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The locally integrable function $E(z)=1/(\pi z)$ on the plane is a fundamental solution for the Wirtinger derivative ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$. Let $L=(4\pi {)}^{-1}\log (x^2+y^2)$ be the fundamental solution of the Laplace equation. Its first distributional derivatives agree with its ordinary ones: deleting a radius-$\epsilon$ disk in integration by parts gives an error $O(\epsilon |\log \epsilon |)$, tending to zero. Hence $E=4{\partial }_{z}L$ andThe analogous fundamental solution for ${\partial }_z$ is $1/(\pi \bar{z})$. The singular support of each kernel is exactly $\{0\}$: they are smooth elsewhere, and smoothness near zero would make their Wirtinger derivatives smooth there, contradicting the Dirac delta distribution identity.

Target: Distributional Cauchy-Riemann lemma. Parent: Cauchy-Riemann equations. Intended file: analysis.bigb.

The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

The locally integrable function $\log |x|$ has distributional derivative equal to the principal-value reciprocal distribution. Symmetric integration by parts makes the boundary term $\log \epsilon [\phi (\epsilon )-\phi (-\epsilon )]$ tend to zero.

Distribution theory extends ordinary functions by letting them act on smooth test functions. A locally integrable function $f$ acts by $⟨f,\varphi ⟩=\int f\varphi$, while the Dirac delta distribution acts by $⟨{\delta }_a,\varphi ⟩=\varphi (a)$. The latter is a point source, rather than an ordinary function with a very large value. The space of test functions and its topology make distributions continuous linear functionals.^[ref]

Integration by parts supplies a distributional derivative even when ordinary differentiability fails:For the Heaviside function $H$, $⟨H^{\prime },\varphi ⟩=-{\int }_0^{\infty }{\varphi }^{\prime }(x)dx=\varphi (0)$, so $H^{\prime }={\delta }_0$. A jump has become a point source under differentiation. This is useful for impulses in differential equations, concentrated loads and interface jump conditions.

The extension has limits: multiplying a distribution by a smooth function is always defined, but arbitrary products of distributions are not. The theory controls which operations are meaningful through their action on test functions, rather than by manipulating symbolic point values.

For a continuous even function $f$ on $\mathbb{R}$, the Cauchy principal valuedefines a distribution of order of a distribution at most one. For test functions supported in $[-R,R]$, the absolute value is at most $2R{\max }_{[0,R]}|f|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, since the numerator is bounded by $2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. When $\varphi (0)=0$, the expression equals the absolutely convergent ordinary integral $\int f\varphi /x$.

It is a regular distribution exactly when ${\int }_0^r|f(x)|/xdx<\infty$ for some $r>0$. Necessity follows by testing away from zero: any representing locally integrable function must equal $f/x$ almost everywhere off zero. Sufficiency follows from absolute convergence. Continuity with $f(0)=0$ alone does not suffice: $f(x)=1/\log (1/|x|)$ near zero gives the divergent integral ${\int }_1^{\infty }ds/s$.

Target: L1-normalized dilation of a distribution. Parent: Dilation of a distribution. Intended file: distribution-theory.bigb.

The normalized Haar scaling functions and wavelets obeyThis orthogonal two-by-two transformation gives the multiresolution analysis decomposition $V_{j+1}=V_j\oplus W_j$. Iterating it expresses a Haar approximation either through level-$j$ cell averages or through coarse averages and all wavelet details at lower levels. The identities remain valid locally for coefficient integrals of locally integrable functions, without a global $L^2$ assumption.

For a locally integrable function $f$ on ${\mathbb{R}}^d$, the centred Hardy-Littlewood maximal function isOn the real line the balls are intervals, giving $Mf(x)={\sup }_{r>0}(2r{)}^{-1}{\int }_{x-r}^{x+r}|f(t)|dt$. It records the largest absolute average visible at any scale around $x$; the absolute value prevents cancellation from hiding large values.

For $f=1_{[0,1]}$ on the line, $Mf(x)\le 1$, and it equals1 at every interior point. Far outside that interval it decays only on the order of $1/|x|$, because a large centred interval can still capture the unit mass. Thus an integrable input need not have an integrable maximal function. The correct endpoint bound is the Hardy-Littlewood maximal inequality of weak type$(1,1)$, while Strong Lp bound for the Hardy-Littlewood maximal function holds for $p>1$. These bounds control the exceptional sets in the Lebesgue differentiation theorem.

For a function smooth on either side of a joining point $a$, integration by parts on the two intervals adds a Dirac mass whose coefficient is the right trace minus the left trace. If the one-sided derivatives are locally integrable, a first weak derivative can be represented by a locally integrable function across the join only when the jump vanishes. Derivative jumps themselves do not obstruct first-order Sobolev membership: a continuous piecewise Lipschitz function still has a bounded weak first derivative.

