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For arbitrary real $s,t$, the convolution of $u\in H^s(\mathbb{R})$ and $v\in H^t(\mathbb{R})$ has a canonical definition as a tempered distribution. With $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$, write $A=⟨\xi {⟩}^s\hat{u}$ and $B=⟨\xi {⟩}^t\hat{v}$, both in the L2 space. Then $AB\in L^1$ by Cauchy-Schwarz inequality, andThus the product defines a tempered distribution, and the displayed inverse Fourier transform defines $u\ast v$. This extends ordinary convolution of Schwartz functions by the convolution theorem and density in the Sobolev norm.

The result need not belong to any Sobolev space, even when both inputs belong to every Sobolev space. For $\hat{u}=\hat{v}={\xi }^{-1/3}{1}_{(0,1)}(\xi )$, each weighted squared input is integrable, but the squared product is ${\xi }^{-4/3}{1}_{(0,1)}(\xi )$, which is not locally integrable at zero. Every weight $(1+{\xi }^2{)}^r$ is bounded above and below there, proving the failure for all real $r$.

For a $2\pi$ periodic distribution $u\in {\mathcal{S}}^{\prime }(\mathbb{R})$, choose a smooth periodic partition of unity $\Psi$, and define $c_n=(2\pi {)}^{-1}⟨u,\Psi e^{-inx}⟩$. A finite Schwartz seminorm bound on the fixed compact support gives $|c_n|\le C(1+|n|{)}^N$. For a Schwartz function $\varphi$, the Poisson summation formula and Fourier inversion imply $P\hat{\varphi }(x)={\sum }_n\varphi (-n)e^{inx}$. Localizing the pairing for $u$ givesproving the displayed Fourier transform formula. The series converges in tempered distributions because its coefficients have polynomial growth while $\varphi (n)$ decreases rapidly. Applying the inverse Fourier transform gives $u={\sum }_n{c}_n{e}^{inx}$ in tempered distributions. These are the smooth-window Fourier coefficients of a periodic distribution, and the transform formula shows independence of $\Psi$.

On the real line, let $T(x)=\max (1-|x|/2,0)$. The Fourier transform of a triangular function gives $\hat{T}(\omega )=2{\sin }^2\omega /{\omega }^2$, with value two at zero. The Gaussian heat kernel has transform $e^{-t{\omega }^2}$. Hence $u=q_t\ast T$ solves the heat equation with initial profile $T$, and the inverse Fourier transform gives the displayed integral. Differentiating for positive time is justified by Gaussian decay. The uniform continuity of $T$ and the vanishing Gaussian mass outside any fixed neighborhood of zero prove uniform recovery of $T$ as $t\downarrow 0$. Convolution preserves total integral two, keeps the solution nonnegative, and smooths the corners for every positive time. Unlike the finite-interval problem in heat flow from a triangular initial temperature, this evolution has no endpoints through which heat can escape.

Use the Fourier transform conventionThe absolute integrability of $v$ makes this integral exist for every real $\omega$, with $|\hat{v}(\omega )|\le \Vert v{\Vert }_1$. With this convention the inverse Fourier transform has factor $1/(2\pi )$; maintaining that normalization is essential in part (e).

Insert part (i) in the inverse Fourier transform. The factor two in $\hat{T}$ cancels half of the inverse normalization:The integrand has a removable singularity at zero. For $t>0$, the Gaussian controls every polynomial factor at infinity, so differentiation under the integral is valid. Differentiation in $t$ and twice in $x$ both multiply the integrand by $-{\omega }^2$, proving $u_t=u_{xx}$.

There is also a direct physical-space representation:Since $q_t$ has integral one, writeThe continuous compactly supported profile $T$ is uniformly continuous. Given $ϵ>0$, choose $\delta >0$ so its change is below $ϵ$ for $|y|<\delta$. That part of the integral is bounded by $ϵ$. The other part is bounded by $2\Vert T{\Vert }_{\infty }{\int }_{|y|\ge \delta }{q}_t(y)dy$, which tends to zero by the substitution $y=2\sqrt{t}z$ and the vanishing tails of the Gaussian integral. This proves $u(t,x)\to T(x)$ uniformly as $t\downarrow 0$ and verifies the initial condition.

Thus the formula supplies the full heat evolution of a triangular profile, including the corner points of the initial profile. Positive time immediately smooths those corners.

Use the Fourier transform convention $\hat{u}(k,t)={\int }_{\mathbb{R}}{e}^{-ikx}u(x,t)dx$. Then the equation and Dirac delta initial condition becomeso $\hat{u}=e^{-(k^2+1)t}$. The inverse Fourier transform givesComplete the square as $-t{k}^2+ikx=-t(k-ix/(2t){)}^2-x^2/(4t)$ and use the permitted complex-shift Gaussian integral. Thus the heat kernel with constant absorption isIts mass is $e^{-t}$, consistent with constant absorption, and it tends to the Dirac delta as a distribution as $t\downarrow 0$.

