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A subset $B$ of a topological vector space is bounded when every neighborhood $U$ of zero absorbs it: $B\subseteq tU$ for all sufficiently large positive $t$. In a locally convex space defined by seminorms $p_{\lambda }$, this is equivalent to ${\sup }_{x\in B}{p}_{\lambda }(x)<\infty$ for every $\lambda$. In particular, a bounded subset of the Schwartz space has a uniform bound for every Schwartz seminorm; it need not have uniformly bounded supports.

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]

For $T\in {\mathcal{S}}^{\prime }$ and $\psi \in \mathcal{S}$, define $(T\ast \psi )(x)=⟨T_y,\psi (x-y)⟩$. Translations are smooth in the Schwartz space, so every derivative is ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T,{\partial }^{\alpha }\psi (x-\cdot )⟩$. The finite-seminorm continuity estimate for $T$ implies $|{\partial }^{\alpha }(T\ast \psi )(x)|\le C_{\alpha }(1+|x|{)}^m$ for one $m$. Thus the result is a smooth function defining a tempered distribution, but it need not be a Schwartz function: $1\ast \psi =1$ when $\int \psi =1$. If both factors are radial, testing the rotation action on the translated $\psi$ shows that the convolution is radial.

For the Fourier transform convention $\mathcal{F}f(\omega )={\int }_{\mathbb{R}}f(t)e^{-i\omega t}dt$, Fourier inversion at $-\omega$ gives $f(-\omega )=(2\pi {)}^{-1}{\int }_{\mathbb{R}}(\mathcal{F}f)(t)e^{-i\omega t}dt$. Thus transforming twice reflects the argument and multiplies by $2\pi$. The identity holds directly on the Schwartz space, and extends to the L2 space and to tempered distributions. It lets an already known transform pair be used in reverse without another integration.

For the angular-frequency Fourier transform and Schwartz seminorms $p_{m,n}(\varphi )={\sup }_x|x^m{\varphi }^{(n)}(x)|$, every output seminorm is bounded by finitely many input seminorms. To prove this, differentiation under the integral sign and integration by parts giveThe generalized Leibniz rule, $\Vert \hat{g}{\Vert }_{\infty }\le \Vert g{\Vert }_1$, and $\Vert x^a{\varphi }^{(b)}{\Vert }_1\le \pi (p_{a,b}+p_{a+2,b})$ implyThis proves continuity of the Fourier transform on the Schwartz space and rapid decay of every transformed derivative.

The Fourier transform with negative exponential and inverse factor $(2\pi {)}^{-n}$ is a continuous isomorphism of the Schwartz space. Weighted frequency derivatives are transforms of derivatives of weighted input functions, so their suprema are controlled by finitely many Schwartz seminorms. Fourier inversion gives ${\mathcal{F}}^2=(2\pi {)}^n\mathcal{R}$, where $\mathcal{R}f(x)=f(-x)$. Transposition gives a continuous isomorphism on tempered distributions as well.

With $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, the Fourier transform of a tempered distribution is defined byContinuity of the Fourier transform on the Schwartz space makes this well defined. For regular integrable functions, Fubini's theorem shows that it agrees with the ordinary transform. The inverse convention has factor $(2\pi {)}^{-n}$ and the positive exponential.

For $\mathcal{H}f(x)={\pi }^{-1}\mathrm{pv}\int f(y)/(x-y)dy$ and $\hat{f}(\xi )=\int e^{-ix\xi }f(x)dx$, the Fourier transform multiplier is $-i\mathrm{sgn}\xi$. The Heaviside function transform and the double-transform reflection identity give the sign. Plancherel theorem proves the $L^2$ isometry, and multiplication by $i\xi$ proves commutation with differentiation. The map need not preserve the Schwartz space.

A tempered distribution is a continuous linear functional $u:\mathcal{S}(\mathbb{R})\to \mathbb{C}$. Equivalently, for some integer $N$ and constant $C$,The Fourier transform of a tempered distribution is defined byThe Fourier bounds for Schwartz functions make the right side a continuous linear functional on the Schwartz space, so $\hat{u}$ is again a tempered distribution.

For an integrable function $f$, the regular distribution $u_f$ acts by $⟨u_f,\varphi ⟩=\int f\varphi$. The estimate $|⟨u_f,\varphi ⟩|\le \Vert f{\Vert }_1{p}_{0,0}(\varphi )$ proves it is a tempered distribution. Since $\iint |f(x)\varphi (\xi )|dxd\xi <\infty$, Fubini's theorem givesThus the distributional Fourier transform agrees with the ordinary Fourier transform.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, with complex-linear distribution pairings throughout. The Riemann-Lebesgue lemma says that, for an integrable function $f\in L^1(\mathbb{R})$, its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. It also satisfies $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ belongs to $L^1(\mathbb{R})$ by Tonelli's theorem, and Fubini's theorem proves the displayed identity. The companion Fourier product formula, for Schwartz functions, is $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

