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At level $J$, a Haar scaling function is the normalized indicator function ${\varphi }_{J,k}=2^{J/2}{1}_{I_{J,k}}$ of the dyadic interval $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$. Its L2 norm is one because its squared height is $2^J$ and its support length is $2^{-J}$.

For a real square-integrable function $f$, its coefficient isMultiplying the coefficient by the Haar scaling function restores the cell average on that interval. The disjoint supports make the functions at one level an orthonormal basis of the cellwise-constant subspace. This normalization is also what turns coefficient variances in the Gaussian white noise model into $1/n$ and cellwise variances into $2^J/n$ in the Haar projection estimator in Gaussian white noise.

The Haar wavelet is a signed contrast between two adjacent half-intervals:It has compact support, zero integral, and unit L2 norm: $\int \psi =1/2-1/2=0$ and $\int |\psi {|}^2=1$. The zero integral is its one vanishing moment. Its Haar scaling function is $\varphi =1_{[0,1)}$, which records a coarse value rather than a contrast. Endpoint conventions do not alter the L2 inner products.

The normalized copies ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ compare the halves of the dyadic interval $[k{2}^{-j},(k+1)2^{-j})$. Narrower copies are taller so that their L2 norms stay one. Thus a coefficient identifies both a location, through $k$, and a scale, through $j$.

For a simple step function $f$ equal to $2$ on $[0,1/2)$ and $6$ on $[1/2,1)$, its coarse coefficient is $⟨f,\varphi ⟩=4$, while its first contrast is $⟨f,\psi ⟩=-2$. Thus $f=4\varphi -2\psi$. The negative coefficient raises the second half and lowers the first. Finer contrasts vanish because $f$ is constant on each half. A jump aligned with this dyadic interval boundary therefore needs only two nonzero coefficients; jumps at other positions can require details at many scales.

Its jump discontinuities make the Haar wavelet particularly transparent for changes between neighbouring regions, while limiting approximation of smooth variation. A Daubechies wavelet uses a more elaborate pattern to obtain additional vanishing moments and, at sufficiently high order, smoother basis functions. The elementary Haar refinement identity has coefficients $h_0=h_1=1/\sqrt{2}$ and explains how the coarse and fine descriptions fit together.

The historical source is Alfred Haar's ^[ref]. Chapter III, beginning on page 361, constructs the constant function and these localized piecewise-constant orthogonal functions. Its original endpoint assignments differ from the half-open convention used here, without changing L2 inner products.

Use the $\{0,1{\}}^n$ convention for the Hamming cube $H_n$. Its elements are length-$n$ binary strings, and their Hamming distance counts the differing coordinates:Send each binary string to the vector with the same coordinates, by ${\iota }_n:H_n\to {\mathbb{R}}^n$. Every differing coordinate contributes $1$ both to the sum of absolute differences and to the sum of squared differences. Therefore the L1 norm and L2 norm give, respectively,This short coordinate calculation proves that the L1 norm gives an isometric embedding, while Euclidean distance grows like the square root of the number of changed binary digits. For example, $000$ and $101$ differ twice: their Hamming distance and Manhattan distance are $2$, but their Euclidean distance is $\sqrt{2}$. Comparing binary strings in these two geometric representations answers different questions even though the points have the same coordinates.

For distinct pairs $k$ ranges over every integer from $1$ to $n$. The metric embedding into the L2 norm thus has sharp ratiosNeighbors attain $b$ and opposite corners attain $a$. This computes the bi-Lipschitz distortion of this coordinate map; it does not prove that every metric embedding into an L2 space has at least this bi-Lipschitz distortion. Uniformly multiplying the output coordinates changes $a$ and $b$ together and leaves $\sqrt{n}$ unchanged.

The important family distinction is that these coordinate maps have no dimension-independent bound on their bi-Lipschitz distortion into the L2 norm, yet they do give a uniform coarse embedding of a family of metric spaces. Pad each output vector with zero coordinates to view all maps in the single l2 sequence space, an L2 space for counting measure. Then the same functionswork for every $n$, and ${\rho }_{-}(t)\to \infty$. Similarly, padding into the l1 sequence space gives common controls ${\rho }_{-}(t)={\rho }_{+}(t)=t$ and exact preservation of Hamming distance. These claims use unnormalized Hamming distance; replacing it by normalized Hamming distance changes the family of source metrics.

The geometry has a practical origin: Hamming's 1950 paper uses the number of differing positions between binary strings to study error-correcting codes. Changing one binary digit moves to a neighboring corner, so a question about corrupted messages becomes a question about separation in the Hamming cube. ^[ref]

Let $0<\lambda <d$ and $1<p,q<\infty$ satisfy $1/p+1/q+\lambda /d=2$. The Hardy–Littlewood–Sobolev inequality states thatHere the right side uses Lp norms, and the constant is independent of $f,g$. The exponents are linked by the displayed dimensional relation; they are not a conjugate exponents pair. This is a bound for a singular convolution rather than a pointwise bound on the kernel.^[ref]

For $d=3$, $\lambda =1$ and $p=q=6/5$, applying it to $f=g=|u{|}^2$ givesThis is the potential-energy estimate used for the Gravitational Hartree equation. A Gagliardo-Nirenberg interpolation inequality then relates that intermediate Lp norm to mass and the gradient L2 norm.

