How can I mathematically prove that the function $f(x) = x^3$ is not convex?
Graphically it's clear. I computed the Hessian to be $6x$, but not sure how I can conclude the positive definiteness of $6x$.
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4Is $f''(x) = 6x$ positive for all $x \in \Bbb R$? – Martin R Dec 15 '19 at 21:33
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Wow, thats very obvious. Thank you – Eisen Dec 15 '19 at 21:34
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1$f''(-1)=-6 \rightarrow$ concave at $x=-1$; $f''(1)=6\rightarrow$convex at $x=1$. Therfore, $f(x)$ is not always convex. – Ty Jensen Dec 15 '19 at 21:37
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You can also use the definition of convexity to show that it is not convex. If it was convex, then we would need to have for all $t\in [0;1]$ and all $x,y\in \mathbb{R}$ $$ f(tx + (1-t)y ) \leq t f(x) + (1-t) f(y). $$ From the graph you would conjecture that we should have problems for negative values. So let us pick $x=-1$ and $y=0$. Then we would have for all $t\in [0;1]$ $$ -t^3 = f(-t) = f(tx + (1-t)0) \leq t f(-1) + (1-t) f(0) = -t. $$ Picking $t=1/2$ shows that it is not true, so our function is not convex.
Severin Schraven
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