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Starting from $f_{-1}=\sin \theta e^{-i\varphi }$, part (i) gives $L^2{f}_{-1}=2{\hslash }^2{f}_{-1}$ and $L_3{f}_{-1}=-\hslash f_{-1}$. Apply the angular momentum raising operator explicitly:A second application givesSince $L_{+}$ commutes with $L^2$, both raised functions retain the eigenvalue $2{\hslash }^2$. Their $L_3$ eigenvalues follow either by differentiating their $\varphi$ dependence or from $[L_3,L_{+}]=\hslash L_{+}$. Thus three valid unnormalized functions, in the requested order, areFor later Born rule calculations, normalize with the solid angle measure $d\Omega =\sin \theta d\theta d\varphi$. Their squared norms are $8\pi /3$, $4\pi /3$, $8\pi /3$, respectively. For example ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. A convenient orthonormal basis is thereforeOrthogonality follows from the distinct Fourier factors in $\varphi$. An overall phase of each basis function is arbitrary; these choices need not impose the usual phase convention for the spherical harmonics.

The common Fock vacuum obeys $N\psi ={\sum }_j{a}_j^{\ast }(a_j\psi )=0$. Put $B=(a_1^{\ast }{)}^{n_1}(a_2^{\ast }{)}^{n_2}$. By the commutator product rule and part (a),Moving $N$ through $B$ therefore givesDividing by $\sqrt{n_1!n_2!}$ provesThe constructed number state is nonzero: the relation $a_j(a_j^{\ast }{)}^r\psi =r(a_j^{\ast }{)}^{r-1}\psi$ follows from $[a_j,a_j^{\ast }]=I$ and $a_j\psi =0$. Commuting different modes and successively moving each creation operator across the inner product yieldsThus the given factorial denominator also gives unit norm, as expected for a number state.

For $\lambda =3/4$, the adjoint operator relation $J_{+}^{\ast }=J_{-}$ and the given product identity implyA vector of zero norm is zero, so $J_{+}\psi =0$. Part (i) givesThe commutator $[J^2,J_{-}]=0$ also gives $J^2\phi =(3/4){\hslash }^2\phi$, while $J_3\phi =-(\hslash /2)\phi$. Applying the other product identity to this unnormalized vector,ThereforeThese are the two angular momentum ladder endpoints of a spin one-half pair. In particular, ${\psi }_2=\phi /\hslash$ has unit norm, and $J_{+}{\psi }_2=\hslash {\psi }_1$ because $J_{+}{J}_{-}\psi ={\hslash }^2\psi$.

The orthonormal basis makes the squared norm equal to the sum of squared expansion coefficients:Choose a positive real wavefunction normalization constant, fixing an irrelevant global phase:As a direct check, ${\int }_0^{2\pi R}{\sin }^4(x/R)dx=3\pi R/4$.

By the Born rule, the probability for a measured energy eigenvalue is the squared norm of the orthogonal projection onto its whole energy eigenspace. ThusEvery other energy eigenvalue has probability zero. These probabilities are independent of $T$: unitary time evolution under the unchanged Hamiltonian operator changes the dynamical phases but not the squared expansion coefficients. The two equal-energy terms describe one measurement outcome.

Use the canonical commutation relationsWriting $c_j=\sqrt{m{\omega }_j/(2\hslash )}$, $d_j=1/\sqrt{2m\hslash {\omega }_j}$, direct expansion of the commutator givesThe Fock vacuum ${\psi }_{0,0}$ is annihilated by both $a_{j,-}$. Successive creation operators generateHence, in the order of part (a), the six energy eigenstates are proportional toIf the Fock vacuum has unit norm, dividing the general expression by $\sqrt{n_1!n_2!}$ gives normalized number states. Normalization is not required here, but using this convention makes the later change of basis unambiguous.

Use the inner product $⟨f,g⟩=\mathbb{E}[f(Z)g(Z)]$ for a standard normal random variable $Z$. The degree-zero monic polynomial is $p_0=1$. Write $p_1=x+a$ and $p_2=x^2+bx+c$. The conditionsuse $\mathbb{E}Z=\mathbb{E}{Z}^3=0$ and $\mathbb{E}{Z}^2=1$. Hence these first three Probabilists' Hermite polynomials areThe Gaussian fourth moment is $\mathbb{E}{Z}^4=3$, so their squared norms are

The squared norm of $p_{\ell }(x)e^{-x^2/4}$ in $L^2(\mathbb{R},dx)$ is $\sqrt{2\pi }{h}_{\ell }$, where $(h_0,h_1,h_2)=(1,1,2)$. Choose positive normalizing constantsThen ${\varphi }_0,{\varphi }_1,{\varphi }_2$ are an orthonormal set for Lebesgue measure. Expanding the finite sums givesThus this is an orthogonal polynomial projection kernel, explicitlyThe positive constants above are forced by the reproducing identity: before normalization, its convolution multiplies each independent term ${\varphi }_{\ell }(x){\varphi }_{\ell }(z)$ by $\sqrt{2\pi }{h}_{\ell }/{\kappa }_{\ell }^2$. Each multiplier must equal one. Changing the sign of a real ${\kappa }_{\ell }$ leaves the kernel unchanged.