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.

Take the indicator function $f=1_{(0,1)}$. Direct integration givesAlthough the function vanishes beyond one, its average remembers the unit area accumulated before one. Squaring and integrating both portions givesThus the tail accounts for half of the squared L2 norm. Averaging does not always decrease the L2 norm with this Lebesgue measure on the half-line. The same indicator function also explains the restriction $p>1$: ${\int }_1^{\infty }{x}^{-1}dx=\infty$, so its average is not in L1 space.

A homogeneous Sobolev space controls derivatives without an independent low-frequency L2 norm: formally $\Vert f{\Vert }_{{\dot{H}}^s}^2=\int |\xi {|}^{2s}|\hat{f}(\xi ){|}^{2}d\xi$, with the normalization chosen according to the Fourier transform. Completions or quotient conventions must be specified when polynomial ambiguities occur. In three dimensions ${\dot{H}}^1$ is represented by $L^6$ functions with gradient in $L^2$, and $\Vert f{\Vert }_6\le C\Vert \nabla f{\Vert }_2$.

For the forward diffusion equation $u_t=\kappa u_{yy}$ on $0<y<L$, with $\kappa >0$ and homogeneous Dirichlet boundary conditions, a solution continuous on the closed initial-boundary domain can take initial data $u(0,y)=\varphi (y)$ only if $\varphi (0)=\varphi (L)=0$. These equalities express the same corner value through the initial condition and the boundary condition.

An incompatible endpoint value does not prevent an interior or L2 space initial-data solution. For example, on $0<y<1$ the initial ramp $\varphi (y)=y$ givesIt satisfies both zero endpoint values for every $t>0$ and tends to $y$ at every fixed $0<y<1$ as $t\downarrow 0$, also in the L2 norm. It cannot extend continuously to $(0,1)$ with both the initial value one and the boundary value zero. Each sine term vanishes at the endpoint, so the series cannot silently repair that pointwise contradiction.

If the equation is written $a{u}_t=u_{yy}$, the intended forward heat coefficient requires $a>0$. For $a<0$, the ramp's coefficients would grow like $e^{n^2{\pi }^{2}t/|a|}/n$, failing square summability for every $t>0$. This example distinguishes an incompatible corner from the separate ill-posedness of backward diffusion.

An input-output gain bounds how large a system's output can be relative to its input, after the input and output measures have been specified. For a zero-initial-state operator $T$ and chosen signal norms, its induced signal gain isA finite gain gives $\Vert y{\Vert }_{\mathrm{out}}\le \gamma \Vert u{\Vert }_{\mathrm{in}}$. The supremum is over the admissible input class, not merely a collection of trial inputs. Initial-state response must be treated separately.

For an original scalar linear time-invariant system, letUsing the time-domain L-infinity norm, the convolution solution obeysA constant input of suitable sign approaches the limiting ratio $|b|/a$, so this bound is sharp. At $a=2$ and $b=3$, the peak-amplitude gain is1.5: an input bounded by2 gives output bounded by3 from zero initial state, and the constant input approaches that value.

The signal norms change the problem. For the same stable scalar system, the peak frequency response magnitude isIt coincides numerically with the preceding gain in this example. For stable linear systems, this frequency-domain quantity is the energy-signal induced gain associated with the time-domain L2 norm; it should not generally be identified with the peak-amplitude induced gain. MathWorks's linear-model norm documentation distinguishes this peak frequency-response quantity from its separate impulse-response norm.^[ref]

The units matter as well: a gain from force input to displacement output has physical units, whereas identical normalized signal units can produce a dimensionless gain. In robust performance, the bound must hold across the allowed control model uncertainty. A nominal gain calculation alone supplies no such universal guarantee.

For a nonnegative symmetric unit-integral kernel for density estimation and $f\in H^2(\mathbb{R})$, the integral Taylor remainder and Minkowski inequality give $\Vert K_h\ast f-f{\Vert }_2\le h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_2/2$. Translation preserves the L2 norm. For a signed kernel the same proof requires the absolute moment $\int u^2|K(u)|du$.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

An arbitrary $L^2$ initial function has no well-defined nodal point samples. Cell averages or a bounded projection to the discrete approximation space supply grid data independently of its pointwise representative. Jensen's inequality bounds the cell-average discrete L2 norm by the continuous L2 norm. Combining consistent bounded initialization, stability, and density of smooth data can extend convergence to rough initial functions, without claiming their nonexistent classical derivatives.

An L2 initial condition specifies a square-integrable initial function up to equality almost everywhere. It need not have pointwise nodal values or classical derivatives. L2-compatible initialization of grid data uses bounded averaging or projection, and evolution estimates use the relevant L2 norm.

