# Legendre transform of log moment function

Here is something I do not understand for my lecture notes. The lemma is this.

Let $\mu$ be a probability measure on $R$, and $\Lambda^*_\mu$ is the Legendre transform of $\mu$. $\Lambda_\mu^*\geq 0$.

If $\int_R |x|\mu(dx)<\infty$ and $p=E(x) = \int_Rx\mu(dx)$, then $\Lambda^*_\mu(p)=0$. It is also non decreasing on $[p,\infty)$, non-increasing on $(-\infty,p]$

I worked through the proof, but got stuck when my lecturer set $p=\infty$. It says

$p=E(X_1)=-\infty$, the $\Lambda_\mu(\lambda)$, that is the log moment generating function $= \infty$ for $\lambda$ negative, $x\geq p$; $\lambda x-\Lambda_\mu(\lambda)\leq \Lambda^*_\mu(p)=0$

I have no clue what the last line is meant to say. It is not very clearly written. For $\lambda$ negative and $x\geq p$. I cannot see why that is true....

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Don't know if you're still looking for an answer, but $\Lambda_{\mu}(\lambda) = \infty$ for $\lambda < 0$ if $EX_1 = -\infty$ because by Jensen's inequality, $$\Lambda_{\mu}(\lambda) = \log Ee^{\lambda X_1} \geq \log e^{\lambda EX_1} = \lambda EX_1,$$ hence if $\lambda < 0$ then $\Lambda_{\mu}(\lambda) \geq \infty$, so $\Lambda_{\mu}(\lambda) = \infty$.
Thanks, I have figured this out, though I have another question, what do you know why $\Lambda'(\infty) = \text{esssup } X$? –  Lost1 Feb 22 '13 at 19:16
$e^{\lambda \text{ esssup } X} \geq Ee^{\lambda X}$, so take the logarithms of both sides, then divide by $\lambda$, and take the limit as $\lambda \rightarrow +\infty$. Since $\frac{\Lambda(\lambda)}{\lambda} = \Lambda'(\lambda)$ you're done. –  august Feb 23 '13 at 16:17
thanks a lot, but this proves $\text{esssup } X\geq \Lambda'(\infty)$, but not equals to? –  Lost1 Feb 26 '13 at 16:21
and I do not see why $\frac{\Lambda(\lambda)}{\lambda}=\Lambda'(\lambda)$ –  Lost1 Feb 26 '13 at 16:28