Let $T>0$, and let $g$ be a locally integrable function with period $T$ on nonnegative time. For $\mathrm{Re}s>0$, its unilateral Laplace transform isSplit the defining integral into successive periods and put $t=u+jT$. Periodicity makes each piece $e^{-sjT}{\int }_0^T{e}^{-su}g(u)du$; summing the absolutely convergent geometric series proves the formula. Local integrability on one period also controls the corresponding absolute integrals.

For a square pulse equal to one on $[0,T/2)$ and zero on $[T/2,T)$, repeated periodically, the result is $(1-e^{-sT/2})/[s(1-e^{-sT})]=1/[s(1+e^{-sT/2})]$ in that half-plane. At $s=0$ the defining improper integral generally fails; continuing the quotient algebraically does not enlarge the defining region of absolute convergence of a Laplace transform.

The Mellin transform is an integral transform suited to changes of scale: replacing $f(x)$ by $f(ax)$ multiplies its transform by $a^{-s}$ for $a>0$. For a locally integrable function $f:(0,\infty )\to \mathbb{C}$, definewhere the improper integral converges. The complex number $x^{s-1}=\exp ((s-1)\log x)$ uses the real natural logarithm because $x>0$. Its absolute value is $x^{\sigma -1}$, so absolute convergence depends on the real part $\sigma$ rather than the imaginary part $\tau$. The Mellin fundamental strip records the open range of such real parts.^[ref]

A familiar example is the exponential function $f(x)=e^{-x}$:This is exactly the Gamma integral. Near zero $e^{-x}$ tends to one, so absolute convergence requires $\sigma >0$; at infinity $x^r{e}^{-x}\to 0$ for every real $r$, so there is no upper boundary. The Gamma function has a meromorphic continuation with simple poles at $0,-1,-2,\dots$, but a finite continued value elsewhere in $\sigma \le 0$ does not make the defining improper integral converge.^[ref]

The Mellin transform of an interval indicator gives an elementary example with absolute convergence for every complex number $s$. The Mellin transform in logarithmic coordinates explains why changes of scale become translations, connecting the scaling property of the Mellin transform to the Fourier transform.

On the whole line, the homogeneous one-dimensional wave equation with constant wave speed $c>0$ isHere $u$ is displacement in the string interpretation; $f$ is initial displacement and $g$ is initial velocity. Its two directions of travel giveThis is D'Alembert's formula, including both Cauchy data. The countertravelling decomposition of the one-dimensional wave equation identifies $F(x-ct)$ as the right-moving travelling wave and $G(x+ct)$ as the left-moving travelling wave. Their sum is permitted by the superposition principle. For $f\in C^2$ and $g\in C^1$, it is a classical solution. More generally, travelling waves with profiles $F$ and $G$ that are locally integrable functions solve the equation as distributions, with derivatives interpreted distributionally.^[ref]

The synthetic diagram uses the smooth bump function $b$ defined in D'Alembert's formula, with $f(x)=b(x)$ and $c=1$. Releasing this displacement from rest, $g=0$, gives $F=G=f/2$. Each emerging travelling wave has half the original peak. Launching the identical initial displacement solely to the right instead requires $g=-c{f}^{\prime }$ and gives $F=f$, $G=0$. At $t=1.5$, the released profiles have peaks $1/2$ at $x=\pm 1.5$, while the launched profile has peak $1$ at $x=1.5$. These are calculated model values. A purely left-moving launch requires $g=c{f}^{\prime }$. The initial velocity therefore selects the balance of directions; the initial shape alone does not.

On each interval $[k,k+1]$, $k\in \mathbb{Z}$, the function is $d(x)=x-k$ on $[k,k+\frac{1}{2}]$ and $d(x)=k+1-x$ on $[k+\frac{1}{2},k+1]$. Its ordinary derivative exists off $\frac{1}{2}\mathbb{Z}$ and is the locally integrable functionAssign arbitrary values at the endpoints. For a test function $\varphi$, split $-\int d{\varphi }^{\prime }$ at these corners and apply integration by parts on each piece. The adjacent boundary terms cancel because $d$ is continuous, and only finitely many pieces meet the compact support of $\varphi$. HenceThe distributional derivative is exactly the regular distribution of the ordinary derivative defined almost everywhere. There are no Dirac delta distributions in this first distributional derivative: corners change slope, while a jump of the function itself would create a point mass.

No. Consider the continuous even functionThe formulas agree at $|x|=e^{-1}$, and $f(x)\to 0$ at zero. Part (i) defines its even-weight principal-value distribution. However,If this were a regular distribution represented by $g\in L_{\mathrm{loc}}^1(\mathbb{R})$, testing on either side of zero would force $g=f/x$ almost everywhere there, by uniqueness of the locally integrable function representing a regular distribution. Changing the value at zero cannot remove the displayed divergence, contradicting local integrability of $g$.

