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The contraction mapping theorem states that a contraction mapping $f$ of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if $d(f(x),f(y))\le qd(x,y)$ with $0\le q<1$, iteration from any $x_0$ converges to that fixed point.

To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$ the triangle inequality givesThus $(x_n)$ is a Cauchy sequence, and completeness supplies a limit $x_{\ast }$. A contraction mapping is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the uniform metric $\rho (f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho (f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\le d(f(x),g(x))+d(g(x),h(x))$ proves the triangle inequality. Hence $\rho$ is a metric.

Suppose $(f_n)$ is a Cauchy sequence for this uniform metric. At each $x$, $d(f_n(x),f_m(x))\le \rho (f_n,f_m)$, so completeness of $X$ supplies a value $f(x)={\lim }_n{f}_n(x)$. Given $\epsilon >0$, choose $N$ with $\rho (f_n,f_m)<\epsilon /2$ for all $n,m\ge N$. Fix $n\ge N$ and let $m\to \infty$ pointwise; then $d(f_n(x),f(x))\le \epsilon /2$ for every $x$. Thus $\rho (f_n,f)\le \epsilon /2<\epsilon$. The uniform limit theorem makes this limit continuous, so $f\in F$. This proves the completeness of the continuous-map space in the uniform metric.

Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta (f)$ and $x_g=\theta (g)$. Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing $\rho (f,g)<(1-q_f)\epsilon$ proves continuity of $\theta$ at every $f$. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.

A photon of wavelength $\lambda$ has energy $E=h\nu =hc/\lambda$ and momentum $\mathbf{p}=(h/\lambda )\hat{\mathbf{n}}$ along its propagation direction.

The specified Compton scattering formula requires the electron to be initially at rest in the frame where the wavelengths and angle are measured. This initial-rest assumption is implicit in the printed request; it is not valid for arbitrary initial electron momentum. Use four-momentum with Minkowski metric signature $(+---)$: initially $P=(mc,0)$, while $K=(E/c,(E/c)\hat{\mathbf{n}})$ and $K^{\prime }=(E^{\prime }/c,(E^{\prime }/c){\hat{\mathbf{n}}}^{\prime })$ are null. Conservation gives $P^{\prime }=P+K-K^{\prime }$, and the outgoing electron satisfies the same mass-shell condition $P^{\prime 2}=P^2=m^2{c}^2$. HenceAfter dividing by $E{E}^{\prime }$ and substituting the photon energy relation,The scale $h/(mc)$ is the electron Compton wavelength. For a moving initial electron, the earlier invariant identity remains valid, but $P\cdot K$ includes its initial velocity and the wavelength shift is different.

A Hermitian operator satisfies $⟨u,Av⟩=⟨Au,v⟩$ for all admissible states in its domain; in finite dimension this is $A^{\ast }=A$. For an unbounded physical observable one specifies a self-adjoint realization, rather than only a formal differential expression.

Here $p=-i\hslash d/dx$ and $H=-{\hslash }^2{d}^2/(2md{x}^2)+V(x)$. For states with sufficient regularity and decay, integration by parts givesThe multiplication term cancels because $V$ is real. Thus the Quantum Hamiltonian is Hermitian on the stated admissible domain. Equivalent boundary conditions must remove this boundary term if the system is on a bounded interval.

The Schrödinger equation $i\hslash \dot{\psi }=H\psi$ and its conjugate imply, for a time-independent observable $A$ and normalized state,This follows by differentiating both factors in $⟨\psi ,A\psi ⟩$ and moving $H$ using Hermiticity. The canonical commutator $[x,p]=i\hslash$ givesConsequently the two Ehrenfest theorem identities areThey require the displayed expectations and domain operations to exist; real-valuedness of a potential alone is not a proof about every possible singular operator domain.

The positive-definite symmetric form $b$ is a real inner product. Choose a $b$-orthonormal basis by the Gram-Schmidt process. The stated identity means that the matrix $K$ of $\psi$ satisfies $K^T=-K$. If $d=\dim U$, thenAn invertible $K$ has nonzero determinant, so $d$ is even.

To cover a possibly singular map, first show $\ker \psi =(\mathrm{im}\psi {)}^{\perp }$. If $x\in \ker \psi$, then $b(x,\psi y)=-b(\psi x,y)=0$ for every $y$, giving one inclusion. Conversely orthogonality to the image gives $b(\psi x,y)=0$ for all $y$, so nondegeneracy of $b$ gives $\psi x=0$. Positivity then implies $\mathrm{im}\psi \cap \ker \psi =\{0\}$.

