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Put $\delta =\exp (-{\epsilon }^{-C})$, with the absolute constant $C$ chosen sufficiently large below, and suppose for a contradiction that no $g_2$ with $0\le g_2\le 1$ satisfies $\Vert g_1-g_2{\Vert }_{\mathcal{F}}\le \epsilon$.

LetThe set $K$ is compact and convex, while $B$ is closed and convex, so $K+B$ is closed and convex. We use the following finite-dimensional form of the Hahn-Banach separation theorem: if a point lies outside a nonempty closed convex set, there is a linear functional whose value at the point is strictly greater than its supremum over that set. Identifying linear functionals on ${\mathcal{H}}_N$ through the inner product, there is therefore a function $\psi$ such thatThe separating functional cannot have zero dual norm, so rescale it to make $\Vert \psi {\Vert }_{\mathcal{F}}^{\ast }=1$. Because $\mathcal{F}$ is closed, convex, and symmetric, the finite-dimensional Bipolar theorem for a dual pair says that the unit ball of $\Vert \cdot {\Vert }_{\mathcal{F}}^{\ast }$ is precisely $\mathcal{F}$. Thus $\psi \in \mathcal{F}$ and $|\psi (n)|\le 1$.

The support function of $K$ is obtained by choosing $g(n)=1$ where $\psi (n)>0$ and $g(n)=0$ where $\psi (n)<0$. In terms of the positive part of a real-valued function ${\psi }_{+}=\max (0,\psi )$, the separating inequality becomesSince $0\le g_1\le \nu$, pointwise we have $g_1\psi \le \nu {\psi }_{+}$, and hence

Apply the supplied polynomial approximation of the positive part to $\psi$. If $P_{\epsilon }(t)={\sum }_{i=0}^m{a}_i{t}^i$, then its uniform approximation error and $\mathbb{E}\nu \le 1$ implyThe constant function $1$ and $\psi$ belong to the dual unit ball. By the assumed submultiplicativity, $\Vert {\psi }^i{\Vert }_{\mathcal{F}}^{\ast }\le 1$ for every $i\ge 0$. Dual seminorm therefore givesThe stated coefficient bound, with $C$ in $\delta$ chosen larger than the absolute constant in that bound, makes this last quantity at most $\epsilon /50$. Together with the polynomial-approximation error, this contradicts $⟨\nu -1,{\psi }_{+}⟩>\epsilon$. The required $g_2$ consequently exists. This proves the dense model theorem for a multiplicative test family.

Write $H={\partial }_x^2-V(x)$. The real potential and periodic boundary condition make $H$ a self-adjoint operator on the periodic domain. Since $u_t=-iHu$,because the inner product $⟨u,Hu⟩$ is real. Thus the Schrodinger equation preserves the $L^2$ norm.

For two nonorthogonal input states $|\alpha ⟩,|\beta ⟩$, preservation of inner products by $G$ givesHence the two output-program states agree up to phase. Any two qubit states can be joined through a state nonorthogonal to both, so after absorbing phases
$|{\tilde{P}}_U⟩$ is independent of the input.

A real matrix is orthogonal when $Q^{T}Q=I$. If $Qv=\lambda v$, norm preservation gives $|\lambda |=1$. If $Qv=\lambda v$ and $Qw=\mu w$, preservation of the Hermitian inner product gives $(1-\overline{\lambda }\mu )v^{†}w=0$; distinct unit-modulus eigenvalues therefore have orthogonal eigenvectors.

The nonreal eigenvalues of a real $3\times 3$ matrix occur in conjugate pairs. Since their product is one and $\det Q=-1$, the remaining real eigenvalue is $-1$. If $x\cdot n=0$, then $(Qx)\cdot (Qn)=x\cdot n=0$, and $Qn=-n$, so $Qx\cdot n=0$: the plane $\Pi$ is invariant.

The restriction to $\Pi$ is a planar orthogonal map whose determinant is $(-1)/(-1)=1$, hence a rotation through some $\theta$. In an orthonormal basis adapted to $n$,Therefore $\mathrm{tr}Q=-1+2\cos \theta$ and

The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Associate to each set $A_i\subseteq [n]$ its characteristic vector of a set $v_i\in {\mathbb{F}}_2^n$. The hypotheses sayIf ${\sum }_i{c}_i{v}_i=0$, taking the dot product with $v_j$ gives $c_j=0$. Thus the vectors are linearly independent, so $m\le \dim {\mathbb{F}}_2^n=n$. Equality is attained by the $n$ singleton sets.