Use the intended finite-dimensional irreducible highest weight module $A^{\varphi }$, with dominant integral highest weight $\varphi$. Its highest weight multiplicity is one. An arbitrary direct sum having that same largest representation weight would not in general satisfy this formula. Set $f={\pi }_{A^{\varphi }}$ and use the corrected denominator $D$ of part (a).

Restriction to each simple root ${\mathfrak{sl}}_2$ decomposes into symmetric finite representation weight strings, so $f$ is Weyl-invariant. Consequently $Df$ is Weyl-antisymmetric. We prove that no lower Weyl alternants occur, rather than assuming the desired character formula.

First establish the denominator identity itself. The finite sum $D$ has support in $\rho -Q_{+}$ and coefficient one at $\rho$. Antisymmetry kills all singular representation weight coefficients, since a reflection fixing the representation weight reverses their sign. Every remaining Weyl orbit has a unique regular dominant representative $\eta$. If it occurs in $D$, then $\eta \le \rho$. Since $\eta$ is integral and regular dominant, $\eta -\rho$ is dominant integral; it also belongs to $-Q_{+}$. A dominant vector $v=-\sum n_i{\alpha }_i$ satisfies $(v,v)=-\sum n_i(v,{\alpha }_i)\le 0$, so it must be zero. Thus the only possible orbit is $W\rho$. The leading coefficient givesFor the formal Laplacian $\mathcal{D}(e^{\lambda })=\Vert \lambda {\Vert }^2{e}^{\lambda }$, this implies $\mathcal{D}D=\Vert \rho {\Vert }^{2}D$.

Now choose root vectors dual under an invariant form, so $[E_{\alpha },E_{-\alpha }]=H_{\alpha }$ with $H_{\alpha }$ corresponding to the vector $\alpha$, and take the quadratic Casimirfor an orthonormal Cartan basis. On a highest vector its scalar is $c_{\varphi }=\Vert \varphi +\rho {\Vert }^2-\Vert \rho {\Vert }^2$, hence it has this scalar throughout the cyclic highest weight module.

Here is the computation of the weighted trace of its radial identity. For a positive root, let $X$ be the formal weighted trace of $E_{-\alpha }{E}_{\alpha }$ and $Y$ that of $E_{\alpha }{E}_{-\alpha }$. Cyclicity of trace and representation weight shifting give $Y=e^{\alpha }X$, while their difference is the weighted trace of $H_{\alpha }$, namely $(\alpha ,\nabla f)$. ThusTaking the trace of $C$ yieldsThe product expression for $D$ givesUsing $2\rho ={\sum }_{\alpha >0}\alpha$ and the product rule for the formal Laplacian therefore givesHence every representation weight in the support of $Df$ has norm $\Vert \varphi +\rho \Vert$. Its support is also contained in $\varphi +\rho -Q_{+}$, and it is antisymmetric. Any occurring regular dominant weight $\eta$ thus satisfies $\eta \le \varphi +\rho$. If the inequality were strict, thenbecause the first factor is a nonzero nonnegative sum of simple roots and the second is strictly dominant. This contradicts the equal-norm condition. Only the orbit $W(\varphi +\rho )$ remains. Its leading coefficient is one, proving

Use a smooth exact solution such as $u(x,t)=e^{-D{q}^{2}t}\sin (qx)$ on a domain with matching boundary conditions, and compare numerical and exact values at a fixed physical time. To measure spatial order, halve $h$ while making temporal error negligible; the error norm should fall by a factor of four. To measure temporal order, halve $k$ on a sufficiently fine fixed spatial grid and again expect a factor of four. Report observed order $p={\log }_2(E_h/E_{h/2})$ and use several refinements to identify the asymptotic regime.

Check more than one smooth mode, boundary implementation and expected conservation or decay properties. A manufactured solution with an added known source can test nontrivial boundaries and forcing. Repeat in maximum and root-mean-square norms, stop before finite-precision or solver tolerances dominate, and separate stability failures from errors in convergence order. A single small error or a single visually smooth solution does not establish accuracy.

Inside the well, ${\psi }^{\prime \prime }+k^2\psi =0$. Vanishing at zero removes the cosine term; vanishing at $L$ requires $k_{n}L=n\pi$. The nonzero states have positive integer $n$ because $n=0$ gives the zero function and negative $n$ only repeats a state up to sign. Normalization follows from ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$. ThusThe wavefunction vanishes outside the infinite barriers, so there is no additional contribution to its norm.

