Prove that $\int \limits_{0}^{\infty} x^p e^{-g(x)/x} dx \leq e^{p+1} \int \limits_{0}^{\infty} x^p e^{-g'(x)}dx$.

Question

For a convex function $g(0)=0$ and for any $-1

It the generalization of the not-so-famous Carleman's Integral Inequality, which states that :

$$\int \limits_0^\infty \text{exp}\left\{\frac{1}{x}\int \limits_{0}^{x }\ln(f(t))\,dt\right\}dx \leq e \int \limits_0 ^{\infty}f(x)\,dx.\,\,\,\,\,(♥)$$

$(♥) $ is a special case of $(♣) $ with $p=0$.

$(♥)$ is "OK-ayish" but $(♣)$ is very tricky to prove, in which I horribly fail. I can not find any suitable reference or method or something which can help me prove this monster.

Any help will be appreciated.

Thanks in Advance ! :-)

Answer 1

A proof can be found here, at section 2, but I'll go over how it works here. First, the convexity condition on $g$ is used, in particular fact 5 here to show that

$$g(k x) \geq g(x) + (k-1) x g'(x)$$ for any $k>1$. Then, consider the integral $$J = \int_0^A x^p \exp\left(-\frac{g(kx)}{kx}\right) dx$$ for $A>0$.

By a substitution, you can show $$\begin{align}J &= k^{-p-1}\int_0^{Ak} x^p \exp\left(-\frac{g(x)}{x}\right) dx \\ &\geq k^{-p-1}\int_0^{A} x^p \exp\left(-\frac{g(x)}{x}\right) dx\end{align}$$

On the other hand, use the convexity inequality on $g$ to show

$$J \leq \int_0^A x^p \exp\left(-\frac{g(x)}{kx} - \frac{(k-1) g'(x)}{k}\right) dx$$ from which you can use Holder's inequality here in integral form to get

$$J \leq \left(\int_0^A x^p \exp\left(-\frac{g(x)}{x}\right) dx \right)^{1/k} \left(\int_0^A x^p \exp\left(-g'(x)\right) dx\right)^{(k-1)/k}$$

which should start to look familiar.

Putting our bounds on $J$ together, we get

$$k^{-p-1}\left(\int_0^A x^p \exp\left(-\frac{g(x)}{x}\right) dx \right)^{(k-1)/k} \leq \left(\int_0^A x^p \exp\left(-g'(x)\right) dx\right)^{(k-1)/k}$$ So take the limit as $A \to \infty$ and rearrange to get

$$\int_0^\infty x^p \exp\left(-\frac{g(x)}{x}\right) dx \leq \left(k^{\frac k{k-1}}\right)^{p+1} \int_0^\infty x^p \exp\left(-g'(x)\right) dx$$

and taking the $k\to 1$ limit finishes off the answer!