let $F$ a closed convex subset of $\mathbb{R}^n$, let $x,y\in F$ and assume that for any $s\in[0,1]$ we have $f(s):=\mid sx+(1-s)y-z\mid\geq \mid y-z\mid$ why is it true that $\frac{\partial}{\partial s} f(s) |_{s=0}\geq 0$?
an estimate for derivative
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calculus
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differential-geometry
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0What is $z$ in your expression? – 2011-04-18
1 Answers
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$g(s):=\left(f(s)-|y-z|\right)\ge0$ for $s\ge0 $ and $g(0)=0$ so the minimum of $g$ happens at $s=0.$ Therefore, after $s=0$, $g$ increases; that's why the derivative must be non-negative.