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A metric space metric $d:X\times X\to [0,\infty )$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the triangle inequality. A subset $U\subseteq X$ is an open set when every $x\in U$ has some $\epsilon >0$ with the open ball $B_d(x,\epsilon )\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the topology axioms.

For $C[0,1]$, the Cauchy-Schwarz inequality gives $d_1(f,g)\le d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikesThenThus $f_n$ converges to zero in the L1 norm but not in the L2 norm. The two metrics induce different topologies.

The L1 norm properties of absolute homogeneity and the triangle inequality follow from the corresponding properties of the absolute value and the integral. If $\Vert f{\Vert }_1=0$, the nonnegative continuous function $|f|$ has zero integral. Were $|f(x_0)|>0$, continuity would make it bounded below by a positive number on a nontrivial interval, contradicting the zero integral. Thus $f=0$, proving positive definiteness.

Put $M=\Vert f{\Vert }_{\infty }$ and choose the continuous cutoff functionThen $g_n(0)=g_n(1)=0$, $\Vert g_n{\Vert }_{\infty }\le M$, and $f-g_n$ is supported in the two endpoint intervals of total length $2/n$. Therefore

Now suppose ${\int }_0^{1}fg=0$ for every $g\in \mathcal{S}$. Apply this to the above $g_n$. By the uniform norm bound,Hence ${\int }_0^1{f}^2=0$, and continuity again gives $f=0$.

The dual vector space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all linear functionals on $V$. For a vector subspace $U\le V$, its annihilator of a vector subspace is

Given a basis $x_1,\dots ,x_n$, define $x_i^{\ast }$ by $x_i^{\ast }(x_j)={\delta }_{ij}$ and extend linearly. If ${\sum }_i{a}_i{x}_i^{\ast }=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^{\ast }$ are linearly independent. Every $\varphi \in V^{\ast }$ satisfiesso they also span $V^{\ast }$ and therefore form its dual basis.

Write $\varphi =a{x}_1^{\ast }+b{x}_2^{\ast }+c{x}_3^{\ast }+d{x}_4^{\ast }$. The condition $\varphi \in U^0$ isGaussian elimination gives $b=-2c-3d$ and $a=c+2d$. Hence

This is the Gram-Schmidt process applied degree by degree. Inductively, subtracting projections onto $p_0,\dots ,p_n$ makes $p_{n+1}$ orthogonal to every earlier polynomial. Since every subtracted polynomial has degree at most $n$, the leading coefficient of the monic $q_{n+1}$ remains one, so $p_{n+1}$ is a real monic orthogonal polynomial of degree exactly $n+1$.

Now take $q_{n+1}=x{p}_n$. For $k\le n-2$, symmetry of the inner product givesbecause $x{p}_k$ has degree at most $n-1$. Only the $p_n$ and $p_{n-1}$ projections remain, yielding the three-term recurrence for monic orthogonal polynomialswhereThe second identity uses $x{p}_{n-1}=p_n$ plus lower-degree terms.

For the symmetric Legendre inner product, parity gives ${\alpha }_n=0$. Starting with $p_0=1$, one obtains $p_1=x$, thenThus the first four monic Legendre polynomials are

Write the jump from side $-$ to side $+$ as $[\mathbf{F}]={\mathbf{F}}_{+}-{\mathbf{F}}_{-}$. A thin rectangular loop in the static Faraday's law and a thin pillbox in Gauss's law for magnetism giveThus the tangential electric field and normal magnetic field are continuous in the plane's rest frame.

For the boost $\mathbf{v}=v\mathbf{n}$, the Lorentz transformation of electromagnetic fields givesThereforeMoreover $[\mathbf{E}]$ is purely normal, so $[{\mathbf{B}}_t^{\prime }]=\gamma [{\mathbf{B}}_t]$. Consequently the transformed tangential boundary condition isUsing the magnetic-field jump across a surface current, $\mathbf{n}\times [\mathbf{B}]={\mu }_0\mathbf{K}$, this can equivalently be writtenThe nonzero jump is consistent with Faraday's law because the magnetic-field discontinuity itself moves through Albert's frame.

In atomic units the time-independent Schrödinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

For a variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, the first variation isTwo applications of integration by parts remove derivatives from $\eta$. The boundary terms vanish, and the fundamental lemma of the calculus of variations gives the higher-order Euler-Lagrange equation

For the stated integrand,so the equation reduces toThe general solution of this linear ordinary differential equation isDecay and finiteness of the integral force $A=B=0$; $y(0)=1$ gives $C=1$, and $y^{\prime }(0)=2$ gives $D=3$. Hence the unique stationary function is