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The space of test functions isA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{D}$ when all supports lie eventually in one compact set $K$ andfor every multi-index $\alpha$. A distribution is a linear functional $u:\mathcal{D}\to \mathbb{C}$ such that, for every compact $K$, there are $C_K>0$ and $m_K\in {\mathbb{N}}_0$ withwhenever $\mathrm{supp}\phi \subseteq K$. Convergence in ${\mathcal{D}}^{\prime }$ is pointwise convergence on test functions.

Continuity plainly implies sequential continuity. Conversely, suppose the linear form is sequentially continuous but the displayed estimate fails for some compact $K$. For every $j$, choose ${\phi }_j$ supported in $K$ such thatThen ${\phi }_j\to 0$ in $\mathcal{D}$, while $u({\phi }_j)\nrightarrow 0$, a contradiction. Hence the seminorm estimate holds on every compact set and $u\in {\mathcal{D}}^{\prime }$.

For a translation vector $h$ and a multi-index $\alpha$, defineThese definitions extend ordinary translation and differentiation to distributions.

If ${\tau }_{t{e}_i}u=u$ for every $t$, differentiating its pairing at $t=0$ gives ${\partial }_{i}u=0$. Conversely, if ${\partial }_{i}u=0$, then for every test functionThe pairing is constant in $t$, hence ${\tau }_{t{e}_i}u=u$. This proves both directions of translation invariance and vanishing distributional derivative.

Finally, the distributional differentiation obeys the linear chain rule under the linear coordinates $s=x-y$ and $t=x+y$. Thusand likewiseas distributions. Adding the two identities givesthe one-dimensional wave equation; this is the low-regularity form of the travelling waves in the D'Alembert formula.

A complex representation of $G$ is a complex vector space $V$ together with a homomorphismIt is faithful when $\ker \rho =\{e\}$, equivalently when $\rho$ is injective. Every finite group has a faithful complex representation: in the regular representation, $G$ permutes the basis $\{e_h:h\in G\}$ byand a group element fixing every basis vector must be the identity.

If $g$ is conjugate to $g^k$, then $\rho (g)$ is similar to $\rho (g^k)=\rho (g{)}^k$. Thus the two matrices have the same spectrum, while an eigenvalue $\lambda$ of $\rho (g)$ gives the eigenvalue ${\lambda }^k$ of $\rho (g{)}^k$. Hence

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\le k<p$. Faithfulness makes $\rho (g)$ have order $p$, so its spectrum contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct valuesin the spectrum. Therefore $\dim V\ge p-1$. This is the spectrum orbit bound for a faithful symmetric-group representation.

Finally let $G$ be the group of all permutations of $\mathbb{N}$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a subgroup $(C_2{)}^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the group of diagonal sign matrices, which has order $2^d$. A faithful restriction would require $2^n\le 2^d$ for every $n$, an impossibility.

The closest point theorem in a Hilbert space states that every nonempty closed convex subset $C$ of a Hilbert space has a unique point nearest to any given $x$.

Let $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The parallelogram identity and convexity giveso $(y_n)$ is Cauchy. Completeness and closedness give a limit $y\in C$ with $\Vert x-y\Vert =d$. If $y,z$ were both minimizers, the same identity with their midpoint would force $\Vert y-z\Vert =0$.

Apply the theorem to a closed subspace $F$. For $x\in H$, let $y\in F$ be closest and put $z=x-y$. For every $u\in F$, minimality of $\Vert z-tu{\Vert }^2$ at $t=0$ for real and imaginary $t$ gives $⟨z,u⟩=0$. HenceThe intersection is zero, so $H=F\oplus F^{\perp }$.

If $T$ is a shift with orthonormal basis $(e_n)$, it is an isometry,Conversely, suppose these three properties hold. Choose a unit vector $e_1$ spanning $(\mathrm{Im}T{)}^{\perp }$ and set $e_n=T^{n-1}{e}_1$. Isometry makes this sequence orthonormal. Iterating the orthogonal decompositiongivesA vector orthogonal to every $e_n$ lies in every $\mathrm{Im}(T^n)$ and is therefore zero. Thus $(e_n)$ is an orthonormal basis and $T{e}_n=e_{n+1}$, so $T$ is a shift. This is the wandering-vector characterization of a unilateral shift.

If $q\in \mathcal{Q}$ and $\varphi \in \mathcal{A}$, thenbut strict positivity of $q$ and the nonzero vector $P\varphi \ge 0$ make the left side positive. Thus the two sets cannot both be nonempty.

Conversely, suppose $\mathcal{A}=\varnothing$. The subspace $L=\mathrm{Im}P$ is disjoint from the compact simplexStrict separation gives a vector $q$ that vanishes on $L$ and is positive on all of $\Delta$. The first property says $P^{T}q=0$; testing the vertices of $\Delta$ shows every $q_i>0$. After normalization ${\sum }_i{q}_i=1$, this gives $q\in \mathcal{Q}$. This is Stiemke theorem in normalized form. Therefore

Let $P$ be the state-by-asset matrix of discounted gains and identify $Y$ with its state vector. Fix $q^0\in \mathcal{Q}$. If $q\ge 0$, $P^{T}q=0$, and $q\ne 0$, thenbelongs to $\mathcal{Q}$. The assumption and passage to the limit imply $q^{T}Y\ge 0$ for every nonnegative $q\in \ker P^T$.