Use $-\pi <{\theta }_s<\pi$, with the crack along the negative radial axis. In the coefficient convention used here, the leading crack-tip expansion isThe square-root term is harmonic and has zero angular derivative on both crack faces. Regular contributions may include a constant and bounded-gradient terms under smooth loading.

If $E(s)$ is the minimized elastic-plus-load potential, the energy release rate is $\boxed{G(s)=-dE(s)/ds}$. With this particular square-root coefficient normalization,For example, the small-circle energy flux is $\int [\mathcal{W}{n}_1-\mu u_r{u}_{x_1}]R_{s}d{\theta }_s$. The singular term gives $\mathcal{W}=\mu k_s^2/(8R_s)$, $u_r=k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$ and $u_{x_1}=-k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$. Integration over $(-\pi ,\pi )$ produces the stated factor. Other stress-intensity normalizations change that factor.

For toughness $G_c>0$, local stability against infinitesimal extension means $\boxed{G(s)\le G_c}$, since the derivative of $E(s)+G_c(s+1)$ is nonnegative. Global energetic stability against every extension is the stronger condition $E(s)\le E(r)+G_c(r-s)$ for all $r\ge s$.

In full isotropic linearized elasticity,For the antiplane field this becomesAhead of the tip, ${\theta }_s=0$, the singular gradient is $(0,k_s/(2\sqrt{R_s}))$. ThusThis is stronger than the printed remainder $o(R_s^{-1})$. If $k_s$ denotes a nonnegative intensity, it has the printed leading coefficient; for a signed singular coefficient its absolute value is essential because the left side is a norm.

Take $f$ in the generator domain $D(L)$. The semigroup commutation with its unbounded generator implies $P_{t}f\in D(L)$ andIndeed, applying the bounded operator $P_t$ to the difference quotient at zero gives the right derivative $P_{t}Lf$; strong continuity gives the full orbit derivative for $t>0$. Now integration by parts in the Bochner integral yieldsThe boundary term at infinity vanishes because $\Vert e^{-\lambda t}{P}_{t}f\Vert \le e^{-\lambda t}\Vert f\Vert$. The derivative integral exists as a Bochner integral, since its integrand is bounded in norm by $e^{-\lambda t}\Vert Lf\Vert$.

Multiplying this identity by $\lambda$ givesPart (b) applies to the fixed vector $Lf\in E$, soThe bounded operators $L_{\lambda }={\lambda }^2{R}_{\lambda }-\lambda I$ are the Yosida approximation of the semigroup generator. The proof only applies the unbounded operator $L$ to $f\in D(L)$; the contraction and limit arguments then act on the ordinary vector $Lf$.

For $0<\alpha <2$, write $z=r{e}^{i\theta }$ with $0<\theta <\pi$. The chosen power map sends this to $r^{\alpha }{e}^{i\alpha \theta }$, soAt $\alpha =2$ this is the plane slit along the nonnegative real axis. For every $0<\alpha \le 2$, the map is conformal on the upper half-plane.

To check finiteness of the physical exit time, start a Brownian motion $B$ at $i$ and let $T$ be its real-axis exit. The time change of a continuous process in conformal invariance of planar Brownian motion isWe know $T<\infty$ almost surely. The Cauchy exit law from a Brownian half-plane assigns probability zero to $B_T=0$. On such a path, $B_s\ne 0$ throughout the compact interval $[0,T]$, because it is in the open upper half-plane before exit and has nonzero endpoint. Its modulus therefore has a positive minimum and finite maximum. The displayed integrand is bounded on that interval, proving $A_T<\infty$. The image Brownian motion exits $D$ at this finite changed time.

The endpoint $\alpha =0$ printed in the source is degenerate: $z^0=1$, so its image is the singleton $\{1\}$, not an open domain, and the map is not conformal. If exit from that set is defined for Brownian motion started at one, its exit infimum is zero. It cannot be included in a nondegenerate conformal-domain argument.

Here is the martingale argument for maximal comparisons, including the starting-value issue. For $P=f+ig$, the Cauchy-Riemann equations and Itô's formula show thatare orthogonal continuous local martingales with the same quadratic variation. Localize before large disks and before exit. The Burkholder-Davis-Gundy inequalities then givewith a constant depending only on $p$; Fatou's lemma removes the localization. These are centered martingales. For the uncentered processes this impliessince $|f(i^{\alpha })|\le \sup |f(Z_t)|$.