For $\epsilon >0$, insert a Gaussian function into the proposed inverse Fourier transform:The Fourier bounds for Schwartz functions show that $\hat{\varphi }$ is an integrable function, so dominated convergence givesOn the other hand, Fubini's theorem applies because $\Vert \varphi {\Vert }_1\int e^{-\epsilon {\xi }^2/2}d\xi <\infty$. The given Fourier transform of a Gaussian, rescaled by $\sqrt{\epsilon }$, yieldsThe last integrand is bounded by $\Vert \varphi {\Vert }_{\infty }{e}^{-z^2/2}/\sqrt{2\pi }$. Since $\varphi$ is continuous, another application of dominated convergence makes this tend to $\varphi (x)$. Equating the limits proves the Fourier inversion theorem:The Gaussian functions above form an approximate identity; no Fourier inversion was used in evaluating their integrals.

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.

Use the smooth periodic partition of unity $\Psi$ above and define the Fourier coefficients of the periodic distribution byThese are the smooth-window Fourier coefficients of a periodic distribution for period $2\pi$. The finite Schwartz seminorm bound for $u$, restricted to the fixed compact support of $\Psi$, gives $|c_n|\le C(1+|n|{)}^N$ for some $C,N$. Indeed, only finitely many derivatives of $\Psi e^{-inx}$ occur, and each contributes at most a fixed power of $|n|$.

For a Schwartz function $\varphi$, the preceding Fourier series formula and $\hat{\hat{\varphi }}(n)=2\pi \varphi (-n)$ yieldAll derivatives of this series converge uniformly. The localization formula in the previous answer therefore givesConsequently the Fourier transform of a periodic distribution isThe polynomial bound on $c_n$ and rapid decay of $\varphi (n)$ make every displayed pairing absolutely convergent. In particular there are no derivatives of Dirac delta distributions in this formula.

The inverse Fourier transform of ${\delta }_n$ is $(2\pi {)}^{-1}{e}^{inx}$. Applying continuous Fourier inversion on tempered distributions to the symmetric partial sums givesExplicitly, for every Schwartz function $\theta$, the pairing of the partial sum with $\theta$ is ${\sum }_{|n|\le M}{c}_n\hat{\theta }(-n)$, which converges absolutely to $⟨u,\theta ⟩$. The Fourier transform formula also proves that $c_n$ is independent of the chosen smooth periodic partition of unity. For a locally integrable $2\pi$ periodic function it reduces, by splitting the integral into periods, to $c_n=(2\pi {)}^{-1}{\int }_0^{2\pi }u(x)e^{-inx}dx$.

Use the conventionThese are the Fourier transform and inverse Fourier transform for suitably integrable data. Assuming the required ordinary transforms exist and the lower derivatives vanish at both ends, integration by parts gives $\widehat{g^{\prime }}=[g{e}^{-ikx}{]}_{-\infty }^{\infty }+ik\hat{g}=ik\hat{g}$. Repeating this argument yieldsThe decay assumptions are used to discard the boundary terms at each step.

For positive diffusivity $\alpha$, transforming the heat equation gives ${\partial }_t\hat{f}=-\alpha k^2\hat{f}$. Thus $\hat{f}(k,t)=e^{-\alpha k^{2}t}{\hat{f}}_0(k)$, where $f_0(x)=f(x,0)$. The convolution theorem and the Gaussian heat kernel giveFor $f_0(x)=e^{-x^2}$, the Fourier transform of a Gaussian is $\sqrt{\pi }{e}^{-k^2/4}$. Multiplying by the heat factor and inverting gives the Gaussian initial data for the heat equation solutionIt recovers the initial profile as $t\downarrow 0$ and preserves its spatial integral $\sqrt{\pi }$.

With the stated angular-frequency convention, the inverse Fourier transform is $f(x)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }\tilde{f}(k)e^{ikx}dk$. Direct integration gives ${\tilde{f}}_0(k)={\int }_0^{\infty }{e}^{-(\alpha +ik)x}dx=1/(\alpha +ik)$. For $n\ge 1$, integration by parts has zero boundary terms and gives ${\tilde{f}}_n=n{\tilde{f}}_{n-1}/(\alpha +ik)$. ThereforeThis also follows from the Fourier transform of a polynomial multiple, since multiplication by $x$ corresponds to $i{\partial }_k$.

To prove Parseval's identity, substitute the inverse transform of $g$ into $\int f^{\ast }g$. Under the assumed integrability, Fubini's theorem permits interchange, and the inner integral is $\int f(x{)}^{\ast }{e}^{ikx}dx=\tilde{f}(k{)}^{\ast }$. HenceApply this with $f=g=f_n$. The spatial Gamma integral is ${\int }_0^{\infty }{x}^{2n}{e}^{-2\alpha x}dx=(2n)!/(2\alpha {)}^{2n+1}$. The transformed squared modulus is $(n!{)}^2/({\alpha }^2+k^2{)}^{n+1}$, giving the scaled even rational Parseval integral

With the stated Fourier transform convention, the inverse Fourier transform and the deterministic autocorrelation of a square-integrable function areAssume initially sufficient decay and smoothness, for example a Schwartz function, to justify interchanging integrations and Fourier inversion. With $s=x+y$, the cross-correlation identity isEvaluating $h$ directly and by its inverse transform at zero proves the Parseval identity:Approximation extends the identity to L2 space, as the Plancherel theorem.

