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From an Lp norm Cauchy sequence choose a subsequence with successive difference norms at most $2^{-j}$. Minkowski inequality and monotone convergence theorem make the sum of the absolute differences an $L^p$ function, hence finite almost everywhere. The telescoping series defines an $L^p$ limit; the geometric bound on its tail proves norm convergence of the subsequence, and the original Cauchy property proves convergence of the whole sequence. No sigma-finiteness assumption is needed.

For a finite interval $(a,b)$ with Lebesgue measure, restrictions of continuous functions on $[a,b]$ form a dense subspace of the Lp space $L^p((a,b))$ for $1\le p<\infty$. One proof first approximates by simple functions. The regularity of Lebesgue measure approximates each of their measurable sets in measure by a finite union of intervals. An interval's indicator function is approximated by continuous functions taking values in $[0,1]$ and differing from that indicator function only in endpoint strips of total length tending to zero. Their Lp norm error is at most the $1/p$ power of that length. The triangle inequality completes the approximation of each simple function.

For a function with bounded derivatives through order $k$, the derivative supremum norm is $|u{|}_{k;U}={\sum }_{j=0}^k{\sup }_U|D^{j}u|$, with any equivalent finite-dimensional norm for each derivative tensor. This is a classical $C^k$ norm; it is different from the Lp norm with exponent $k$.

A dual norm measures how strongly a bounded linear functional can respond to a unit-sized input. If $p$ is a bounded linear functional on a normed vector space $X$, write $⟨p,x⟩=p(x)$ and defineIt is the operator norm of $p$ when its scalar output is measured by absolute value. On ${\mathbb{R}}^n$, represent $p(x)=a^{T}x$ using a coefficient vector $a$. The unit ball is symmetric, so maximizing $|a^{T}x|$ gives the same value as maximizing $a^{T}x$: negate an input whenever its output is negative.^[ref]

For an original two-coordinate example, fix $a=(2,-1{)}^T$, so $p(x)=2x_1-x_2$. The input norm changes the allowable unit ball and hence the dual norm:Under the L1 norm, $|2x_1-x_2|\le 2(|x_1|+|x_2|)\le 2$, and $(1,0{)}^T$ attains the bound. Under the Euclidean norm, the Cauchy-Schwarz inequality gives $|a^{T}x|\le \Vert a{\Vert }_2\Vert x{\Vert }_2\le \sqrt{5}$, with equality at the displayed normalized vector. Under the maximum norm, each coordinate can reach unit absolute value, so $|2x_1-x_2|\le 2+1=3$; $(1,-1{)}^T$ attains it. Thus the dual norms of these three vector norms are respectively the maximum norm, the Euclidean norm, and the L1 norm. For $1\le p\le \infty$, Lp norms on a finite coordinate space pair with their Holder conjugate exponents.^[ref]

This geometry explains the usefulness of a dual norm: it translates a bound on inputs into the sharp bound on a linear output. The support function of the unit ball gives its value in the coefficient direction $a$. Changing the input budget changes the set of possible inputs; the three different answers do not conflict. In finite dimensions a closed unit ball is compact, so a continuous linear functional achieves its maximum absolute value. In an infinite-dimensional normed vector space, the supremum remains the definition and need not be achieved.

For a function on the dyadic rationals in $[0,1]$, let $A_n$ be its largest absolute adjacent increment on the level-$n$ grid. If $K_{\alpha }<\infty$ for $\alpha >0$, the function extends uniquely to a Hölder continuous function withChoose $n$ with $2^{-n}\le |t-s|<2^{1-n}$. The level-$n$ left approximations differ by at most two grid steps, and each finer approximation adds at most one level increment. Thus the difference is bounded by $2{\sum }_{j\ge n}{A}_j\le 2^{-n\alpha }{K}_{\alpha }$. For a stochastic process with $\Vert {\xi }_t-{\xi }_s{\Vert }_p\le C|t-s{|}^{\beta }$, the bound $\Vert A_n{\Vert }_p\le C{2}^{-n(\beta -1/p)}$ makes this series converge in the Lp norm for $0<\alpha <\beta -1/p$.

