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the problem is: Is the function $\phi(f)=\frac{\sqrt{f}}{2}\cdot\log f$ for $f\in L^2(\mathbb{R}_{>0})$ convex or concave. My idee or presumption is, that the function $\phi$ is concave, because we have the inequalitiy\[\sqrt{\alpha\cdot f+(1-\alpha)\cdot g}\geq \alpha\cdot \sqrt{f}+(1-\alpha)\cdot\sqrt{g}\quad\text{for}~\alpha\in(0,1).\]

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  • $\begingroup$ How do you define convexity of $\phi$? $\endgroup$ – gerw Jan 22 at 15:20
  • $\begingroup$ We call a map convex, if \[\phi(\alpha\cdot f+(1-\alpha)\cdot g)\leq \alpha\cdot \phi(f)+(1-\alpha)\cdot \phi(f)\] for $\alpha\in (0,1)$ $\endgroup$ – FuncAna09 Jan 22 at 17:59
  • $\begingroup$ But $\phi( something )$ is a function. Is $\ge$ to be understood pointwise? $\endgroup$ – gerw Jan 22 at 18:00
  • $\begingroup$ Sorry, the correct definition is $\phi(f(x)).$ f is also a map. $\endgroup$ – FuncAna09 Jan 22 at 18:06
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Waht i have so far for $\phi:\mathbb{R}_{>0}\to\mathbb{R}$ with $f\mapsto \phi(f)$ $\begin{align*} -\phi(\alpha\cdot f+(1-\alpha)\cdot g)&=-\sqrt{\alpha\cdot f+(1-\alpha)\cdot g}\log(\alpha\cdot f+(1-\alpha)\cdot g)\\ &\leq -(\alpha\sqrt f+(1-\alpha) \sqrt{g})\log(f^{\alpha}g^{(1-\alpha)})\\ &=-\left[(\alpha\sqrt f+(1-\alpha) \sqrt{g})(\alpha\log(f)+(1-\alpha)\log(g))\right]\\ &=-[\alpha^2\sqrt{f}\log(f)+\alpha(1-\alpha)\sqrt{f}\log(g)+\alpha(1-\alpha)\sqrt{g}\log(f)\\ &+(1-\alpha)^2\sqrt{g}\log(g)]\\ &=... \end{align*}$

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