I am a worker and I am trying to understand a proof. I am trying to understand a step in a longer proof. The hypothesis are:
- $X$ is a stochastic process $\{X_n\}_{n\in\mathbb{N}}$ on a probability space $(\Omega,\mathcal{F}, P)$, where $\mathcal{F}$ is a $\sigma$-algebra and $P$ is a probability measure.
- $\sup_{n\in\mathbb{N}}||X||_p < \infty$, with $1<p<\infty$ where:$$||X||_p = E[|X|^p]^{\frac{1}{p}} $$ I would like to show to that: \begin{equation} \sup_{n\in\mathbb{N}}||X_n||_1\le \sup_{n\in\mathbb{N}}||X_n||_p. \end{equation}
My attempt:
Since $\sup_{n\in\mathbb{N}}||X_n||_p < \infty$ then $||X_n||_p < \infty$ for all $n\in\mathbb{N}$, that is $X_n \in L^p$. Now I wish to use Jensen inequality to prove that $||X||_1\le ||X||_p $ if $X \in L^p$; this will be sufficient to complete the step.
In general if $1\le r < p \le \infty$ I wish to show that, if $X\in L^p$ $||X||_r \le ||X||_p $.
Define $X_k = (\min\{|X|,k\})^r$ for all $k\in\mathbb{N}$. I consider the function $\phi(x) = x ^\frac{p}{r}$ with $x\in[0;\infty[$ which is convex. I can use the Jensen inequality:
\begin{equation}
\phi(E[X_k])\le E[ \phi(X_k) ]
\end{equation}
because:
1. $X_k$ is integrable:
\begin{equation}
E[|X_k|] = E[(\min\{|X|,k\})^r] \le E[k^r] = k^r
\end{equation}
since we are dealing with a space with finite measure ($P(\Omega)=1$),
- $\phi(X_k)$ is integrable: \begin{equation} E[|\phi(X_k)|] = E[(\min\{|X|,k\})^p] \le E[|X|^p] \le \infty \end{equation} Hence we get: \begin{equation} \phi(E[X_k])=E[X_k ]^{\frac{p}{r}} \le E[|\phi(X_k)|] \le E[|X|^p] \Rightarrow E [X_k]^{\frac{1}{r}} \le ||X ||_p \end{equation} We take the limit for $k \rightarrow \infty$ and using Monotone Convergence Theorem we get: $ ||X ||_r \le ||X ||_p$.
Is everything correct? Thank you!
This follows immediately from Holder's inequality. Take $q$ such that $\frac{1}{p}+\frac{1}{q}=1$. Then:
$||X||_1=\int_{\Omega} |X|\leq (\int_{\Omega} |X|^p)^{\frac{1}{p}}(\int_{\Omega} 1^q)^{\frac{1}{q}}=||X||_p$
Edit: the general case follows from the special case. Assume $1\leq r<p<\infty$. We can define $Y=X^r$. By the special case we proved we know that $||Y||_1\leq ||Y||_{\frac{p}{r}}$, this is because $1<\frac{p}{r}$. But now note that $||Y||_1=(||X||_r)^r$ and $||Y||_{\frac{p}{r}}=(||X||_p)^r$.