The finite-state superhedging alternative now gives $\theta$ withThis residual cannot vanish identically, since then every $Q\in \mathcal{Q}$ would have $E_{Q}Y=q^{T}P\theta =0$. Hence it is strictly positive in at least one state, which has positive physical probability. This is the required inequality.

The same mixing argument with a fixed $q^0\in \mathcal{Q}$ shows thatfor every $q\ge 0$ in $\ker P^T$, including boundary vectors. By the finite-state superhedging alternative, there is $\varphi \in {\mathbb{R}}^d$ such thatwhich is the asserted almost-sure inequality. No strict residual is required here.

Let ${\epsilon }_1,\dots ,{\epsilon }_n$ be independent Rademacher signs. For the set of evaluation vectorsthe empirical Rademacher complexity isFor an independent sample $Z_1,\dots ,Z_n$,

Condition on the sample. The set of vectorshas at most $s(H,n)$ distinct members, each with Euclidean norm at most $M\sqrt{n}$. The Massart finite-class lemma, the Sauer-Shelah growth bound, and $VC(H)\le p$ giveTaking expectations and substituting this into part (i) yields

Set $c=⟨\psi |\varphi ⟩$. The off-diagonal Hermiticity identity holds automatically, whilemust be real. Thusis necessary and sufficient. For $g\ne 0$, put $r=gc/|g|\in \mathbb{R}$. Since the two normalized spanning vectors are linearly independent, $|r|<1$.

The normalized eigenvectors and eigenvalues areDirect application of $H$ verifies the eigenvalue equations. Their unnormalized inner product isso the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If $g=0$, $H=0$ and every vector is an eigenvector.

The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ give conserved specific energy and angular momentum,The normalization $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$ then rearranges toso the displayed radial energy is also constant. This is the effective potential for timelike Schwarzschild geodesics.

Let $\theta$ be the angle from the downward vertical to the hoop's centre, measured at the suspension point, and let $\varphi$ be the absolute angle from the downward vertical to the bead's radius from the hoop centre. The bead position is the sum of two vectors of length $a$. Including the hoop's moment of inertia about the pivot giveswhileThus one suitable Lagrangian is

The first register isFor $j,j^{\prime }$,The geometric sum is one when $j=j^{\prime }$ and zero otherwise. There are $N$ such orthonormal vectors in an $N$-dimensional space, so they form an orthonormal basis.

For a mixed spin-half state, the Bloch vector formula givesPhysical consistency is equivalent to the resulting Bloch vector having length at most one.

Three linearly independent Hermitian operators do not always suffice. The operatorsare linearly independent, but the first expectation is always one and none detects the $y$ component. States with Bloch vectors $(0,c,0)$ and $(0,-c,0)$ have identical expectations for all three. The correct criterion for informationally complete three-observable qubit tomography is that the three traceless parts span the Pauli space.

The original binary Hamming code has parameters $[7,4,3]$, so it has $2^4=16$ codewords. Its minimum distance three makes the radius-one balls about its codewords disjoint. Each ball hasvectors, and $16\cdot 8=128=2^7$, the size of the ambient space. Equality in the Hamming bound therefore shows that these balls partition ${\mathbb{F}}_2^7$: the code is perfect.

Orthogonal projection gives the unique componentsChooseThen $a,b\ge 0$, $a^2+b^2=1$, and the required decomposition follows. If one projection vanishes, its coefficient is zero and the corresponding normalized vector may be chosen arbitrarily in that subspace; the nonzero projected components remain unique.

Conjugating a reflection by a unitary reflects in the transported vector. Since $A|0^n⟩=|\Omega ⟩$,and every vector orthogonal to $|\Omega ⟩$ is fixed. Hencein particular both sides act identically on $|\Omega ⟩$.

Both operators map linear combinations of $|{\psi }_g⟩$ and $|{\psi }_b⟩$ to linear combinations of the same two vectors: $R_g$ merely changes one coefficient, whilehas $|\Omega ⟩\in P$. Therefore each leaves $P$ invariant.

A vector field $V=V_1{\partial }_x+\varphi {\partial }_u$ is a Lie symmetry whenon the equation manifold $\Delta =0$. With total derivative $D_x$, the prolongation coefficients obeyFor $V_1=f(x)$, direct iteration givesThus $\alpha =\beta =3$.

For $u^{\prime \prime \prime }=u^{-3}$, invariance requires ${\varphi }_3+3u^{-4}\varphi =0$ after substituting $u^{\prime \prime \prime }=u^{-3}$. Equating coefficients of the independent jet variables givesand the remaining terms force $\varphi =(3a/4)u$ and $f=ax+b$. Hence the Lie algebra is spanned by

Differentiating $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, substituting Maxwell equations, and using the stated vector identity gives the Maxwell stress tensor conservation lawMoving the divergence to the left identifies

The vectors $T={\partial }_{\tau }x$ and $S={\partial }_{\sigma }x$ are coordinate vector fields on the parameter surface, so their ordinary commutator vanishes. Torsion-freeness then givesSince each $\tau$-curve is geodesic, ${\nabla }_{T}T=0$. Apply the Ricci identity to $V=T$ and use ${\nabla }_{S}T={\nabla }_{T}S$:Thus the desired form holds with