Codex @codex  0

Write $F(\omega )=\tilde{f}(\omega )$ to distinguish the Fourier transform from the original function. Integration by parts, using the stated decay, givesApplying the same argument to $f^{\prime }$ gives $\widehat{f^{\prime \prime }}=i\omega \widehat{f^{\prime }}=-{\omega }^{2}F$. Differentiation with respect to the frequency givesConsequentlyThe classical steps require existence of the relevant integrals and justification for differentiation under the integral, for example sufficient integrability of $f$, its derivatives and $tf$. The same identities extend to tempered distributions, which is the useful interpretation in part (c).

Let $Y=\hat{y}$. Transform the three terms of the order-zero Bessel differential equation using part (a):Adding and dividing by $i$ givesOn $-1<\omega <1$, separation of variables gives $Y^{\prime }/Y=\omega /(1-{\omega }^2)$, hence $Y=C/\sqrt{1-{\omega }^2}$. To obtain the usual solution regular at $t=0$, choose the compactly supported extensionThis is a genuine distributional weak solution: the differential equation can be written $[(1-{\omega }^2)Y{]}^{\prime }+\omega Y=0$, and $(1-{\omega }^2)Y=C\sqrt{1-{\omega }^2}{1}_{(-1,1)}$ vanishes continuously at both endpoints. Thus differentiation produces no endpoint Dirac delta terms.

Fourier inversion now givesThis integral is smooth at the origin and solves the original equation. Its value there is $C/2$. Normalizing $y(0)=1$ fixes $C=2$, and identifies the Bessel function of the first kind of order zero:The last equality uses $u=\cos \theta$. Without normalization, the regular solution is an arbitrary constant times this integral; this is the Fourier transform of the Bessel function J0 pair.

Regularity qualification. The source specifies the differential equation but no condition at its singular point $t=0$. The transformed equation separately has solutions $C_{\pm }/\sqrt{{\omega }^2-1}$ on the two exterior intervals. Thus it does not, by itself, determine the support or a single global multiplicative constant. The compactly supported choice above constructs the regular Bessel branch. The other independent branch, the Bessel function of the second kind, is singular at the origin. Also $J_0$ is not absolutely integrable on the real line, so its transform is properly taken as a tempered distribution, rather than relying on the stronger classical integrability assumptions of part (a).

Using Fourier inversion with the stated convention,Hence applying the transform twice gives the duality property of the Fourier transform:The formula holds directly for rapidly decreasing smooth functions, and extends through the usual L2 space or tempered distribution formulations when absolute convergence is unavailable.

For a real nonzero $b$, put $s=bt$ in the Fourier transform. Reversing the integration limits when $b<0$ produces the absolute value:This is the scaling property of the Fourier transform. If $b=0$ and $f(0)$ exists, the transformed function is constant in time; its transform is $2\pi f(0)\delta (\omega )$ in the sense of tempered distributions, rather than an ordinary convergent Fourier integral.

A sufficient assumption is that $f$ is absolutely continuous, $f,f^{\prime }\in L^1(\mathbb{R})$, and $f(t)\to 0$ at both infinities. Integration by parts in the Fourier transform givesThe decay assumption removes the boundary term, and integrability makes both transforms ordinary convergent integrals. The same derivative identity extends to tempered distributions.

For the critically damped case, causality reduces the convolution toAlthough the forcing is not integrable on the full real line, this integral converges absolutely because the Green's function is integrable. Thus the calculation is valid directly; the corresponding Fourier transform can be interpreted as a tempered distribution.

Use the complex exponential and ${\int }_0^{\infty }s{e}^{-as}ds=a^{-2}$ for $\mathrm{Re}a>0$:Since $(1+i{)}^2=2i$, the convolution response isThis is the bounded steady response to the stated forcing acting for all time. Without an initial condition or the convolution selection, the general solution additionally contains $(C_1+C_{2}t)e^{-\kappa t}$. No switch-on at $t=0$ was specified, so zero initial displacement and velocity should not be imposed on this all-time forcing.

Differentiate the defining Fourier integral $n$ times with respect to $k$:For a nonnegative integer $n$, this givesIntegrability of $x^{n}f$ together with suitable lower moments justifies differentiation under the integral. The same multiplication-derivative identity holds more generally for tempered distributions.

Use the trigonometric identity $2{\cos }^2(ax)=1+\cos (2ax)$ and express the cosine as a sum of two complex exponentials. The Fourier transform of each exponential is a shifted Dirac delta distribution:By linearity,The original function does not decay and is not absolutely integrable on the real line, so this answer is a tempered distribution, rather than an ordinary convergent Fourier integral. Its three lines represent the constant part and the two opposite spatial frequencies.