For $1<p<\infty$ and nonzero elements $f,g$ of the same real or complex Lp space, equality in the Minkowski inequality holds if and only ifEquivalently, $g=cf$ almost everywhere for a single positive real number $c$. The multiplier must be constant across $X$; matching signs at each input alone is insufficient. For complex numbers, an arbitrary complex scalar multiple is also insufficient: $g=if$ gives $\Vert f+g{\Vert }_p=\sqrt{2}\Vert f{\Vert }_p<2\Vert f{\Vert }_p$ for nonzero $f$. If either element is zero in Lp space, equality holds automatically, even if a representative has nonzero values on a null set.

The reason is strictly convex normed space geometry. Normalize $f,g$ to unit Lp norm and use weights $\Vert f{\Vert }_p/(\Vert f{\Vert }_p+\Vert g{\Vert }_p)$ and $\Vert g{\Vert }_p/(\Vert f{\Vert }_p+\Vert g{\Vert }_p)$. Distinct normalized functions give a convex combination of Lp norm strictly below one. This follows by integrating the strict convexity of $z\mapsto |z{|}^p$ on the real numbers or complex numbers; equality therefore forces the normalized functions to agree almost everywhere.

At $p=1$, equality instead means $|f+g|=|f|+|g|$ almost everywhere: the nonnegative difference has zero integral. For real functions, this is $fg\ge 0$ almost everywhere. For complex functions, their nonzero values must have the same argument wherever both are nonzero, but their magnitude ratio may vary. The disjoint-coordinate vectors $(1,0)$ and $(0,1)$ illustrate equality without constant proportionality.

At $p=\infty$, no such constant proportionality condition is necessary: $(1,0)$ and $(1,1)$ have L-infinity norms one and a sum of L-infinity norm two. Nor must an essential supremum be attained: under Lebesgue measure on $(0,1)$, $f(x)=g(x)=x$ give equality $2=1+1$, although neither function takes the value one. The endpoint Lp spaces need not be strictly convex normed spaces.

The essential range of a real or complex measurable function $m$ consists of the values $z$ whose every neighborhood receives inputs of positive measure:It depends on the measure as well as the function. This definition concerns values arbitrarily near $z$, not necessarily inputs at which $m=z$ exactly. It therefore retains meaningful limiting values while ignoring values occurring only on null sets. ^[ref].

For $m(x)=x$ on $(0,1)$ with Lebesgue measure, the ordinary image of a function is $(0,1)$, but the essential range is $[0,1]$. Neither endpoint is attained, yet inputs with $|m(x)|<\epsilon$ or $|m(x)-1|<\epsilon$ have positive measure for every $\epsilon >0$. Every interior value also belongs to the essential range, although its exact fiber is a singleton of zero measure. Positive measure of a fiber is sufficient, but is not necessary, for a value to belong to the essential range.

The two-sided example $g$ from essential infimum and essential supremum illustrates what disappears. It follows $g(x)=x$ on $[0,1]$, except $g(1/4)=-3$ and $g(3/4)=4$. Its ordinary image of a function is $([0,1]\setminus \{1/4,3/4\})\cup \{-3,4\}$, whereas its essential range remains $[0,1]$. The missing ordinary values $1/4$ and $3/4$ still have arbitrarily close neighbors with positive measure; the exceptional values $-3$ and $4$ can each be isolated by a sufficiently small neighborhood whose preimage is a null set.

For the three-point example in L-infinity norm, the values are $1,3,10$. With weights $(0.5,0.5,0)$ the essential range is $\{1,3\}$; with weights $(0.499,0.5,0.001)$ it is $\{1,3,10\}$. A small but positive probability retains ten, so an essential range does not discard a value merely because it is rare. For a real function on a space of positive total measure, the essential infimum and essential supremum are the lower and upper edges of its essential range, allowing infinite bounds. The L-infinity norm instead takes the upper edge of the essential range of $|m|$.

A useful operator connection is the spectrum of a real multiplication operator: on a nonzero L2 space over a sigma-finite measure, multiplication by a real measurable function $m$ has spectrum equal to its essential range, with the natural domain when $m$ is unbounded. Outside that range, $|m-z|$ has a positive lower bound almost everywhere, so multiplication by $1/(m-z)$ gives an inverse function that is a bounded linear operator. Near an essential value, this lower bound fails. Thus discarding null sets is precisely what makes the range match an operator acting on almost everywhere equivalence classes. ^[ref].