More generally, the exact condition isNecessity is the preceding argument; sufficiency follows because $f/x$ is then a locally integrable function and its truncated integrals converge absolutely. The value $f(0)=0$ alone provides no such rate of decay.

If $F$ is any other fundamental solution, then $H=F-E$ satisfies ${\partial }_{\bar{z}}H=0$ as a distribution. Applying $4{\partial }_z$ gives $\Delta H=0$. Elliptic regularity for the Laplacian implies that the initially arbitrary distribution $H$ is a smooth function. Its equation is then the ordinary Cauchy-Riemann equation, so $H$ is an entire function.

For a direct justification of the needed regularity, choose a radial mollifier ${\eta }_r$. Locally, $H\ast {\eta }_{\epsilon }$ is a smooth harmonic function. On a ball with sufficient room for both radii, the mean value property gives $(H\ast {\eta }_{\epsilon })\ast {\eta }_r=H\ast {\eta }_{\epsilon }$. Passing to the weak convergence of distributions limit $\epsilon \to 0$ yields $H\ast {\eta }_r=H$ on a smaller ball. The left side is smooth, proving the claim locally. This establishes the distributional Cauchy-Riemann lemma without assuming in advance that $H$ is a locally integrable function.

Conversely every entire function $h$ satisfies ${\partial }_{\bar{z}}h=0$ and can be added to $E$. Hence all fundamental solutions areNo growth condition at infinity was imposed, so $h$ need not be a polynomial.

Write $z=x+iy$, $r=|z|$, and $L=(4\pi {)}^{-1}\log (x^2+y^2)$. The Wirtinger derivatives satisfy $4{\partial }_{\bar{z}}{\partial }_z=\Delta$. Away from zero, ordinary differentiation givesThis equality also holds for distributional derivatives. To check that no Dirac delta distribution was lost at zero, integrate by parts outside the disk $r\le \epsilon$ against a test function. The new boundary term has absolute value at most $C\epsilon |\log \epsilon |\Vert \varphi {\Vert }_{\infty }\to 0$. Both $L$ and its displayed first derivatives, of size $O(r^{-1})$, are locally integrable functions in two dimensions, so the remaining integrals converge to their full-plane values.

Using the supplied fundamental solution of the Laplace equation identity $\Delta L={\delta }_0$, and commutation of distributional derivatives, now givesThus the Cauchy kernel fundamental solution has exactly the required normalization.

Write $U=\hat{u}$, $V=\hat{v}$, and $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$. By the Sobolev space hypotheses, $A=⟨\xi {⟩}^{s}U$ and $B=⟨\xi {⟩}^{t}V$ belong to the L2 space. Their product is an integrable function by Cauchy-Schwarz inequality. Thus $UV=⟨\xi {⟩}^{-s-t}AB$ is a locally integrable function defining a tempered distribution, with estimateThe supremum is controlled by finitely many Schwartz seminorms. Define the convolution of Sobolev distributions byIt is independent of the representatives chosen for $U,V$, and the convolution theorem shows that it agrees with ordinary convolution for Schwartz functions. More generally, approximate $u,v$ by Schwartz functions in their respective Sobolev norms; the displayed estimate shows convergence of the products in tempered distributions, so their ordinary convolutions converge to this definition. For $u,v\in L^1$ that also satisfy the stated Sobolev space hypotheses, the ordinary convolution theorem gives the same result.

There need not be any real $r$ with $u\ast v\in H^r(\mathbb{R})$. ChooseHere $1_{(0,1)}$ is an indicator function. Since ${\int }_0^1{\xi }^{-2/3}d\xi =3$ and every weight $(1+{\xi }^2{)}^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every Sobolev space. But their convolution has Fourier transform $UV={\xi }^{-2/3}{1}_{(0,1)}$, and for every real $r$,Thus $\boxed{u,v\in \bigcap _{q\in \mathbb{R}}{H}^q(\mathbb{R})\text{but}u\ast v\notin \bigcup _{r\in \mathbb{R}}{H}^r(\mathbb{R})}$. The obstruction is failure of local square integrability at zero; a weight controlling large frequencies cannot fix it.

A locally integrable function $u$ has weak derivative $g_i$ in direction $i$ when $g_i$ is locally integrable andThe test functions have compact support inside the domain, so no boundary values enter this definition. Such $g_i$ is unique almost everywhere and is denoted ${\partial }_{i}u$.

For $1\le p<\infty$, the first-order Sobolev space consists of the Lp space classes whose first weak derivatives are also in that Lp space:This Sobolev norm is equivalent to $\Vert u{\Vert }_p+\Vert \nabla u{\Vert }_p$; either convention is acceptable. When $p=2$, this is the Hilbert space $H^1(B_1)$.

