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Choose a star centre $a$ of the open star-shaped set $D$ and define the candidate antiderivative by a straight-segment contour integral:For each fixed $z\in D$ and all sufficiently small $h$, the filled triangle with vertices $a,z,z+h$ lies in $D$. Indeed, the compact segment $[a,z]$ has a neighbourhood of some positive radius contained in the open set $D$, and every point of this triangle lies within $|h|$ of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to $f(z)$ as $h\to 0$. HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.

On a general domain, the conclusion can fail. Take $D=\mathbb{C}\setminus \{0\}$ and $f(z)=1/z$. Every filled triangle contained in $D$ has zero boundary integral by Cauchy integral theorem, but the unit circle has integral $2\pi i$. An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.

Conversely, suppose $F$ is an entire function with the displayed bound. Its real restriction has polynomial growth, so its inverse Fourier transform defines a tempered distribution byRapid decay of the Schwartz function $\hat{\phi }$ makes this integral absolutely convergent and continuous in the Schwartz space topology.

To determine the support of a distribution, take a test function supported in a half-space $x\cdot \omega \ge H_K(\omega )+\delta$, where $|\omega |=1$ and $\delta >0$. The contour-shift proof of the Paley–Wiener–Schwartz theorem gives, for each $t\ge 0$,Here is the estimate that justifies both the shift and its limiting use. Since $\hat{\phi }(-\xi -it\omega )=\int e^{ix\cdot \xi }{e}^{-tx\cdot \omega }\phi (x)dx$, repeated integration by parts with $1-{\Delta }_x$ yieldsFor fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the Cauchy integral theorem on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega )=t{H}_K(\omega )$ now bounds the pairing bywhich tends to zero. A separating hyperplane exists at every point outside the compact convex set $K$. A finite partition of unity on the support of any exterior test function reduces it to such half-spaces, so $\mathrm{supp}u\subseteq K$. Fourier inversion gives uniqueness; the entire transform of this $u$ agrees with $F$ on ${\mathbb{R}}^n$ and hence everywhere, by applying the one-variable identity theorem successively in each coordinate. This completes both directions.

Now apply this theorem to compact support solvability for a constant-coefficient ordinary differential equation. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise writeThe solution space of $P(-D)\phi =0$ isIndeed these exponentials solve the equation, their initial derivative vectors have a nonzero Vandermonde determinant, and uniqueness for the order-$m$ ordinary differential equation makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:

For sufficiency, choose $R$ with $\mathrm{supp}v\subseteq [-R,R]$ and let $V(\zeta )=⟨v,e^{-ix\zeta }⟩$. The assumed annihilation is precisely $V({\alpha }_j)=0$ for every $j$. Since all roots of a polynomial are simple, $U=V/P$ has removable singularities at every root and is an entire function. Outside one disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^m$. Inside that disk, the extended $U$ is bounded. The polynomial division preservation of exponential type therefore givesThe Paley–Wiener–Schwartz theorem produces a compactly supported distribution $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so Fourier inversion gives $P(D)u=v$. ThusThe compactly supported solution is unique: $P\hat{u}=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha )=0$, or pairing with $x^k{e}^{-i\alpha x}$, up to one less than each multiplicity.

Again use $D_j=-i{\partial }_j$ and distribution dual pairings with complex bilinearity. The Malgrange–Ehrenpreis theorem asserts that every nonzero constant-coefficient differential operator defined by a polynomial has a fundamental solution of a linear differential operator:If $P=a\ne 0$ is constant, take $E={\delta }_0/a$. For positive degree $m$, write $P_m$ for the highest homogeneous part. A nonzero polynomial cannot vanish on all of ${\mathbb{R}}^n$, so choose a real unit vector $e$ with $P_m(e)\ne 0$. A change of coordinates by an orthogonal matrix makes it the last coordinate direction. In these coordinates,Crucially the leading coefficient $a$ is constant and never vanishes as ${\xi }^{\prime }\in {\mathbb{R}}^{n-1}$ varies.

The finite-height polynomial root avoidance construction gives the needed Hörmander staircase. Set $h_{\ell }=3\ell$ for $0\le \ell \le m$. For each fixed ${\xi }^{\prime }$, the fundamental theorem of algebra supplies $m$ roots $z_1,\dots ,z_m$, with multiplicity. A root's imaginary part can be within distance strictly less than one of at most one candidate height, since those heights are separated by three. The pigeonhole principle therefore leaves at least one height with $|h_{\ell }-\mathrm{Im}{z}_j|\ge 1$ for all roots. At that height, for every real $s$,No continuous labelling of the roots is required. To select the heights measurably, defineEach $B_{\ell }$ is closed, since the coefficients are continuous in ${\xi }^{\prime }$. Continuity in $s$ makes the rational test equivalent to the bound for all $s\in \mathbb{R}$. The $A_{\ell }$ are therefore disjoint Borel sets covering ${\mathbb{R}}^{n-1}$. The union of horizontal fibresis a Hörmander staircase, with bounded imaginary height and a uniform nonzero denominator. For $n=1$, the transverse space has one point, and this is simply the choice of one horizontal line.

