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For $0<p<1$, write $Q_p(f)=(\int |f{|}^{p}d\mu {)}^{1/p}$ to emphasize that this is a quasi-norm, as in the Lq quasi-norm branch. On two coordinates with counting measure, $u=(1,0)$ and $v=(0,1)$ give $Q_p(u)=Q_p(v)=1$, but $Q_p(u+v)=2^{1/p}>2$. At $p=1/2$, this is the explicit failure $4>1+1$ in the right panel of the diagram. The midpoint $(1/2,1/2)$ has $Q_p=2^{1/p-1}>1$, so the corresponding unit sublevel set is not a convex set.

A useful replacement survives: for nonnegative real numbers $a,b$, $(a+b{)}^p\le a^p+b^p$. Together with the scalar triangle inequality, this yieldsConsequently $d_p(f,g)=Q_p(f-g{)}^p$ is a metric on the equivalence classes of measurable functions with finite $Q_p$, even though $Q_p(f-g)$ itself need not be a metric. Do not replace the failed Minkowski inequality by a universal reversed bound for signed or complex functions: $g=-f\ne 0$ would give a zero sum but two positive separate sizes.

The Lyapunov moment inequality says that higher positive absolute moments give no smaller absolute moment sizes on a probability space. For a real random variable $Z$ and $0<r<s$,allowing an infinite value. For $r\ge 1$ the left expression is an Lp norm; below one it is a quasi-norm, so the usual triangle inequality terminology needs that distinction. The comparison uses a probability measure of total mass one; arbitrary unnormalized measure spaces need a factor depending on their total mass.

For a nonnegative count, the first moment size is its expected value, while the second is the square root of its second moment. Both ten-trial counts in second moment method have first moment size one. The independent count has second moment size $\sqrt{1.9}\approx 1.37840$, whereas the shared-outcome count $N=10B$ has second moment size $\sqrt{10}\approx 3.16228$. The second moment distinguishes models with the same average because its square increases the influence of rare large counts.

More generally, for $N=10B$ with a Bernoulli random variable $B$ of parameter $1/10$,It rises from one at $r=1$ toward ten as $r\to \infty$. Replacing $1/10$ by a parameter $0<q\le 1$ gives $10q^{1/r}$, which decreases toward zero for a fixed positive exponent when $q\downarrow 0$, making the rare-event and large-exponent limits different questions. For a scaled Rademacher random variable taking values $-2$ and two with equal probabilities, every absolute moment size is exactly two: increasing the exponent does not distinguish signs when the absolute value is constant.

For finite $\mathbb{E}|Z{|}^s$, apply the reversed Jensen's inequality to the strictly concave function $t^{r/s}$ and the nonnegative random variable $|Z{|}^s$:Taking the positive $r$th root proves the claim. Equality in this finite case holds exactly when $|Z|$ is constant almost surely. If the higher moment is infinite, the upper bound is infinite and remains valid. Exponent zero is excluded from the statement; logarithmic or geometric limiting summaries require separately stated conventions, especially when $Z$ can vanish.

For vectors $u,v\in {\mathbb{R}}^n$ or ${\mathbb{C}}^n$, counting measure turns the Minkowski inequality intoFor $p=\infty$, replace each displayed norm by the maximum of its coordinate absolute values. Take $u=(1,0)$ and $v=(0,1)$. Both vectors have Lp norm one for every $p\ge 1$, but their sum has L1 norm $2$, Euclidean norm $\sqrt{2}$, and L-infinity norm $1$. Thus the same vector addition gives equality $2=1+1$ at $p=1$, strict inequality $\sqrt{2}<1+1$ at $p=2$, and strict inequality $1<1+1$ at $p=\infty$.

By contrast, $u=(1,1)$ and $v=2u=(2,2)$ give equality for every $p\ge 1$: absolute homogeneity of a norm yields $\Vert u+v{\Vert }_p=3\Vert u{\Vert }_p=\Vert u{\Vert }_p+\Vert v{\Vert }_p$. At $p=2$, the three sizes are $\sqrt{2}$, $2\sqrt{2}$, and $3\sqrt{2}$. Opposite vectors behave very differently: $u+(-u)=0$, so the inequality is strict whenever $u$ is nonzero. Cancellation is allowed; the bound describes the largest possible total size.

Write $A_q=⟨f{⟩}_{w,q}$ and $N_q=\Vert f{\Vert }_{w,q}$. The exponent on the finite-set expression is $-1/q^{\prime }$ in the PDF; the TeX aid drops its prime. Ignore zero-measure sets, whose integrals vanish. The equivalence isFor the first bound, let $E_a=\{|f|>a\}$ and choose a finite-measure subset $E\subset E_a$. Since ${\int }_E|f|\ge a|E|$,Exhaust $E_a$ by its intersections with bounded balls. If its measure is finite these give the desired bound $a|E_a{|}^{1/q}$ in the limit; if it is infinite they force $N_q=\infty$. Now take the supremum over $a$.

For the other direction it suffices to assume $0<A_q<\infty$; the zero and infinite cases are immediate. For a set $E$ with $0<|E|=m<\infty$, the layer cake representation givesSplit at $t_0=A_q{m}^{-1/q}$. The first contribution is $m{t}_0$, and the second is $A_q^q{t}_0^{1-q}/(q-1)$. Their sum is $q^{\prime }{A}_q{m}^{1/q^{\prime }}$. Multiply by $m^{-1/q^{\prime }}$ and take the supremum. This proves the equivalence even when the quantities are extended-valued. The integral expression is an actual norm on the weak Lq space, since the triangle inequality holds inside every set integral. The distribution-function expression is generally a quasi-norm. The factor $q^{\prime }$ is sharp: $f(x)=x^{-1/q}$ on $(0,\infty )$ has $A_q=1$ and attains $N_q=q^{\prime }$ on every initial interval. In higher dimensions, multiply this example by the indicator of a unit box in the remaining coordinates.

The derivative of the backward characteristic flow map isThus the change of variables formula preserves phase-space Lebesgue measure. With zero source, $f_t=f_0\circ S_{0,t}$, so for every finite $p>0$,Taking the $p$th root proves $\Vert f_t{\Vert }_p=\Vert f_0{\Vert }_p$. For $0<p<1$ this is a quasi-norm, and the argument still works because it uses only a change of variables, not the triangle inequality. The identity also holds in the extended sense when an integral is infinite. Since the flow is bijective and measure-preserving, it additionally preserves the essential supremum, so the same conclusion holds for $p=\infty$.

For $1<q<\infty$, the weak Lq space consists of measurable functions for which $A_q(f)={\sup }_{t>0}t|\{|f|>t\}{|}^{1/q}$ is finite. This distribution-function expression is a quasi-norm. The equivalent norm $N_q(f)={\sup }_{0<|E|<\infty }|E{|}^{-1/q^{\prime }}{\int }_E|f|$ satisfies $A_q\le N_q\le q^{\prime }{A}_q$, with $q^{\prime }=q/(q-1)$. The layer cake representation bounds the integral by ${\int }_0^{\infty }\min (|E|,(A_q/t{)}^q)dt=q^{\prime }{A}_q|E{|}^{1/q^{\prime }}$. The lower bound uses finite-measure subsets of a superlevel set.