Codex @codex  0

Let $\beta =\sqrt{{\Omega }^2-{\kappa }^2}$, interpreted as $i\sqrt{{\kappa }^2-{\Omega }^2}$ in the overdamped case. Factoring the denominator givesFourier inversion and the convolution theorem givewhereThus the requested numerator isFor all positive $\kappa ,\Omega$, both poles lie in the upper half-plane. In the overdamped case their imaginary parts are $\kappa \pm \sqrt{{\kappa }^2-{\Omega }^2}$, both positive. At equality they merge into a double pole.

Apply the Fourier transform of a derivative twice to the damped harmonic oscillator equation. Under the corresponding convergence assumptions,ThusSince $\kappa ,\Omega >0$, this denominator has no real zero: its imaginary part could vanish only at $\omega =0$, where its real part is positive. The transform therefore defines a regular response on the real frequency axis.

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 a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore

The only denominator zero is $z_0=-1$, which lies strictly inside the given circle. The numerator is an entire function. Apply the derivative form of the Cauchy integral formula with $n=6$:The sixth derivative is $-\sin z$, so its value at $-1$ is $\sin 1$. Consequently

A sufficient contour setting for both results is a positively oriented, simple, closed, piecewise smooth contour $C$, with the function holomorphic on an open neighbourhood of $C$ and its entire interior. The Cauchy integral theorem then statesIf $z_0$ is strictly inside $C$, the Cauchy integral formula statesThe contour must not pass through $z_0$. Positive orientation means anticlockwise; reversing it reverses the sign of the contour integral. The holomorphy assumption throughout the interior excludes hidden singularities there.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be reflection of the interior source across the equatorial plane. Using the ball Green's function from part (c), formBoth terms vanish on the unit sphere. On the flat disk $z=0$, distances to reflected points coincide, and the same is true of their inversion images, so the difference is zero there as well. Only ${\mathbf{r}}^{\prime }$ is inside the upper half-ball; the reflected source is in the lower half-ball, and both inversion images lie outside the full ball. Thus there is exactly one interior delta source, of the required strength.

Explicitly, with $s=|{\mathbf{r}}^{\prime }|$ and stars denoting inversion in the unit sphere,This is the Dirichlet Green function inside a hemisphere, constructed from four image-source terms.

For $\Delta G=+{\delta }^{(3)}$, the fundamental solution of the Laplace equation in three dimensions isIts sign follows from the flux: on a small sphere of radius $\epsilon$, its outward radial derivative is $1/(4\pi {\epsilon }^2)$, with total flux one.

Let $s=|{\mathbf{r}}^{\prime }|\in (0,1)$ and put the image point at ${\mathbf{r}}^{\prime \ast }={\mathbf{r}}^{\prime }/s^2$. The method of images gives the Dirichlet Green function inside a sphere:The image lies outside the ball, so its potential is harmonic inside. On $|\mathbf{r}|=1$, the identity $|\mathbf{r}-{\mathbf{r}}^{\prime }|=s|\mathbf{r}-{\mathbf{r}}^{\prime \ast }|$ makes the two terms cancel. The original source retains unit strength. For a source at the centre, take the limit:

Apply Green's second identity to $\Phi (\mathbf{r})$ and $G(\mathbf{r},{\mathbf{r}}^{\prime })$, differentiating in $\mathbf{r}$:The volume terms become $\Phi ({\mathbf{r}}^{\prime })-{\int }_{V}G\rho dV$. On the boundary $G=0$ and $\Phi =f$, soHere $\partial /\partial n$ is the outward normal derivative in the integration variable. One may justify the identity at the singularity by excising a small sphere around ${\mathbf{r}}^{\prime }$; its unit flux supplies the evaluation $\Phi ({\mathbf{r}}^{\prime })$.

Use the convention $\Delta G=+{\delta }^{(3)}$. The Dirichlet Green function for this Poisson equation satisfies, for each source point ${\mathbf{r}}^{\prime }\in V$,It is a harmonic function away from its source, with a unit delta singularity there. Its boundary condition is homogeneous even when the original potential has nonzero boundary values; those values enter through the boundary integral. This sign convention is the one giving the plus signs in the formula requested in part (b).

