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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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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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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