I have a convex function (since its double derivative is greater than zero). The function is defined over the real values The first derivative of the function is a polynomial of 3rd degree (that is it involves $x^3$ term). Since the first order derivative has three possible roots. Is it possible that more than one root is real? I think it is not possible because the first derivative is an increasing function on the real values and hence it can not have more than one real roots. Is this right reasoning? Any help in this regard will be much appreciated. Thanks in advance.
2026-03-27 23:23:20.1774653800
Is the following logic right or wrong?
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One characterization of convexity is that the derivative is nondecreasing; see Rockafellar's Convex Analysis. So you cannot have two or three roots, your reasoning is correct.
Edit: Some details added.
Your cubic can only have 1 or 3 real roots (because they come in complex conjugate pairs). Now suppose you would have three real roots, $a<b<c$. If $f$ is convex, then $f'$ is increasing (a.k.a. "nondecreasing") and so $f'(a)\leq f'(u)\leq f'(b) \leq f'(v)\leq f'(c)$, where $a<u<b<v<c$. But then $f'$ is identically $0$ on the interval $[a,c]$, which is absurd since a cubic has only 1 or 3 real roots.