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Suppose $\mathcal{M}_1$ represents the space of smooth probability density functions with unit mean, whose support is contained in $[0,\infty)$ (or $\mathbb{R}_+$). Define the following functional $$\mathrm{J}(f):= \int_0^\infty x\frac{(f')^2}{f} \mathrm{d}x$$ for $f \in \mathcal{M}_1.$ I am conjecturing that $$\mathrm{J}(\rho) \leq \mathrm{J}(f) \quad \text{with} \quad \rho(x): = \int_{z\geq x} \frac{(f*f)(z)}{z} \mathrm{d}z,$$ in which $f*f$ denotes the self-convolution of $f$. I have tried some specific examples (even though not too many) and I did not find any counter-examples (analytically or numerically), so I am wondering whether there exists a proof of this conjecture/inequality, if not, a counter-example is welcome!

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    $\begingroup$ certainly not. Take f to be the uniform distribution. $\endgroup$ May 22, 2021 at 10:42
  • $\begingroup$ Thanks.... this seems to be a very good example... $\endgroup$
    – Fei Cao
    May 22, 2021 at 16:47
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    $\begingroup$ I have added a top-level tag following a suggestion from chat. I will add that possible improvements to the title were mentioned in the same message. $\endgroup$ May 23, 2021 at 4:27

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