I'm confusing with inequality of p-norm 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.
 A: 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}
