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As far as I know, for $p \ge 1$, $||X||_p \equiv (E|X|^p)^{1/p}$ becomes a norm in probability space.

If this is right, those two inequalities on each link seem to contradict with each other.


Lyapunov's inequality says:

For $0<p<q$, $(E|X|^p)^{1/p} \le (E|X|^q)^{1/q}$ (I don't use $||\cdot||_p$ notation since it just satisfy pseudo norm property when $p < 1$.

On the other hand,

$L_p$-norm says:

$1 \leq p \leq q \lt \infty$ then $\|x\|_{q} \leq \|x\|_{p}$


What do I miss here?

Thanks for any comments.

inmybrain
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    The second is for the $\ell^p$ sequence spaces. Which are $L^p$ spaces for the counting measure. Lyapunov's inequality is for spaces of finite measure (if the measure is not $= 1$, there appears a factor containing the measure of the space), the inequality in the other direction for spaces where there don't exist subsets of arbitrarily small positive measure. If you have a space of infinite measure where there are subsets of arbitrarily small positive measure, you don't have an inequality either way. – Daniel Fischer Dec 06 '14 at 11:56
  • Ok, I got it. You mean those inequalities could be different depending on what measures I assume. – inmybrain Dec 06 '14 at 12:07
  • Meanwhile, I couldn't get what you wrote from the third line, which starts with 'the inequality in the other direction for spaces...' Could you explain a little bit specifically? – inmybrain Dec 06 '14 at 12:09
  • and thanks so much. I was really confused. – inmybrain Dec 06 '14 at 12:09
  • @DanielFischer: Nice comment. I think you should post that (maybe answering the OPs additional question) as an answer. If you do not want to, I would post it as a community wiki answer this evening (if you don't mind). – PhoemueX Dec 06 '14 at 12:14
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    Not could be, are. For finite measures $\mu$, we have $q \leqslant p \implies L^p(\mu) \subset L^q(\mu)$, and the inclusion is continuous, there is a $C_{p,q}$ with $\lVert f\rVert_q \leqslant C_{p,q}\lVert f\rVert_p$. For the counting measure $\zeta$ (and similar enough measures, don't know the exact conditions off the top of my head), we have $q \leqslant p \implies L^q(\zeta) \subset L^p(\zeta)$ and the inclusion is continuous, there is a $K_{p,q}$ with $\lVert f\rVert_p \leqslant K_{p,q}\lVert f\rVert_q$. For measures like Lebesgue measure, you have neither inclusion. – Daniel Fischer Dec 06 '14 at 12:14

1 Answers1

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The inequalities hold in different cases, depending on the characteristics of the measure space.

If we have a finite measure $\mu$ (on some $\sigma$-algebra $\mathcal{A}$ on some set $\Omega$), then for $0 < p < q$ Hölder's inequality gives us

$$\int_\Omega \lvert f\rvert^p\,d\mu = \int_\Omega 1\cdot \lvert f\rvert^p\,d\mu \leqslant \Biggl(\int_\Omega 1^{q/(q-p)}\,d\mu\Biggr)^{1-p/q}\cdot\Biggl(\int_\Omega \bigl(\lvert f\rvert^{p}\bigr)^{q/p}\,d\mu\Biggr)^{p/q},$$

and taking $p$-th roots,

$$\Biggl(\int_\Omega \lvert f\rvert^p\,d\mu\Biggr)^{1/p} \leqslant \mu(\Omega)^{\large\frac{1}{p}-\frac{1}{q}}\Biggl(\int_\Omega \lvert f\rvert^q\,d\mu\Biggr)^{1/q}.$$

Thus we have an inclusion $L^q(\Omega,\mu) \hookrightarrow L^p(\Omega,\mu)$ in that case, and the inclusion is continuous. If $\mu$ is a probability measure - $\mu(\Omega) = 1$ - this is precisely Lyapunov's inequality, just written in a different way (using $f$ for the function instead of $X$ for the random variable, and writing the integral down instead of an expected value).

The sense of the inclusion - and the inequality - is reversed if we look at a different type of measures. The second link is specifically concerned with the $\ell^p$ sequence spaces, but it generalises to (at least) purely atomic measures with a lower bound on the measure of the atoms. If $T$ is a set, and $w\colon T \to (0,+\infty)$ is a weight function with $w(t)\geqslant \varepsilon$ for some $\varepsilon > 0$, then we can consider the weighted counting measure

$$\nu(A) = \sum_{t\in A} w(t),$$

and for the $L^p(T,\nu)$ spaces, we have the reverse inclusions and inequalities. If $0 < p < q$, and $f\in L^p(T,\nu)$, then we have $f\in L^q(T,\nu)$ and

$$\Biggl(\sum_{t\in T} \lvert f(t)\rvert^qw(t)\Biggr)^{1/q} \leqslant \varepsilon^{\large 1 - \frac{q}{p}} \Biggl(\sum_{t\in T} \lvert f(t)\rvert^pw(t)\Biggr)^{1/p}.$$

In particular, for the classical $\ell^p(\mathbb{N})$ spaces, $\lVert x\rVert_q \leqslant \lVert x\rVert_p$ - also for $p < 1$ (and $q < 1$).

When we have a measure like the Lebesgue measure $\lambda$ on $\mathbb{R}^n$, where neither the space has finite measure nor there are atoms, we have neither inclusion, and there is no inequality of the (pseudo-) norms either way, for $p\neq q$ we have

\begin{gather} L^p(\mathbb{R}^n,\lambda) \supsetneqq L^p(\mathbb{R}^n,\lambda)\cap L^q(\mathbb{R}^n,\lambda) \subsetneqq L^q(\mathbb{R}^n,\lambda),\\ \sup_{f\in L^p\cap L^q\setminus \{0\}} \frac{\lVert f\rVert_p}{\lVert f\rVert_q} = +\infty = \sup_{f\in L^p\cap L^q\setminus \{0\}} \frac{\lVert f\rVert_q}{\lVert f\rVert_p}. \end{gather}

Daniel Fischer
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