Set $\tilde{\varphi }(x)=\overline{\varphi (-x)}$. This is a Schwartz function and its Fourier transform is $\hat{\tilde{\varphi }}(\xi )=\overline{\hat{\varphi }(\xi )}$. The convolution theorem givesThe convolution of two Schwartz functions is again a Schwartz function: differentiating under the integral and expanding $x^m=(y+(x-y){)}^m$ bounds every weighted derivative by products of integrable and bounded weighted derivatives. Evaluate Fourier inversion for this convolution at zero:The Schwartz space is dense in the L2 space. For $f\in L^2(\mathbb{R})$ choose Schwartz functions ${\varphi }_j\to f$ in the L2 norm. The identity makes ${\hat{\varphi }}_j$ a Cauchy sequence in the L2 space, and its limit defines $\mathcal{F}f$, independently of the approximation. It agrees with the distributional Fourier transform, because Cauchy-Schwarz inequality makes convergence in the L2 norm imply convergence when paired with every Schwartz function. For $f\in L^1\cap L^2$ it therefore agrees almost everywhere with the ordinary Fourier transform.

The extension is onto: its image is closed by the norm identity and contains the Schwartz space, since Fourier inversion maps that space to itself. This proves Plancherel theorem:Its inverse is the extension of the inverse Fourier transform, with limits understood in the L2 norm.

Let $R\varphi (x)=\varphi (-x)$, and define reflection of a tempered distribution by $⟨Ru,\varphi ⟩=⟨u,R\varphi ⟩$. The Fourier inversion theorem on the Schwartz space says $\hat{\hat{\varphi }}=2\pi R\varphi$. ThusReflection commutes with the Fourier transform, since $\widehat{R\varphi }=R\hat{\varphi }$ and the same relation follows by duality for tempered distributions. Define the inverse Fourier transform on tempered distributions byIt follows thatIn particular $u=(2\pi {)}^{-1}R\hat{\hat{u}}$ is the general Fourier inversion formula. When $\hat{u}$ is an integrable function, this is the regular distribution of $(2\pi {)}^{-1}\int \hat{u}(\xi )e^{ix\xi }d\xi$; an arbitrary tempered distribution need not have an ordinary integral representation.

A distribution $u$ is a $2\pi$ periodic distribution whenIteration gives invariance under all integer multiples of $2\pi$. For a tempered distribution the identity also holds for every Schwartz function, since test functions are dense in the Schwartz space: multiplying by cutoff functions $\chi (x/R)$ equal to one near zero gives such an approximation in every Schwartz seminorm.

For $\varphi \in \mathcal{S}(\mathbb{R})$ define $a_k(x)=\Psi (x+2\pi k)\varphi (x)$. The smooth periodic partition of unity gives $\varphi ={\sum }_k{a}_k$ with convergence in the Schwartz space. Indeed, each $a_k$ is supported where $|x+2\pi k|\le 2\pi$, and the generalized Leibniz rule and arbitrarily rapid decay of the derivatives of $\varphi$ give, for every $m,n,M$,Consequently continuity and periodicity of the tempered distribution giveFor the last interchange, all summands have the fixed compact support of $\Psi$ and their derivatives decrease faster than any power of $|k|$. Hence the series converges in the space of test functions, as well as in the Schwartz space. Notice that $P\varphi$ itself need not be a Schwartz function; the factor $\Psi$ makes the pairing legitimate.

Use $\hat{f}(\lambda )=(2\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-i\lambda x}f(x)dx$. The transform statement requires sufficient decay and regularity for the transform and its derivatives; functions in the Schwartz space suffice, and all the functions constructed below belong to that space. Integration by parts twice and differentiation under the integral giveTransforming the differential equation therefore givesThe differential equation by itself does not ensure that an ordinary Fourier transform exists: $f=e^{x^2/2}$ solves it with $\mu =1$ but grows too rapidly. Thus the usual transformability assumptions are implicit in this part.

For definiteness take $f$ in the Schwartz space; sufficiently smooth compactly supported data are also adequate. These hypotheses make the line integrals, Cauchy principal values and decay estimates below well defined. Write $\lambda =r{e}^{i\theta }$ with $r>0$, $r\ne 1$, and introduce orthonormal local coordinatesLet $F(\tau ,\rho ,\theta )=f(x_1,x_2)$, $a_r=(r+r^{-1})/2$ and $b_r=(r-r^{-1})/2$. Substitution into the directional transport equation givesNow make the invertible real-linear change of coordinatesSince $\mathrm{Im}z=-a_r\rho /b_r$, its Wirtinger derivative is ${\partial }_{\bar{z}}=({\partial }_{\tau }-i{b}_r{\partial }_{\rho }/a_r)/2$. Hence $L_{\lambda }=\nu (r){\partial }_{\bar{z}}$, as required. The change becomes singular at $r=1$, where the complex operator degenerates to a real directional derivative. Also the original equation contains ${\lambda }^{-1}$, so $\lambda =0$ is excluded from the differential equation; the normalized solution will extend analytically there.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

Use $e(t)=e^{2\pi it}$ and the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e(-x\xi )dx$, initially for an integrable function. For the interval indicator function, direct integration givesThe integral bound is one, and the displayed numerator has modulus at most two. By the convolution theorem, the Fourier transform of the $m$-fold convolution is its $m$th power. Thus the Fourier decay of repeated interval convolutions givesAt zero, interpret the second term of the minimum as infinity.