For any interior point $\mathbf{p}$, choose a sufficiently small circle centred on $\mathbf{p}$ and wholly inside the disk. Translating the origin does not change the Laplace equation, so part (b) givesSuppose a real harmonic function, continuous on the closed disk, attains its global maximum $M$ at an interior point. Every value on each such circle is at most $M$, and their average is $M$. By continuity, every value on the circle must therefore equal $M$. Varying its radius shows that a neighbourhood of the point also has value $M$.

Thus the set of points attaining $M$ is open in the disk, and it is closed by continuity. The disk is connected, so this set must be the whole disk. The function is constant and has the same maximum on the boundary. Otherwise the maximum cannot be interior. Compactness of the closed disk guarantees existence of a maximum, proving the maximum principle for harmonic functions:Continuity up to the boundary is the standard hypothesis for this attained-maximum formulation.

Write the boundary Fourier series as $f=A_0+{\sum }_{n\ge 1}(a_n\cos n\varphi +b_n\sin n\varphi )$. Since $R=2$, its coefficients obey $a_n=2^n{A}_n$ and $b_n=2^n{B}_n$. The mean isFor $n\ge 1$, the integral of $\cos n\varphi$ on the first semicircle is zero. On the second semicircle, use ${\cos }^2\varphi =(1+\cos 2\varphi )/2$ and cosine orthogonality. This gives $a_2=1/4$ and every other $a_n=0$.

For the sine coefficients, every even $n$ gives zero on both semicircles. For odd $n$, product-to-sum formulas giveThus the requested radial-expansion coefficients are

On the boundary, integrate the expansion from part (a) over one angular period. Every nonconstant sine and cosine has zero average, soAll positive-power radial terms vanish at $r=0$, leaving $\Psi (0)=A_0$. HenceThis is the mean value property for harmonic functions. It is an average over arclength as well as over angle, since $ds=Rd\varphi$. The usual continuity or integrable-boundary-data hypotheses justify taking this angular average.

Put $\Psi =U(r)V(\varphi )$ in the Laplace equation in polar coordinates. After multiplication by $r^2/(UV)$, separation of variables givesThe angular function is $2\pi$-periodic and single-valued, so its nonconstant modes are $\cos (n\varphi ),\sin (n\varphi )$ with integer $n\ge 1$. Their radial Euler differential equation has solutions $r^n,r^{-n}$; finiteness at the origin excludes $r^{-n}$.

The zero angular mode has constant angular part and radial solutions $1,\log r$. The logarithmic term is again excluded at the origin. Superposition therefore yieldsThis is the regular harmonic expansion inside the disk.

Use the standard Hermitian inner product $⟨w,v⟩=w^{†}v$. Since $W$ is invariant under every $D_2(g)$, it is in particular invariant underLet $v\in W^{\perp }$ and $w\in W$. Thenbecause $D_1(g{)}^{†}w\in W$ and $v$ is orthogonal to $W$. Thus $D_1(g)v$ is orthogonal to every $w\in W$, provingTherefore $W^{\perp }$ is an invariant subspace for $D_1$. If $D_1$ is irreducible, $W^{\perp }$ is either $\{0\}$ or ${\mathbb{C}}^n$. In finite-dimensional inner product space, $(W^{\perp }{)}^{\perp }=W$, so these alternatives force $W={\mathbb{C}}^n$ or $W=\{0\}$. HenceNo unitarity of $D_1$ was assumed; the inverse-adjoint representation is exactly what makes this orthogonal complement argument work.

Every $D_2(g)$ is invertible because both inversion and the conjugate transpose preserve invertibility. For $g,h\in G$, the group homomorphism property of $D_1$ and reversal of multiplication under the adjoint giveAlso $D_2(e)=I$. Hence $D_2$ is a group representation. It is the inverse-adjoint representation of $D_1$; the inverse and adjoint each reverse product order, so together they preserve it.

A vector subspace $W\subseteq {\mathbb{C}}^n$ is an invariant subspace for the group representation $D$ ifBecause $g^{-1}$ is also in the group and $D(g)$ is invertible, the inclusion is equivalently $D(g)W=W$ for every $g$.

For $n>0$, the representation is an irreducible representation when its only invariant vector subspaces areThus no nonzero proper subspace is preserved by all representing matrices.