For the Volterra integration operatorthe adjoint operator isSince $1/{\sigma }_n=(2n-1)\pi /2$, direct integration givesSimilarly,because $\cos ((2n-1)\pi /2)=0$. The half-integer sine and cosine families are orthonormal bases of $L^2([0,1])$, so $({\sigma }_n,u_n,v_n)$ is a singular system of a compact operator.

The Tikhonov solution for noisy data $g^{(\delta )}$ is thereforeWriting out the inner product and the singular functions makes this explicit:The factors ${\sigma }_n/({\sigma }_n^2+\alpha )$ suppress the unstable reciprocal growth $1/{\sigma }_n$ at high index.

Use the sign conventionfor the Lagrangian function in constrained optimization. The Lagrangian dual problem iswhere $f^{\ast }$ is the convex conjugate. For this convex problem with affine equality constraints, the stationarity and feasibility parts of the Karush-Kuhn-Tucker conditions areThey say exactly that the displayed operator satisfiesThus its zeros are precisely the primal-dual optimal points, subject to the usual attainment assumptions.

For $u=(x,z)$ and $\tilde{u}=(y,s)$, the Euclidean inner product givesThe last two terms cancel by the defining property of the matrix transpose, and the first is nonnegative by part a. Hence $F$ is a monotone operator.

Set $\overline{\mathbf{y}}=({\mathbf{y}}_n+{\mathbf{y}}_{n+1})/2$. The implicit midpoint rule readsTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n=2\overline{\mathbf{y}}$ givesThe last equality again follows from skew-symmetry. Mathematical induction therefore yields

Write $A_n=A({\mathbf{y}}_n)$. The trapezoidal rule isTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n$ givesThe two quadratic terms vanish because each $A_j$ is skew-symmetric. Moreover ${\mathbf{y}}_{n+1}^T{A}_n{\mathbf{y}}_n=-{\mathbf{y}}_n^T{A}_n{\mathbf{y}}_{n+1}$. HenceUnlike the implicit midpoint rule, the trapezoidal rule evaluates $A$ at two different states, so the mixed terms need not cancel.

Take the real $L^2$ inner product of the equation with $u$. The homogeneous Dirichlet boundary conditions and integration by parts giveThus $\Vert u(t){\Vert }_{L^2}\le \Vert u_0{\Vert }_{L^2}$, and applying the same estimate to the difference of two solutions gives continuous dependence on the initial data. The constant-advection term is skew-symmetric under these boundary conditions and contributes no energy. Hence

Every term in the free energy contracts the vector components with the Euclidean inner product: it depends only on $|\mathbf{p}{|}^2$, $|\mathbf{p}{|}^4$, $({\nabla }_i{p}_j)({\nabla }_i{p}_j)$, and $|{\nabla }^2\mathbf{p}{|}^2$. A constant $\mathcal{R}\in SO(3)$ preserves all these contractions and commutes with spatial differentiation. Thereforeso every constant rotation of candidate (ii) has the same free energy. The orientation of the rotation plane of the polar helical smectic is thus continuously degenerate.

Suppose first that the stated unitary and environment states exist. Preservation of the inner product givesThe environment states are normalized, so the Cauchy-Schwarz inequality gives $|⟨e_0|e_1⟩|\le 1$. Therefore

Conversely, writeand assume $|a|\le |b|$. If $b\ne 0$, put $c=a/b$, so $|c|\le 1$, and chooseThen $⟨e_0|e_1⟩=c$ and the desired output vectors have inner product $bc=a$, exactly matching the input vectors. If $b=0$, the inequality forces $a=0$, and one may take $|e_0⟩=|e_1⟩=|0⟩$.

In either case the input pair and output pair have the same Gram matrix. The isometry taking one pair to the other extends to a unitary matrix, so the required $U$ exists. Hencewhich is the environment-assisted two-state pure-state transformation criterion.

Condition on the information available before time $n$. The current wealth is then fixed, while ${\xi }_n\sim N(b,\Sigma )$ is independent of that information. PutFor any portfolio $\theta$, use the inner product induced by the positive-definite matrix $\Sigma$ to writeThe linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, andThus independence of uncorrelated jointly normal variables makes the residual return ${\eta }^T{\xi }_n$ independent of the aligned return; it also has expected value zero.

The continuation value $V(n,\cdot )$ is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda <0$, the portfolios $\lambda {\Sigma }^{-1}b$ and $-\lambda {\Sigma }^{-1}b$ have the same return variance, namely ${\lambda }^{2}B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot )$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda \ge 0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables ${\lambda }_n$ such thatThis is the Gaussian one-fund theorem.

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.