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An Lp norm measures the size of a measurable function by averaging a power of its absolute value with respect to a specified measure. For $1\le p<\infty$,when the integral is finite. The exponent $p$ controls how strongly large values influence the result. The Lp space identifies functions equal almost everywhere, so this becomes a norm on the corresponding equivalence classes. For $p=\infty$, use the essential supremum of $|f|$, as defined by the L-infinity norm. ^[ref].

For a two-coordinate vector $(3,4)$, treat the coordinates as a function on two inputs with counting measure. Then $\Vert (3,4){\Vert }_1=3+4=7$, $\Vert (3,4){\Vert }_2=\sqrt{9+16}=5$, and $\Vert (3,4){\Vert }_{\infty }=4$. The middle case is the Euclidean norm; the first adds the coordinate magnitudes, and the last takes their largest value. These are different answers to different questions about size, all satisfying the triangle inequality.

If instead each input has probability $1/2$, the same values give $\Vert f{\Vert }_1=3.5$ and $\Vert f{\Vert }_2=\sqrt{12.5}$, while $\Vert f{\Vert }_{\infty }=4$ remains unchanged. Thus the measure is not an optional label. On a probability space, the finite Lp norms are nondecreasing in $p$ whenever they are finite, by Hölder's inequality; the unnormalized two-coordinate example above decreases instead. Statements comparing different exponents must specify their measure convention.

Why are these sizes useful? If $e$ is an error function, its L1 norm measures aggregate absolute error, its L2 norm measures the square root of aggregate quadratic loss, and its L-infinity norm bounds error almost everywhere. Concentrating a large error on a set of small positive measure can give a modest finite Lp norm but a large L-infinity norm. The three-point example and diagram in L-infinity norm show this explicitly, including why a null set behaves differently from a rare event.

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 a prime number $p$ and $p\nmid a$, multiplication by $a$ gives a permutation of the $p-1$ nonzero residue classes. For $a=3$ modulo seven, the representatives $1,2,3,4,5,6$ map to $3,6,2,5,1,4$: the same collection in a different order. Repeating this permutation from one visits all six representatives before returning.

For $a=2$ modulo seven, the permutation cycles are instead $(1,2,4)$ and $(3,6,5)$. The powers of two visit only the permutation cycle containing one, so their multiplicative order is three. Both permutation cycles return after three steps; applying the permutation six times still fixes every nonzero residue class. Fermat's little theorem guarantees a return at six, not that every base visits every nonzero residue class.

More generally, every permutation cycle of this map has length equal to the multiplicative order of $a$: starting at a nonzero residue class $x$, a return means $a^{k}x\equiv x(\mathrm{mod}p)$, and its modular multiplicative inverse allows cancellation of $x$. Thus the $p-1$ nonzero residue classes split into equal-length permutation cycles, which also explains the divisibility of the multiplicative order into $p-1$.

The product argument gives a short proof outline. Since multiplication by $a$ is a permutation of the nonzero residue classes,Every factor is a unit, so their product has a modular multiplicative inverse and can be cancelled. This leaves $a^{p-1}\equiv 1(\mathrm{mod}p)$. The whole argument works for $p=2$, where the product is just one.^[ref]

In this positive integer context, a power of two is $2^k$ for a nonnegative integer $k$, giving $1,2,4,8,\dots$. Allowing negative or real exponents gives a broader meaning of exponentiation, which is not intended in the integer divisibility criteria here.

The indicator function example reaches a squared L2 norm ratio of two. To approach four, concentrate more strongly near zero. For $0\le a<1/2$, use the measurable functionAll integrals use Lebesgue measure $dx$. Keeping both portions of the squared L2 norm givesConsequently,Every allowed member has ratio strictly below four. For any proposed constant $C<4$, a sufficiently large $a<1/2$ has ratio greater than $C$. Combined with Hardy's integral inequality, this proves that four is the smallest possible squared L2 norm constant.^[ref]

The endpoint is excluded: at $a=1/2$, ${\int }_0^1|f_a{|}^{2}dx={\int }_0^1{x}^{-1}dx=\infty$, and the inner portion of $\Vert H{f}_a{\Vert }_2^2$ also diverges. Its tail still has finite integral four. There is no finite L2 norm ratio at this endpoint, hence no equality example obtained by substituting it. The limit four is unattained by this family. Multiplying each $f_a$ by $\sqrt{1-2a}$ makes its L2 norm one without changing the ratio.

For a real exponent $s>1$ and a positive integer $N$, the series truncation error in the defining Riemann zeta function series isThe integral comparison for the positive decreasing function $x^{-s}$ gives the explicit boundsOn each unit interval, the area under $x^{-s}$ lies between its endpoint heights. Summing from the first omitted term gives the lower integral from $N+1$ to infinity and the upper integral from $N$ to infinity. This uses positivity and monotonicity, not an assumption that the last computed term equals the remaining error.

At $N=10$, the bounds are $1/11\le R_{10}(2)\le 1/10$, whereas $1/(3\cdot 11^3)\le R_{10}(4)\le 1/(3\cdot 10^3)$. The errors above fit both intervals. Merely looking at the eleventh term would give $1/121\approx 0.008264$ for squares, far smaller than the actual remaining $0.095166$.

For a chosen absolute error target $\epsilon >0$, the sufficient condition is $N\ge ((s-1)\epsilon {)}^{-1/(s-1)}$, rounded upward to an integer. Thus $\epsilon =10^{-4}$ requires $N=10000$ by this bound for squares but only $N=15$ for fourth powers. This is a guarantee from an upper bound, not a claim that those are the smallest successful term counts.