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Define the gradient $\nabla f(x(t))$ as the vector with components $\partial f/\partial x_i$. If $x$ is a function of $t$ with derivative x'(t)=v(t), how can I show that

$ v \cdot \nabla f(x(t)) \geq 0 $

when $t=0$ if $f(x(0))=0$

and if $f(x(t))\geq0$ when $t \gt 0$ ?

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    Sorry, it's the function x.2011-10-28

1 Answers 1

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This in general will only hold if $f$ is differentiable. When $f$ is differentiable, the chain rule gives

\frac{df}{dt} = x'(t)\cdot\nabla f(x(t))

if $f(t) \ge 0$ for $t > 0$ then consider $\frac{df}{dt}$ at $t=0$ given by

$\frac{df}{dt}(0) = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} = \lim_{h\rightarrow 0^{+}}\frac{f(h)}{h} \ge 0$

where the second last step is due to the differentiability of $f$ at $0$ and the last step is due to the fact that $f(h) \ge 0$ when $h > 0$.