An exponent specifies the power in exponentiation: in $a^b$, $a$ is the base and $b$ is the exponent. For a positive integer exponent, $2^3=2\cdot 2\cdot 2=8$. For a positive real number base, a real exponent is defined by $a^b=\exp (b\log a)$ using the exponential function and natural logarithm. The Lp norm uses its exponent $p$ to determine how strongly large absolute values affect the result.

Take $p=3$, $q=3/2$, and vectors $x=(1,2,-1)$ and $y=(2,-1,3)$. Their coordinate products are $(2,-2,-3)$, soTheir Lp norms give the larger boundThus cancellation explains $3<7$, and Hölder's inequality explains the remaining gap between $7$ and the norm product. Small signed sums need not imply small individual coordinate products.

Now keep $x$ and choose $y=(1,4,-1)$. The coordinate products are $(1,8,1)$, with no cancellation, andThe matching powers are $|x_i{|}^3=|y_i{|}^{3/2}$, namely $(1,8,1)$.

A Fourier extension operator forms an oscillatory integral from a density on a frequency surface with measure $\sigma$. The opposite sign in the phase merely reflects the output variable, so it gives identical Lp norms. In the usual dual formulation it is the adjoint of restricting the Fourier transform to that surface.

Particles keep their velocity and move along straight trajectories. The method of characteristics gives $f(t,x,v)=f_0(x-tv,v)$. The shear preserves Lebesgue measure and every Lp norm. Compact initial support remains compact at each finite time, but need not remain in one compact set for all times.

Let $1<p<\infty$ and let $f$ be a real or complex measurable function in Lp space $L^p((0,\infty ),dx)$, with Lebesgue measure $dx$. The Hardy operator, denoted here by $H$, takes the running averageHölder's inequality ensures ${\int }_0^x|f(t)|dt\le x^{1-1/p}\Vert f{\Vert }_p<\infty$, so every such average exists. Hardy's integral inequality statesThe real signed and complex extensions of the nonnegative statement follow from the triangle inequality $|Hf|\le H|f|$. Changing $f$ on a null set changes neither average nor Lp norm.^[ref]

Averaging over an expanding interval spreads a localized input across the whole positive half-line. Hardy's integral inequality bounds the total Lp norm of that spread. The Hardy operator is consequently a bounded linear operator on this Lp space. The factor $p/(p-1)$ bounds its operator norm; its $p$th power is the constant in the displayed integral bound. These are different constants: at $p=2$, the L2 norm bound is two and the squared L2 norm bound is four.

Hölder's inequality controls a product using separate information about its two factors. If one factor is large on part of a domain, the other factor's complementary Lp norm limits how much their product can accumulate there. The same statement bounds finite sums, integrals and expected values; no independence is required in the probabilistic version.

Let $p,q\in [1,\infty ]$ be conjugate exponents, meaningOn a measure space, if $f\in L^p$ and $g\in L^q$, then $fg\in L^1$ andHere the Lp norms are finite; at exponent $\infty$ the norm uses the essential supremum. For a finite vector, integration with counting measure givesThe first step is the triangle inequality; the second is Hölder's inequality.^[ref]

For finite vectors, the endpoint case $p=1$, $q=\infty$ is $|x^{T}y|\le \Vert x{\Vert }_1\Vert y{\Vert }_{\infty }$: bound every $|y_i|$ by its maximum, then sum the $|x_i|$. Swapping the factors gives the other endpoint. The choice $p=q=2$ gives the Cauchy-Schwarz inequality.

Ladyzhenskaya's inequality controls the Lp norm of a function by its Lp norm and the Lp norm of its gradient: in dimension two, $\Vert u{\Vert }_4\le C\Vert u{\Vert }_2^{1/2}\Vert \nabla u{\Vert }_2^{1/2}$; in dimension three the exponents are $1/4$ and $3/4$. These estimates hold for smooth compactly supported functions and extend using density of smooth functions in a Sobolev space.