The L2 inner product is the inner product $⟨f,g⟩=\int f\overline{g}d\mu$ on L2 space. Its induced norm is the L2 norm, and the mass matrix represents it in a finite element basis.

A sequence in L2 space converges in its L2 norm when $\int |f_n-f{|}^{2}d\mu \to 0$. This concerns the integral of the squared error and need not imply pointwise convergence. In particular a complete Fourier sine basis reconstructs each odd L2 function in this norm, while pointwise claims about its Fourier series require additional hypotheses.

Take the polynomial $f(x)=1+x+x^2+x^3$ on $[-1,1]$ with unit weight. Using the four Legendre polynomials above, direct integration givesFor example, $⟨f,P_2⟩=\frac{1}{2}{\int }_{-1}^1(3x^4+2x^2-1)dx=4/15$: the omitted odd terms have zero integral. Dividing by $h_2=2/5$ gives $c_2=2/3$. Substituting all four coefficients proves the exact identity

The successive orthogonal projections onto degrees at most $m$ are thereforeHere $\Vert g{\Vert }_2^2={\int }_{-1}^{1}g(x{)}^{2}dx$ is the squared L2 norm, without division by the interval length. Each row retains the previous Legendre polynomial coefficients and appends one new term. The ordinary power coefficients can change: adding $\frac{2}{3}{P}_2$ changes the constant from $4/3$ to $1$, and adding $\frac{2}{5}{P}_3$ changes the linear coefficient from $8/5$ to $1$. Independence belongs to the chosen orthogonal polynomial coordinates.

The full squared L2 norm is $\Vert f{\Vert }_2^2=192/35$, equal to $32/9+128/75+8/45+8/175$. The residual budgets are the omitted terms:For a direct check independent of summing the budgets, $f-q_2=x^3-3x/5$, and integrating its square gives $2/7-12/25+6/25=8/175$. This residual is orthogonal to $1,x,x^2$, exactly the conditions for the best quadratic least-squares solution. An orthogonal projection need not be a polynomial interpolant: for example, $q_2(1)=18/5$ while $f(1)=4$. The approximation allocates error across the interval according to the chosen weighted inner product.

For normalized Legendre polynomials, this kernel for density estimation reproduces evaluation at zero on every polynomial of degree below $\ell$: expand $p$ in the orthonormal basis to obtain $\int K_{\ell }p=p(0)$. Taking monomials proves integral one and zero moments of orders $1,\dots ,\ell -1$; compact support gives finite absolute moments. If exact order requires a nonzero moment of order $\ell$, add $({\varphi }_{\ell }(0)+1){\varphi }_{\ell }{1}_{[-1,1]}$. Lower moments remain unchanged. Writing $a_{\ell }=⟨u^{\ell },{\varphi }_{\ell }⟩\ne 0$ and evaluating the full expansion of $u^{\ell }$ at zero shows that the new moment is $a_{\ell }$. A useful three-derivative example is $K_3=(9-15u^2)1_{[-1,1]}/8$, whose first two moments vanish and whose squared L2 norm is $9/8$.

The Lévy-Itô decomposition builds a real Lévy process from drift, continuous Gaussian noise and Poisson jumps. For the triplet $(a,b,K)$ in the Lévy-Khintchine formula with small-jump cutoff one, take independent standard Brownian motion $B$ and Poisson random measure $N$ of intensity $dsK(dx)$. ThenHere $b\ge 0$, $K$ is a Lévy measure and $\tilde{N}=N-dsK(dx)$ denotes compensation. The large-jump term has finitely many jumps on each bounded time interval because $K(|x|>1)<\infty$.

The small-jump integral is defined by limits, rather than by subtracting two potentially infinite sums. For $\epsilon >0$, let $J_t^{\epsilon }={\int }_0^t{\int }_{\epsilon <|x|\le 1}x\tilde{N}(ds,dx)$. These are square-integrable martingales. The compensated Poisson isometry givesas $0<\delta <\epsilon \downarrow 0$. Doob's L2 maximal inequality also controls the supremum on $[0,T]$, so the limit exists locally uniformly in probability, indeed in the L2 norm of that supremum. This uses exactly the small-jump second-moment condition of a Lévy measure.^[ref]

The components have independent increments; multiplying their characteristic functions recovers the Lévy-Khintchine exponent. The decomposition separates continuous fluctuations from large jumps and possibly infinitely many small jumps. A finite small-jump Second moment does not require a finite first absolute jump moment, which explains why compensation and its limiting construction are essential.

Suppose $u_j\to u$ strongly in $L^2(B_R)$ for every $R>0$, and assume uniform tail controlThen $u\in L^2({\mathbb{R}}^n)$ and $u_j\to u$ strongly in L2 space on the whole space. Local convergence and the tail bound first bound the global L2 norms; Fatou's lemma gives the same tail bound for $u$. Splitting the error givesChoose $R$ to make the last two terms uniformly small, then increase $j$. Local Rellich-Kondrachov compactness theorem arguments and a diagonal subsequence supply the first hypothesis in many Sobolev space problems; the tail condition rules out loss of compactness at infinity.

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.