For example, $P({\xi }_1,z)={\xi }_1^2+z^2$ has roots $z=\pm i{\xi }_1$. One may select $h=0$ for $|{\xi }_1|\ge 1$, and $h=3$ for $|{\xi }_1|<1$. Then the first region has $|P|\ge 1$, while in the second each root is at least two units from the chosen height. The figure shows the height projection of these horizontal fibres, rather than additional connecting contours.

Define a linear functional on test functions byTo establish that this is a distribution, fix a compact set containing the support of $\phi$, inside a radius-$R$ ball. The Fourier decay in a bounded complex strip estimate follows by applying $(1-\Delta {)}^L$ to $e^{-h_{\ell }{x}_n}\phi (x)$:The finite bound on the heights absorbs all exponential factors and powers of $h_{\ell }$ into $C_{R,L}$. Taking $2L>n$ and using $|P|\ge |a|$ proves absolute convergence and the required finite-order continuity on each fixed support set. Hence $E\in {\mathcal{D}}^{\prime }$.

Now the formal transpose identity and the Fourier derivative identity giveThe polynomial denominator cancels in the pairing for $P(D)E$. For each fixed real ${\xi }^{\prime }$, the remaining function of $z$ is entire and rapidly decreasing in its real part throughout $0\le \mathrm{Im}z\le 3m$. The Cauchy integral theorem therefore allows each horizontal fibre to be shifted to the real axis; its truncated vertical end integrals tend to zero. Absolute convergence justifies Fubini's theorem and the finite sum over the measurable partition. ThusThe last equality is Fourier inversion. Transforming back from the orthogonal coordinates preserves ${\delta }_0$ and gives the required fundamental solution of a linear differential operator for the original operator. The staircase is allowed to jump: after cancellation the shift is performed on each one-dimensional fibre, so no assertion that $\Sigma$ is a smooth contour is used. This construction proves a distributional fundamental solution, without claiming a tempered-growth estimate.

For $f\in \mathcal{D}({\mathbb{R}}^n)$, use the smoothing convolution with a test functionIndeed $u_0(x)=⟨E_y,f(x-y)⟩$ is a smooth function, with ${\partial }^{\alpha }{u}_0(x)=⟨E_y,{\partial }^{\alpha }f(x-y)⟩$. On each compact set of $x$ values, all translated test functions have supports in one fixed compact set, so differentiation is justified by the continuity of the distribution. Compact support of $f$ makes the convolution well defined even when $E$ is not tempered. It does not imply compact support of $u_0$.

Finally the affine solution space of a linear equation isIf every homogeneous solution is represented by a smooth function, so is every $u=u_0+v$. Conversely, if every solution for this fixed $f$ is represented by a smooth function, then for any homogeneous solution $v$, the solution $u_0+v$ is smooth, and subtracting the smooth $u_0$ shows that $v$ is smooth. Hence All inhomogeneous solutions are smooth functions exactly when all homogeneous solutions are smooth functions. This is a global statement on ${\mathbb{R}}^n$ for the stated data; no additional local regularity theorem is being assumed.

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

On the slit plane $\mathbb{C}\setminus [-a,a]$, chooseso that $F(z)\sim z$ as $z\to \infty$. This branch cut gives, for $-a<x<a$, the boundary values

Integrate $F$ over the boundary of a large disk with a thin slit removed. The upper bank is traversed from $-a$ to $a$ and the lower bank from $a$ to $-a$, so together they contributeAt infinity the Laurent series isand hence the counterclockwise large-circle integral tends to $-\pi i{a}^2$. By the Cauchy integral theorem, the large-circle and slit-boundary contributions sum to zero. Thereforewhich yields

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

Putand take the branch cut $[-1,1]$. On $\mathbb{C}\setminus [-1,1]$, choose the analytic branch of a square root whose upper boundary value isIts lower boundary value is $s(x-i0)=-\sqrt{1-x^2}$. This fixes the integrand to be $+1$ at $0^{+}$ as required.

For $0\le \arg z\le \pi$, start at $0^{+}$ and use any path to $z$ lying above the cut; path deformations within that region do not alter the integral by the Cauchy integral theorem. For $\pi <\arg z<2\pi$, start in the same way, continue counterclockwise around the branch point $-1$ from the upper bank to the lower bank, and then continue to $z$ through the lower half-plane. This second prescription is the continuation obtained as $argz$ increases through $\pi$.