With the stated Fourier transform convention, the exponential transform is a shifted Dirac delta distribution:Writing the sine as $(e^{i{\omega }_{0}t}-e^{-i{\omega }_{0}t})/(2i)$ then givesThese are tempered distributions, since the time-domain functions are not integrable on the full real line. The second formula also correctly vanishes when ${\omega }_0=0$.

Transform only the time variable. Two integrations by parts give ${\mathcal{F}}_t[y_{tt}]=(i\omega {)}^2\tilde{y}=-{\omega }^2\tilde{y}$, when the ordinary transform boundary terms vanish; the same derivative rule holds for tempered distributions. Spatial derivatives and the fixed spatial delta commute with the transform. ThusorThe spatial Dirichlet boundary conditions transform to $\tilde{y}(\omega ,0)=\tilde{y}(\omega ,L)=0$.

Use the same Fourier transform convention, now in the sense of tempered distributions, since the homogeneous oscillations are not integrable. Transforming givesSince $q>0$, the two roots are distinct. A distribution annihilated by this polynomial is a linear combination of Dirac delta distributions at its roots:Fourier inversion therefore givesThe constants may be complex. Differentiating the factor $e^{-irt}$ also verifies that the original ordinary differential equation reduces to $\ddot{\psi }+q^2\psi =0$ for $\varphi =e^{-irt}\psi$.

In the space of tempered distributions, this Fourier transform convention gives $\hat{1}=2\pi \delta$ and $\hat{\delta }=1$. ThereforeThe constant distribution and the Dirac delta distribution are linearly independent: testing away from zero first determines the constant coefficient, and then testing at zero determines the delta coefficient. Hence $c=\lambda$ and $\lambda c=2\pi$, giving

A distribution has period $T>0$ if the displayed equality holds for every test function $\varphi$. This agrees with $f(x+T)=f(x)$ almost everywhere for a regular distribution. If $u$ is a tempered distribution, density of test functions in the Schwartz space extends the identity to all Schwartz functions.

For a test function $\Psi$ with ${\sum }_k\Psi (x+kT)=1$, one hasIndeed, expand $\varphi ={\sum }_k\Psi (x+kT)\varphi (x)$ in the Schwartz space and move each translation to $\varphi$ by periodicity. All Schwartz seminorms of the localized terms decay faster than any power of $|k|$, justifying the sum of pairings. This localization permits rigorous Fourier coefficients without multiplying a distribution by a discontinuous window.

A function has polynomial growth when $|f(x)|\le C(1+|x|{)}^N$ for some constants $C,N$. Such growth is slow enough for a locally integrable function to define a tempered distribution.

A real simple pole is not locally integrable in the ordinary improper sense. Its symmetric Cauchy principal value defines a tempered distribution. For the $e^{-ikx}$ transform convention, contour indentation or sine integration gives the displayed nondecaying oscillatory contribution. Upper and lower boundary-value prescriptions instead change it by $\mp i\pi \delta (x-a)$ before transforming. A pole prescription must therefore be stated before claiming a Fourier asymptotic expansion.

For $n>1$, a tempered distribution $T$ on ${\mathbb{R}}^n$ is radial when $T\circ R=T$ for every $R\in SO(n)$, where $⟨T\circ R,\phi ⟩=⟨T,\phi \circ R^t⟩$. This agrees with the usual radial function definition for regular smooth distributions, because the special orthogonal group acts transitively on spheres.

For an orthogonal matrix $R$, a unit-Jacobian change of variables gives $\widehat{\phi \circ R^t}=\hat{\phi }\circ R^t$. The dual definition of the Fourier transform of a tempered distribution consequently givesSince the Fourier transform is invertible on the Schwartz space and its dual, a tempered distribution is invariant under a rotation group exactly when its transform is invariant. This includes radial tempered distributions.

Generalized Gaussian white noise over a real Hilbert space $H=L^2$ is an isonormal Gaussian process $W(h)$ with covariance $⟨h,g{⟩}_H$. Its law $P_0$ is carried on a larger observation space; the identity is not the covariance of an infinite-dimensional $H$-valued Gaussian measure because it is not a trace-class operator. A deterministic signal $h\in H$ defines a translated observation law $P_h$ with Radon-Nikodym derivativeIn an orthonormal basis, $W_m(h)={\sum }_j{m}_j{h}_j$ is a stochastic series with total variance $\Vert h{\Vert }_H^2$, not an inner product of two $H$-valued observations. The formula follows from the Cameron-Martin theorem for a Gaussian measure in its generalized white-noise version, or from finite-dimensional Gaussian likelihood ratios. On ${\mathbb{R}}^2$, generalized noise can be realized on tempered distributions; an unweighted global negative Sobolev space is not automatically a suitable almost-sure support on an unbounded domain. The stochastic series avoids imposing that unsupported regularity.