On a measure space of positive total measure, the L-infinity norm of an essentially bounded measurable function is $\Vert f{\Vert }_{\infty }={\mathrm{esssup}}_{\mu }|f|$. The absolute value matters: a constant function $f=-3$ has L-infinity norm three, even though its essential supremum is $-3$. The norm belongs to the equivalence class in Lp space; different representatives equal almost everywhere give the same answer. ^[ref].

For the function $g$ in essential infimum and essential supremum, $\Vert g{\Vert }_{\infty }=1$ under Lebesgue measure, whereas its supremum norm is four. Replacing the exceptional values by the ordinary values $x$ leaves every finite Lp norm unchanged as well: $\Vert g{\Vert }_p=({\int }_0^1{x}^{p}dx{)}^{1/p}=(p+1{)}^{-1/p}$. This tends to one as $p\to \infty$. That agreement explains why the endpoint with exponent $p=\infty$ uses an essential bound rather than a pointwise bound.

A finite probability space makes the distinction between zero and small measure explicit. Let $f(a)=1$, $f(b)=3$, and $f(c)=10$. Assign weights $\mu (a)=1/2-\epsilon$, $\mu (b)=1/2$, $\mu (c)=\epsilon$, where $0\le \epsilon <1/2$. For $1\le p<\infty$,When $\epsilon =0$, the singleton $\{c\}$ is a null set and $\Vert f{\Vert }_{\infty }=3$. For every $\epsilon >0$, however small, $\Vert f{\Vert }_{\infty }=10$. For example, $\epsilon =0.001$ gives $\Vert f{\Vert }_1=2.009$ and $\Vert f{\Vert }_2=\sqrt{5.099}$, yet the large-$p$ Lp norms approach ten. The ordinary pointwise supremum norm is ten in both cases.

Why can a rare value dominate? For a bounded function on a probability space, an almost everywhere bound $|f|\le M$ gives $\Vert f{\Vert }_p\le M$. If $0<t<\Vert f{\Vert }_{\infty }$, the set $A_t=\{|f|>t\}$ has positive probability, so $\Vert f{\Vert }_p\ge t\mu (A_t{)}^{1/p}$. As $p\to \infty$, the factor $\mu (A_t{)}^{1/p}$ tends to one. These two bounds squeeze the limit to $\Vert f{\Vert }_{\infty }$. The probability measure assumption keeps total measure finite; this explanation does not assert the same finite-$p$ integrability on arbitrary infinite-measure spaces.

For a random variable with finite fourth moment, apply Holder inequality with conjugate exponents $3/2$ and $3$ to $|X{|}^{2/3}|X{|}^{4/3}$. This gives $\mathbb{E}|X{|}^2\le (\mathbb{E}|X|{)}^{2/3}(\mathbb{E}|X{|}^4{)}^{1/3}$. Raising to the power $3/2$ proves the displayed Lp norm interpolation inequality.

For a uniformly elliptic divergence-form elliptic operator with lower-order coefficients in the supercritical Lebesgue spaces, truncation tests and Moser iteration bound the positive part of a weak subsolution on an interior ball by its Lp norm on a larger ball and the forcing norms. In particular, every weak subsolution of the homogeneous equation is locally bounded above. This is the preliminary estimate needed when a strong maximum principle is phrased using essential suprema.

Scalar advection preserves every finite Lp norm if its complete differentiable characteristic flow preserves volume and the initial datum belongs to the corresponding Lp space. In general the $p$th power of the norm is the initial $|f_0{|}^p$ weighted by the phase-space flow Jacobian. Completeness alone preserves the essential supremum, but does not preserve finite-$p$ norms for compressible transport.

A martingale satisfying ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$, for $1<p<\infty$, has almost sure convergence and convergence in Lp to a limit $M_{\infty }\in L^p$. Also $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. The Doob Lp maximal inequality makes ${\sup }_n|M_n|$ integrable to the power $p$; the martingale convergence theorem gives the almost-sure limit, and the dominated convergence theorem gives convergence in the Lp norm. Boundedness in $L^1$ alone does not imply convergence in L1.

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.