For a real locally integrable function $f$, the distributional definition of a subharmonic function is $\Delta f\ge 0$: for every nonnegative $\phi \in C_c^{\infty }(\Omega )$,A superharmonic function has the opposite inequality, equivalently $-f$ is subharmonic. A harmonic function has $\Delta f=0$ distributionally, equivalently equality for every test function. These definitions concern almost-everywhere classes; the given upper semicontinuous representative supplies the pointwise version for subharmonic functions.

Use half-open intervals and setThe Haar scaling functions and Haar wavelets areFor fixed $j$, each scaling function has squared integral $2^j{2}^{-j}=1$, and different $k$ have disjoint supports, proving that this family is orthonormal. Each wavelet also has norm one and integral zero. At a fixed level, distinct wavelets have disjoint supports. At different levels, their dyadic supports are either disjoint or the finer support lies in a half on which the coarser wavelet is constant. The finer wavelet has integral zero, so their inner product vanishes. Finally, a wavelet with $l\ge 0$ is supported inside a unit interval; it is orthogonal to its containing ${\phi }_{0,k}$ because its integral is zero, and to all other unit scaling functions because their supports are disjoint. Thus both the scaling family at each fixed level and the complete stated Haar family are orthonormal sets.

The Haar refinement identity isIt shows that the closed scaling spaces $V_j$ satisfy $V_{j+1}=V_j\oplus W_j$, where $W_j$ is the closed span of the level-$j$ wavelets. The union of the $V_j$ is dense in $L^2(\mathbb{R})$: continuous compactly supported functions can be approximated in $L^2$ by their dyadic cell averages, and such continuous functions are dense in $L^2$. Hence the orthonormal Haar family is also a complete orthonormal basis.

For a locally integrable function, define the compact-support coefficient integralsThese exist without requiring $f\in L^2$. The two formulas for the Haar approximation areFor $j=0$ the wavelet sum is empty. For every fixed $j$, these sums are locally finite, so they make sense pointwise and locally in $L^1$. The refinement identities and their orthogonal two-by-two coefficient transformation establish equality by induction, even for merely locally integrable $f$. If $f\in L^2$, this is also the orthogonal projection onto $V_j$.

In particular, for $\delta =2^{-j}$ and $x\in I_k=[k\delta ,(k+1)\delta )$,Now impose the symmetry and monotonicity conditions. The limit zero and decrease on $[0,\infty )$ imply $0\le f(x)\le f(0)$ for $x\ge 0$. For $k\ge 0$, the cell average and every value inside $I_k$ lie between $f((k+1)\delta )$ and $f(k\delta )$. ThereforeSumming over the positive half-line telescopes and gives a bound $\delta f(0)$. Reflection maps each dyadic cell to another dyadic cell up to endpoints. Since $f$ is even, its Haar approximation is even almost everywhere as well, so the negative half-line has the same error. ConsequentlyThe argument proves integrability of the difference, even if $f$ itself is not integrable on the whole line. It is the symmetric monotone case of the Haar approximation error for a function of bounded variation.

The shear $(t,x,v)\mapsto (t,x+tv,v)$ is a smooth diffeomorphism with determinant one and sends compact sets to compact sets. Thus $F,G$ are locally integrable functions. On a compact time interval $I$ and compact label set $K\subset {\mathbb{R}}^6$, the integral ${\int }_K{\int }_I|G|dtdxdv$ is finite. The Fubini theorem, followed by a countable exhaustion of time intervals and label sets, shows that $G(\cdot ,x,v)\in L_{\mathrm{loc}}^1(\mathbb{R})$ for almost every $(x,v)$.

To transform the weak equation, choose $\varphi (t,x,v)=\rho (x-tv,v)\psi (t)$, where $\rho \in C_c^{\infty }({\mathbb{R}}^6)$ and $\psi \in C_c^{\infty }(\mathbb{R})$. Including a velocity cutoff in $\rho$ ensures the required compact support. The transport derivative of this test function is $\rho (x-tv,v){\psi }^{\prime }(t)$. Change variables to the characteristic labels to obtainThe fundamental test-function identity makes the bracket zero almost everywhere. Choose a countable dense family of time test functions on each bounded interval and extend by continuity; there is then a single negligible label set outside which ${\partial }_{t}F=G$ as one-dimensional distributions.

For each remaining label, subtract the primitive $H(t)={\int }_0^{t}G(s)ds$. The distributional derivative of $F-H$ is zero, so it is a constant almost everywhere. Choose the resulting absolutely continuous representative along free characteristics, $\tilde{F}(t)=C+H(t)$. The fundamental theorem of calculus givesThe representative qualification is essential: an arbitrary locally integrable representative can be modified on one time slice without changing the distributional equation, and would not satisfy the identity for every pair of times.