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Knowning that $f(z)=z+a_2z^2+a_3z^3+...$ is a convex function, is it the derivative of f(z) is also a convex function?

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Take $f(x) = - \log x$. –  Qiaochu Yuan Aug 5 '11 at 4:23
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up vote 5 down vote accepted

If $f$ is smooth and both $f$ and $f'$ are convex on all of $\mathbb R$, then $f$ is of the form $f(x) = a + b x + c x^2 + \int_{-\infty}^x \frac{(x-s)^2}{2} g(s)\, ds$ where $c \ge 0$, $g(x) \ge 0$ is continuous and $x^2 g(x)$ is integrable at $-\infty$.

If $f$ is a polynomial, it must be of degree 0, 1 or 2. A non-polynomial example is $e^x$.

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If the function $f$ itself is $z+a_2z^2+a_3z^3+...$, is it the derivative, $f'$ is convex? –  DRN Aug 5 '11 at 5:27
    
Maybe, maybe not. Any function analytic at $0$ whose derivative at $0$ is nonzero can be translated so the constant term is 0, scaled and, if necessary, reflected across the $y$ axis so the coefficient of $z$ is 1, without affecting the convexity. So $z + a_2 z^2 + a_3 z^3 + \ldots$ is not saying much. –  Robert Israel Aug 5 '11 at 7:12
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The functions you're looking at appear to be power series centered at $0$, with $f(0)=0$ and $f'(0)=1$, and they are infinitely differentiable on their intervals of convergence. Such an $f$ is convex if and only if $f''(x)\geq 0$ for all $x$ in the domain, and $f'$ is convex if and only if $f'''(x)\geq 0$ for all $x$ in the domain. There are examples where both $f$ and $f'$ are convex (e.g., $z$, $z+z^2$), and there are many examples where $f$ is convex but $f'$ is not (e.g., $z+z^4$). To do further experimentation, you can continue to apply the second derivative criterion to examples.

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Thank you Sir for giving me very clear answers, really appreciated.. If you want to share with me, may I know the references? I have to learn more about this topic.. –  DRN Aug 5 '11 at 5:37
    
Norlyda: You could start at the Wikipedia article, which has further references. –  Jonas Meyer Aug 5 '11 at 15:49
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