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Let

$z_1=a_1+t_1*v_1$,

$z_2=a_2+t_2*v_2$,

$z_3=a_3+t_3*v_3$

with $a_1,a_2,a_3,v_1,v_2,v_3\in\mathbb{R}^n $ known parameters and $v_1,v_2,v_3$ vectors linearly independent.

I define $D:=||z_1-z_2||+||z_2-z_3||+||z_3-z_1||$, then I am asked to prove that the function defined by $D=f(t_1,t_2,t_3)$ is a strictly convex function.

Lecter
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1 Answers1

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For the euclidean norm:

$f$ is convex since the norm of a linear function is convex. Assume this is not strictly convex. Then, there is equality in the convex relation for some non trivial convex combination of two distinct points. This gives, with the Cauchy-Schwarz equality case, a linear relation between $v_1,v_2,v_3$.

Pierre
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  • what you mean with "there is equality in the convex relation". – Lecter Mar 05 '19 at 18:39
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    I mean there is two distinct points $\mathbf{t},\mathbf{t'}\in \mathbb{R}^3$ and some $\alpha\in(0,1)$ such that $$f(\alpha\mathbf{t} + (1-\alpha)\mathbf{t'})=\alpha f(\mathbf{t}) + (1-\alpha)f(\mathbf{t'})$$ – Pierre Mar 05 '19 at 19:13