For the compact cosine profile, the Fourier transform of a truncated cosine follows from $\cos x=(e^{ix}+e^{-ix})/2$:At $k=\pm 1$ the apparent singularities are removable, with value $\pi /2$. The direct squared norm is ${\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx=\pi /2$. Hence Parseval identity and evenness givePutting $t=\pi k/2$ makes the left side ${\pi }^3/8$ times the requested integral. ThereforeAt $t=\pi /2$ the integrand's continuous extension is $1/{\pi }^2$, so no principal-value prescription is required.

Complete the square in the Fourier transform exponent:The supplied shifted Gaussian integral therefore givesThis is the Fourier transform of a Gaussian, with angular-wavenumber convention $e^{-ikx}$.

Take the spatial Fourier transform of the heat equation. Two integrations by parts, with decaying fields and first derivatives at infinity, give $\mathcal{F}[{\psi }_{xx}]=-k^2\tilde{\psi }$. Thus each mode satisfies the ordinary differential equationInserting the transformed initial condition and integrating the time equation givesFinally set $A=t+1/(4\lambda )>0$. The inverse Fourier transform, evaluated by the same Gaussian identity, givesEquivalently the Gaussian initial data for the heat equation evolve asThe profile broadens and its peak decreases, while its total integral remains $\sqrt{\pi /\lambda }$, as required by diffusion without loss on the whole line. Direct differentiation verifies the equation and the initial limit.

Set $g=f$. The cross-correlation at zero is $h(0)={\int }_{\mathbb{R}}|f(x){|}^{2}dx$, while part (i) gives $\hat{h}=|\hat{f}{|}^2$. Evaluating the inverse Fourier transform at zero yields the Parseval identityThis derivation is immediate for Schwartz functions, for which all integrations and Fourier inversion are justified. Approximation by such functions and completeness then extend the identity to L2 space, defining its Fourier transform by the corresponding norm limit. This is the Plancherel theorem extension; no absolute integrability of an arbitrary square-integrable function is presumed.

With the stated negative sign in the forward exponential, the inverse Fourier transform isFor example, this holds pointwise for Schwartz functions; more general square-integrable functions use the corresponding mean-square interpretation.

For sufficiently rapidly decaying functions, substitute the inverse Fourier transform of $g$ and interchange the integrals:The bracketed integral is the complex conjugate of $\tilde{f}(k)$. This is Parseval's identity for the chosen normalization. One may first prove it for Schwartz functions, where the interchanges are justified, and then extend it to $L^2$ by the Plancherel theorem.

When $a=0$, the explicit amplification factor becomesA single Fourier mode has constant modulus in space and is not in the l2 sequence space. Instead choose a whole interval of frequencies on which there is definite growth. LetThe inverse Fourier transform here gives the real sequenceBy the Parseval identity, its squared norm is $\Vert U^0{\Vert }_{{\ell }_2}^2=\Delta x|E|/(2\pi )=\Delta x/2$, finite and strictly positive. These data are also compatible with sampling a bounded continuous function: useThe value at zero is the limit of the displayed quotient, and $u_0(j\Delta x)=U_j^0$.

On $E$ we have ${\sin }^2\theta \ge 1/2$. Therefore the transformed solution and the Parseval identity giveFor each fixed pair $\Delta x,\Delta t>0$, takeTaking square roots proves $\Vert U^m{\Vert }_{{\ell }_2}\ge (1+c{)}^m\Vert U^0{\Vert }_{{\ell }_2}$ for every $m\ge 1$. In particular the norm already increases at the first step, so no positive mesh pair gives practical stability in the contraction sense of (b).

The native PDF places the existence of $c>0$ before the universal quantifier over mesh pairs. If this means a single positive $c$ independent of $\Delta x,\Delta t$, that assertion is false for nonzero data. Indeed, for every datum the same Fourier symbol givesFor any proposed fixed $c>0$, choose $0<\nu <\sqrt{(1+c{)}^2-1}$. The upper bound then contradicts the requested lower bound at $m=1$. The correct quantifiers are $\forall \Delta x,\Delta t>0\exists c=c(\Delta x,\Delta t)>0$, with nonzero square-summable data as constructed above.

Finally, failure of this stronger practical stability of a finite difference scheme should not be confused with every possible refinement notion of numerical stability. If $\Delta t/(\Delta x{)}^2\le K$, then $m\Delta t\le T$ givesThus a finite-time mesh-uniform bound can hold under this severe refinement even though contraction fails on every positive mesh. This is the distinction recorded in centered-advection refinement-dependent stability.