or of the mean value theorem, but the estimate that are given are somehow in the wrong direction. Could someone provide any hints on that? The result is clear on the set where $f=1$.
I add where this is comming from and post an updated question, since the previous one was answered by Willie Wong.
We have an upper estimate, where we find the integral $$... \leq ... + C\int_{B_R} |u(x)-u(x_0)| g^2 f|\nabla f|$$ for some $u,g\in H^1(B_R)$ and such that $|u(x)-u(x_0)|\leq CR^\alpha$. This term (and several others), is estimated from above by $$ CR^{2\alpha} \left(\int_{B_R} |\nabla g|^2f^2 +\frac{C}{(\sigma R)^2} \int_{B_R} f^2g^2\right)$$
So as i see it: (and here is the first mistake I made), we use the estimate I suggested (this would only lead to $R^\alpha$ instead of $R^{2\alpha}$. Since this does not work, we use Cauchy's inequality and obtain $$C R^{2\alpha} \left(\int_{B_R} f^2g^2 + \int_{B_R} (|\nabla f|)^2g^2\right).$$ The first term can be multiplied by $|B_R|^2/(\sigma R)^2\geq 1$ but for the second term we need now: $$|\nabla f|\leq f (\sigma R)^{-1}$$
But this also seems to fail if we consider the example given by Willie Wong