The Schwartz space $\mathcal{S}(\mathbb{R})$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^a{f}^{(b)}(x)|$ is finite, for nonnegative integers $a,b$. Differentiation under the integral sign and integration by parts giveAll differentiated integrands are integrable, and their boundary terms vanish because of rapid decay. The last transform is bounded by its integrand's $L^1$ norm. The same argument applies at every derivative order, so $\hat{f}\in \mathcal{S}(\mathbb{R})$.

For $f\in \mathcal{S}(\mathbb{R})$, the Poisson summation formula isTo prove it, form the periodization of a Schwartz function $F(x)={\sum }_{n}f(x+n)$. Rapid decay gives uniform convergence of this series and of every differentiated series on $[0,1]$, so $F$ is a smooth periodic function. Its Fourier coefficients areThe series ${\sum }_k\hat{f}(k)e(kx)$ converges absolutely and uniformly. Its Fourier coefficients equal those of $F$. The uniqueness theorem for Fourier series says that two integrable functions with the same coefficients agree almost everywhere; as both functions here are continuous, they agree everywhere. Evaluating at zero proves Poisson summation.

The Gaussian Fourier transform of $e^{-\pi t{x}^2}$, $t>0$, is $t^{-1/2}{e}^{-\pi {\xi }^2/t}$. Apply Poisson summation to obtain the transformation of the Jacobi theta functionFor $\Re s>1$, termwise integration in the Mellin transform gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $t=1/u$ in the lower integral. The theta transformation givesThe integral is an entire function of $s$, since $\vartheta (t)-1$ decays exponentially. This pole-subtracted theta integral for the completed zeta function supplies a meromorphic continuation with simple poles at zero and one. Both the rational term and the integral are unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionEquivalently the Riemann xi function $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication identities for the Gamma function also give $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$ by meromorphic continuation.

Use the Fourier convention $\hat{f}(\xi )={\int }_{{\mathbb{R}}^3}{e}^{-ix\cdot \xi }f(x)dx$, with inverse $(2\pi {)}^{-3}\int e^{ix\cdot \xi }\hat{f}(\xi )d\xi$. For a tempered distribution $T$, define its Fourier transform by duality:The Fourier transform maps the Schwartz space continuously onto itself, making the right side a continuous linear functional there. The use of the same negative exponential in the test-function transform is consistent with bilinear distributional pairing; other unitary or sign conventions change the normalization accordingly.

The heat kernel is $H_t(x)=(4\pi t{)}^{-1/2}{e}^{-x^2/(4t)}$, so the homogeneous solution for smooth bounded data is $u(t,x)=(H_t\ast g)(x)$. Differentiating under the integral gives the heat equation, and the approximate-identity property gives the initial value.

For a tempered distribution, its derivative is defined by $⟨T^{\prime },\psi ⟩=-⟨T,{\psi }^{\prime }⟩$ for every Schwartz test function. A fundamental solution of a constant-coefficient operator $P$ is a distribution $E$ satisfying $PE={\delta }_0$. Put $E(x)=e^{-|x|}/2$. Away from zero, $E^{\prime \prime }=E$; its first derivative has jump $-1$ at zero. Distributional differentiation therefore gives $E^{\prime \prime }=E-{\delta }_0$, andFor an ansatz $u=e^{t}v$, the forced equation becomes $(1-D^2)v=\varphi$. With Fourier convention $\hat{v}(k)=\int e^{-ikx}v(x)dx$, its unique solution in the Schwartz space isThe smooth multiplier and its derivatives have polynomial bounds, so this quotient is Schwartz. For uniqueness, multiplication of the difference's Fourier transform by $1+k^2$ gives zero, forcing the difference to vanish.

Subtract this particular solution and evolve the remaining initial data by the heat kernel:For smooth bounded $f$, the second term remains uniformly bounded, so $e^{-t}u(t,\cdot )\to v$ uniformly. If additionally $f-v$ is integrable, its uniform norm is $O(t^{-1/2})$, so the remainder itself decays. General bounded data need not have a decaying heat remainder, and no such stronger claim follows without a condition on the data. The homogeneous uniqueness assertion is understood in the bounded or usual tempered-growth solution class; the uniqueness of the particular Schwartz ansatz is unconditional within that stated class.

Choose the Fourier transform conventionIt maps the Schwartz space to itself. For a tempered distribution $u$, define the transform by bilinear distributional duality:For an ordinary integrable function this agrees with the first formula by interchanging its integrals. The absence of complex conjugation is the distribution-pairing convention.