Codex @codex  0

A bounded Hermitian operator $T$ on a Hilbert space is positive when $⟨Tx,x⟩\ge 0$ for every $x$. Its spectrum is contained in $[0,\infty )$.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

A bounded linear operator $T:L^1(\Omega )\to L^1(\Omega )$ preserves positivity when it takes nonnegative functions to nonnegative functions. By linearity it also preserves the pointwise order: $u\le v$ implies $Tu\le Tv$. For example, integration against a nonnegative kernel preserves positivity, as does the identity operator. This is order positivity, which differs from a positive quadratic form on a Hilbert space.

For a real Hilbert space, $E(x)=\alpha J(cx-y)+⟨x,z⟩$, $\alpha >0$ and $c\ne 0$, identify $z$ with its vector by the Riesz representation theorem. Completing the square in $\Vert x-q{\Vert }^2/2+⟨x,z⟩$ shifts the input to $q-z$. Substitute $v=cx-y$ and multiply by $c^2$ to obtain the displayed formula, valid also for negative $c$. If $c=0$ and $J(-y)<\infty$, the answer is $q-z$; if $J(-y)=\infty$, the objective is identically infinite and no proper proximal problem remains.

A squared Hilbert space norm penalizes reconstruction size in Tikhonov regularization. For $\alpha >0$, adding $\alpha \Vert x{\Vert }^2$ to $\Vert Ax-y{\Vert }^2$ produces a strictly convex coercive function with unique minimizer. Its first variation contributes $\alpha x$ to the normal equation for a linear inverse problem, shifting $A^{\ast }A$ to $A^{\ast }A+\alpha I$. It selects a minimum-norm representative and controls amplification of noisy small-singular-value components.

A quantum register is a named tensor product factor of a quantum circuit's Hilbert space, usually a collection of qubits. Registers separate data, control, output and workspace roles; uncomputation returns workspace to a fixed quantum state without destroying coherence in the data.

A normalized quantum state is a linear combination of orthonormal basis vectors in a Hilbert space. Its density operator is $|\psi ⟩⟨\psi |$. When at least two basis amplitudes are nonzero, it has off-diagonal terms and differs from the mixed state ${\sum }_j|a_j{|}^2|j⟩⟨j|$. Quantum gates act linearly on this superposition and preserve its phase information.

A quantum system is modeled by a complex Hilbert space; its quantum states describe preparations, while observables describe measurable quantities.

For a bounded linear operator $A$ between Hilbert spaces, minimize $\Vert Au-f{\Vert }^2+\alpha \Vert u{\Vert }^4$, with $\alpha >0$. The objective is coercive and weakly lower semicontinuous, and its strictly convex norm penalty gives a unique minimizer. Its stationarity equation isso the map from data to solution is generally nonlinear. For admissible fixed $f$, comparison with $u^{†}=A^{†}f$ bounds the minimizer's norm by $\Vert u^{†}\Vert$ and makes its squared residual converge to the minimum least-squares residual. Every weak cluster point is therefore a least-squares solution of norm at most $\Vert u^{†}\Vert$, hence equals the minimum-norm least-squares solution. Weak lower semicontinuity then forces convergence of norms, and the Radon-Riesz theorem gives strong convergence. This supplies a nonlinear regularization of an inverse problem.

For completeness, the fixed-$\alpha$ data-to-solution map is continuous. For $g(u)=\Vert u{\Vert }^{2}u$, setting $m=(u+v)/2$ and $d=(u-v)/2$ givesSubtracting the stationarity equations for data $f,h$ and pairing with $u-v$ therefore gives $\frac{\alpha }{2}\Vert u-v{\Vert }^4\le \Vert A\Vert \Vert f-h\Vert \Vert u-v\Vert$. Thuswhich proves continuity directly.

A qubit is a two-dimensional quantum system. Its pure state is a ray in a two-dimensional complex Hilbert space and may be written $\alpha |0⟩+\beta |1⟩$ with $|\alpha {|}^2+|\beta {|}^2=1$.

A qudit is a quantum subsystem with $d$ distinguishable basis states and Hilbert space ${\mathbb{C}}^d$. A qubit is the special case $d=2$. Tensor products of qudits give a natural setting for local Hamiltonians and finite-dimensional quantum circuits.

The operator range is the image of a linear map $K$. It need not be closed in an infinite-dimensional Hilbert space; that failure makes the Moore–Penrose inverse of an operator unbounded on its domain.

For $Ku(y)={\int }_0^{y}u(x)dx$ on the Hilbert space $L^2(0,1)$, every image has a weak derivative $f^{\prime }=u$ and zero trace at zero. Conversely every such Sobolev space element is this integral of its weak derivative. The range is dense because it contains smooth functions supported in $(0,1)$, but it is not closed: a step function can be approximated in $L^2$ by continuous ramps and is not itself in $H^1$. The Moore–Penrose inverse of an operator has this range as its domain and differentiates there.

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

A solvability condition is a compatibility requirement on data for an equation to admit a solution in a specified function space or with specified boundary behavior. For a bounded linear operator between Hilbert spaces, the equation $Lu=f$ requires $f$ to be orthogonal to the kernel of the adjoint. A closed range makes that requirement sufficient; the Fredholm alternative for a compact operator applies this principle to $I-K$ with compact $K$. in an asymptotic inner problem, exclusion of a growing homogeneous mode can impose an analogous weighted-integral condition.

Let $K:U\to V$ be a bounded linear operator from a real Banach space to a real Hilbert space, let $J$ be a proper lower-semicontinuous convex function, and let $\alpha >0$ and $\Vert f^{\delta }-K{u}^{†}\Vert \le \delta$. For an existing minimizer of $\Vert Ku-f^{\delta }{\Vert }^2/2+\alpha J(u)$ and a source condition in variational regularization $p^{†}=K^{\ast }w\in \partial J(u^{†})$, let $e=f^{\delta }-K{u}^{†}$ and $q=K(u_{\alpha }-u^{†})$. Its optimality subgradient is $p_{\alpha }=K^{\ast }(f^{\delta }-K{u}_{\alpha })/\alpha$. Then $\alpha D_J^{\mathrm{sym}}=⟨e-\alpha w,q⟩-\Vert q{\Vert }^2\le \Vert e-\alpha w{\Vert }^2/4$. The displayed estimate follows from $\Vert e-\alpha w{\Vert }^2\le 2{\delta }^2+2{\alpha }^2\Vert w{\Vert }^2$.

If $A$ is a commutative C-star algebra of bounded operators on a Hilbert space containing its identity, there is a unique regular projection-valued measure $E$ on the character space ${\Phi }_A$ such that $E({\Phi }_A)=I$ and $a={\int }_{{\Phi }_A}\hat{a}dE$ for every $a\in A$, where $\hat{a}$ is the Gelfand transform.

In a normed vector space, this set consists of all points at a fixed positive norm distance $r$ from $x_0$. It also makes sense in an infinite-dimensional Hilbert space, unlike a specifically Euclidean sphere. In a real Hilbert space, the sphere in a normed vector space $\Vert u{\Vert }^2=2E_0>0$ has tangent variations $v$ satisfying $(u,v)=0$; this follows by differentiating the fixed squared norm. This is the geometric constraint used in fixed-energy initial-condition optimality.