The source normalizes $P(0)=0$ at a boundary point, rather than the value of $g$ at the starting point. Consequently its displayed constant cannot be uniform in $\alpha$ with only $p$ dependence. To see this, take $P(z)=iz$ and $0<\alpha <\min (1/2,1/p)$. Then $f=-\mathrm{Im}z$ and $g=\mathrm{Re}z$. Inside this narrow wedge,Both $p$th moments are finite and the right one is positive; finiteness follows from the wedge radial-maximum tail of order $R^{-1/\alpha }$, justified below. The source inequality would force $C_p\ge {\cot }^p(\alpha \pi )$, unbounded as $\alpha \downarrow 0$. At $\alpha =0$, the immediate-exit interpretation gives the direct contradiction $1\le C_{p}0$ for the same polynomial.

For a fixed nondegenerate wedge, an uncentered inequality does hold with a constant allowed to depend on $\alpha$ as well as $p$. One rigorous outline for the polynomial class in this question is as follows. Separation of variables in a sector of radius $R$ gives, for exit through its circular arc, the expansionThus at the specified starting point the radial-maximum tail is comparable to $R^{-1/\alpha }$. Hitting any fixed interior subarc has that same power, with positive leading coefficient. If a nonzero polynomial has degree $d$, its real part has magnitude at least $c{R}^d$ on some fixed interior subarc for sufficiently large $R$. Therefore its maximal real-part $p$th moment is infinite whenever $dp\ge 1/\alpha$. In that case the desired extended-real inequality is automatic.

The remaining polynomials have $d<1/(\alpha p)$, so belong to a finite-dimensional vector space space. On that spaceis a genuine norm: if it is zero, Brownian motion's positive probability of reaching any interior neighborhood forces $\mathrm{Re}P=0$ throughout $D$, so analyticity makes $P$ a purely imaginary constant, and $P(0)=0$ makes it zero. The corresponding imaginary-part functional is bounded on this finite-dimensional vector space normed space; its finiteness follows from $dp<1/\alpha$ and the radial tail bound. equivalence of norms in finite dimensions givesThis establishes the fixed-domain interpretation and identifies exactly why the universal centered martingale inequality cannot silently be applied to the printed uncentered boundary normalization. It is the distinction in starting-value obstruction in wedge conjugate maximal inequalities.

Write $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$ and let $|\Lambda |={\max }_i|{\lambda }_i|$. The stated Euclidean norm givesTaking nonnegative square roots proves $\Vert f_n\Vert \le |\Lambda {|}^n\Vert f_0\Vert$. At $n=0$ this is equality. This is a contraction bound in the transformed coordinates; an arbitrary nonorthogonal $S$ does not give the same unit-constant bound for the original error norm.

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

Fix one common eigenvector $v\ne 0$, and write $A{A}^{\top }v=\lambda v$ and $B{B}^{\top }v=\mu v$. By (d),so $\lambda \mu v=0$ and at least one of the two eigenvalues is zero.

To exclude both being zero, use the kernel of a Gram product identity:If both products annihilate $v$, then $A^{\top }v=B^{\top }v=0$. By (b), $v\in \mathrm{im}L$, and by (e), this is $\mathrm{im}B$. On the other hand, $B^{\top }v=0$ makes $v$ orthogonal to every vector in $\mathrm{im}B$. In particular $v$ is orthogonal to itself, so $\Vert v{\Vert }^2=0$ and $v=0$, a contradiction.

Therefore each specified common eigenvector belongs to exactly one of $\ker (A{A}^{\top })$ and $\ker (B{B}^{\top })$. Both eigenvalues are nonnegative by the displayed squared-norm identities, so the other eigenvalue is strictly positive. This is the spectral mechanism behind orthogonal null-space completion of a square matrix.

A penalized least-squares estimator balances squared error with a nonnegative penalty $J$. The Lasso uses the coefficient ${\ell }_1$ norm; graph total variation denoising uses the ${\ell }_1$ norm of endpoint differences. The choice of normalization of the squared loss changes the numerical tuning parameter.

For $R(\theta )=\mathrm{diag}(1,e^{i\theta })$, the difference vanishes on $|0⟩$ and multiplies $|1⟩$ by $e^{i\theta }(e^{i\delta }-1)$. Its operator norm is therefore $|e^{i\delta }-1|=2|\sin (\delta /2)|$. Choosing the nearest angle on the grid $2\pi m/2^n$, with circular distance, gives $|\delta |\le \pi /2^n$ and a uniform bound of this size on the output-state norm error. This bound concerns state vectors, not an asserted equality with measurement-probability error.

The convention in which the anyon fusion trees form an orthonormal basis and a splitting vertex followed by its adjoint fusion vertex gives the identity on the total-charge strand. A closed normalized state has norm one. It differs from the isotopy normalization of anyon diagrams, in which an unknotted closed charge-$a$ strand evaluates to $d_a$.