The image is invariant under $\psi$, and its restricted map $\psi {|}_{\mathrm{im}\psi }$ is injective by that trivial intersection. In finite dimension it is consequently invertible. The restricted inner product is positive definite and the same skew-adjoint identity holds, so the first argument applied to this restriction makes $\dim \mathrm{im}\psi$ even. Thusincluding rank zero. This is the even rank of a skew-adjoint linear map.

Associate to $c=(c_1,\dots ,c_n{)}^T$ the polynomial $p(t)={\sum }_{j=1}^n{c}_j{t}^{j-1}$. The $i$th component of $Ac$ is exactly $p(a_i)$. If $Ac=0$, this polynomial of degree at most $n-1$ has $n$ distinct roots. Repeated use of the factor theorem would make ${\prod }_{i=1}^n(t-a_i)$ divide any nonzero such polynomial, contradicting its degree. Hence $p$ is the zero polynomial and all $c_j=0$. The converse is immediate, proving $\ker A=\{0\}$.

A square matrix has full column rank if and only if its kernel is zero, by rank-nullity theorem. Its row rank equals its column rank. Therefore this Vandermonde matrix has row rank $n$, and its rows $v_1,\dots ,v_n$ are a basis and span ${\mathbb{R}}^n$.

Eliminate the first entries of rows two, three and four using multipliers $a^3,a^2,a$. All remaining subdiagonal entries are then already zero. The unit-lower-triangular LU decomposition isMultiplication verifies $LU=A$. Since $\gamma \ne 0$, all pivots are nonzero. Forward substitution in $Ly=b$ givesBackward substitution in $Ux=y$ gives $x_4=0$, $x_3=-a^2$, $x_2=0$, and $x_1=\gamma +a^4=1$. ThusDirect substitution yields $Ax=(\gamma ,0,0,a\gamma {)}^T$ as a check.

Assume the known variances are positive and write $\bar{x}=n^{-1}{\sum }_i{x}_i$. The likelihood times the normal prior distribution is proportional toCollecting the quadratic and linear terms in $\theta$ completes the square with precision $v^{-1}=n/{\sigma }^2+1/{\tau }^2$ and mean $m=v(n\bar{x}/{\sigma }^2+\mu /{\tau }^2)$. Thus normal-normal conjugacy givesFor quadratic loss, the posterior risk of reporting $a$ is $\mathbb{E}[(a-\theta {)}^2\mid x]=v+(a-m{)}^2$, uniquely minimized at $a=m$. For absolute-error loss, differentiating the posterior risk gives $2F_{\theta \mid x}(a)-1$, so its minimizer is a posterior median. The normal posterior is symmetric about $m$ and has a strictly increasing distribution function, hence that median is uniquely $m$. Therefore both optimal point estimates equal $m$, the posterior mean.

Let $M_{ij}={\int }_S{x}_i{x}_{j}dS$. A rotation matrix $Q$ maps the unit sphere and its surface measure to themselves. Changing variables therefore gives $Q_{ip}{Q}_{jq}{M}_{pq}=M_{ij}$, so $M$ is an isotropic second-rank tensor. Reflection symmetry also directly makes its off-diagonal entries zero, while interchange of coordinates makes its three diagonal entries equal. Its trace ishence $M_{ij}=(4\pi /3){\delta }_{ij}$. Oddness gives ${\int }_S{x}_{i}dS=0$. Expanding the defining product for $T$ now givesThe action on a vector is $Tv=\lambda v+\mu y(y\cdot v)$. For $y\ne 0$, its eigenvalues and eigenspaces are thereforeThe latter has multiplicity two. If $y=0$, the single eigenvalue $4\pi /3$ has multiplicity three and every vector is an eigenvector apart from the excluded zero vector.

Differentiability at $(a,b)$ means that there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatNecessarily $L(h,k)=f_x(a,b)h+f_y(a,b)k$ when those partial derivatives exist. Under the stated assumptions, apply the one-variable mean value theorem separately to the two coordinate increments:for some intermediate $\theta ,\eta \in (0,1)$, omitting a term if its increment is zero. Existence of the partial derivatives in a neighborhood supplies the differentiability along each segment required for that theorem. Their continuity at $(a,b)$ makes the difference from $f_x(a,b)h+f_y(a,b)k$ bounded by $(|h|+|k|)\epsilon (h,k)$, where $\epsilon (h,k)\to 0$. Since $|h|+|k|\le \sqrt{2}\sqrt{h^2+k^2}$, this is the required little-o remainder and proves Fréchet differentiability.

For the particular function, $f(h,0)=f(0,k)=0$, so the defining difference quotients give $f_x(0,0)=f_y(0,0)=0$. However $f(t,t)=1/2$ for every $t\ne 0$. It is not continuous at the origin, and therefore not differentiable there. This illustrates why existence of partial derivatives alone is insufficient.