Lagrange multiplier?

Question

Minimize $\int\limits_1^{+\infty} p\log x\,dx$, s.t. $\int\limits_1^{+\infty} p\,dx=1$ and $\int\limits_1^{+\infty} p x\,dx = 2$.

$p$ is the density function of $x\in [1, +\infty)$. If it were $\int\limits_1^{+\infty} p\log p\,dx$ it would be easy.

Answer 1

The infimum is $0$, but there is no probability measure, let alone one with a density function, which can achieve it. Let $\ \epsilon\ $ be any real number satisfying $\ 0<\epsilon<2\log2\ $, $\ b\ $ the unique positive real number such that $$ \log(1+b)=\frac{\epsilon}{2}\ , $$ and $\ a\ $ the unique real number such that $\ a>e\ $ and $$ \frac{\log a}{a}= \frac{\epsilon}{2} $$ (I'm assuming here that the logarithm is the natural logarithm). Then $\ 0<b<1\ $ and $\ 0< \frac{2-b}{2a-b-1} <\frac{1}{a+1}<1\ $.

Let $\ \pi= \frac{2-b}{2a-b-1}\ $ and $\ p_\epsilon\ $ be the density function defined by $$ p_\epsilon(x)=\cases{\frac{1-\pi}{b}&for $\ 0\le x\le1+b$\\ \pi&for $\ a\le x\le1+a$\\ 0&otherwise.} $$ Then \begin{align} \int_1^\infty xp_\epsilon(x)dx&=\frac{(1-\pi)}{b}\int_1^{1+b}xdx+\pi\int_a^{1+a}xdx\\ &= \frac{(1-\pi)\left(b^2+2b\right)}{2b}+\frac{\pi(2a+1)}{2}\\ &=\frac{(2a-3)(b+2)}{2(2a-b-1)}+\frac{(2-b)(2a+1)}{2(2a-b-1)}\\ &=2\ \text{, and}\\ \int_1^\infty\log xp_\epsilon(x)dx&=\frac{(1-\pi)}{b}\int_1^{1+b}\log xdx+ \pi\int_a^{1+a}\log xdx\\ &\le(1-\pi)\log(1+b)+\pi\log(1+a)\\ &\le\frac{(1-\pi)\epsilon}{2}+\frac{\log a}{a}\\ &\le\epsilon\ . \end{align} In short, by concentrating nearly all of the weight of the density function near $1$, and the rest over a suitably chosen interval, it's possible to satisfy the the required constraints and make the integral $\ \displaystyle\int_1^\infty\log x\,p(x)dx\ $ (which must be positive) as close to zero as you please.