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

If $\mathrm{Im}z>0$, the same upper-half-plane contour calculation givesIf $\mathrm{Im}z<0$, the pole lies below the real integration contour. Closing upward encloses no pole, so Jordan lemma and the Cauchy integral theorem giveHence at a real point $x$ the upper and lower boundary values satisfyThus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.

An analytic continuation must be holomorphic, hence continuous, across every point where it is defined. The undeformed real-axis formula for $G$ has the nonzero jump above, while the continuation constructed in part (c) retains $2\pi i^n{e}^{iz}/(n-1)!$ below the axis. Therefore $G$ cannot be the analytic continuation of $F$.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear partial differential operator $P(D)$ on ${\mathbb{R}}^n$ has a fundamental solution of a linear differential operator: there is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ such that $P(D)E={\delta }_0$.

Write $D=-i\partial$. After an orthogonal change of coordinates and multiplication by a nonzero constant, its polynomial symbol may be written as a monic polynomial in the last frequency,For each real ${\mu }^{\prime }$, this polynomial has $M$ complex roots counted with multiplicity. Among a fixed finite collection of horizontal lines at bounded heights, one can choose a line that stays a positive distance from all those roots. Continuity of the roots preserves the choice on a neighborhood $N({\mu }^{\prime })$. Take a countable locally finite cover by such neighborhoods, refine it to a measurable disjoint partition ${\mathbb{R}}^{n-1}={⨆}_j{\Delta }_j$, and let $c_j$ be the chosen height on ${\Delta }_j$. The resulting Hörmander staircasehas bounded heights and may be chosen so that $|P({\xi }^{\prime },s+i{c}_j)|\ge 1$ on each step.

For a test function $\phi$, defineThe Paley–Wiener–Schwartz theorem gives rapid decay in the real frequency directions and at most a fixed exponential factor in the bounded imaginary direction. Together with $|P|\ge 1$, this proves that the integral defines a continuous distribution. Applying $P(D)$ cancels the denominator. The remaining integrand is entire in $z$, so the Cauchy integral theorem shifts every horizontal contour to the real axis; the partition then recombines into ${\mathbb{R}}^{n-1}$. The Fourier inversion theorem giveswhich proves the theorem.

Statement (iii) is false. Let $\Omega$ be a thin open annulus about the unit circle, let $f(z)=1/z$, and split the circle into two proper closed arcs $K_1,K_2$ whose union is the whole circle. Each arc has connected complement, so polynomial approximation of the reciprocal on a proper circular arc gives uniform polynomial approximations to $f$ on each $K_j$ separately.

If polynomials $P_n$ converged uniformly to $1/z$ on $K_1\cup K_2$, then uniform convergence on the unit circle and the Cauchy integral theorem would givea contradiction.

Write the phase and amplitude asThe saddle point relevant to the contour is $z=0$, since ${\varphi }^{\prime }(z)=\sinh z$ andThe local steepest-descent direction is therefore vertical: with $z=iy$,

The given contour approaches the lines $\mathrm{Im}z=-\pi$ at its left end and $\mathrm{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm 4i$, outside the closed strip $|\mathrm{Im}z|\le \pi$. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty -i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty +i\pi$. On either horizontal ray,so those two integrals are exponentially small, of order $e^{-x}{x}^{-1/2}$.

The vertical segment, oriented upward, contributesIts real phase has a unique maximum at $y=0$. Sincethe simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, givesThe factor $i$ comes from the upward tangent $dz=idy$ of the deformed contour.

The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain, then its integral around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square givesAfter the change of variable $z=x-iq/(2a)$, the integral runs along the horizontal line $\mathrm{Im}z=-q/(2a)$. The integrand $e^{-a{z}^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical integrals tend to zero as their real parts tend to $\pm \infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the Fourier transform of a Gaussian gives

Every holomorphic function $f$ on a simply connected domain has a single-valued antiderivative. Fixing $z_0$ and settinggives a path-independent function with $F^{\prime }=f$, because the Cauchy integral theorem makes the integral around every closed path zero.

If $P(k)$ is analytic in the upper half-plane with suitable growth, paired integrals along the positive real and imaginary rays cancel when their exponential is $e^{ik(x+iy)}$ with $x,y>0$. For $0<y<\ell$, the analogous pair along the imaginary and negative real rays cancels with $e^{ik(x+i(y-\ell ))}$. These two applications of the Cauchy integral theorem eliminate reflected unknown transforms from a strip global relation; an individual ray integral need not vanish.