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A Hamiltonian path preserving the history subspace can be analysed using its restricted spectral gap. If this gap is $\Omega (T^{-2})$ and the derivative has operator norm $O(1)$, a runtime criterion proportional to $\Vert \dot{H}{\Vert }^2/(ϵ{\Delta }^3)$ permits computational history state preparation in time $O(T^6/ϵ)$. Other invariant sectors do not cause leakage from a state initialized in the history subspace.

For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

A bounded linear operator between normed spaces is approximable if it is an operator norm limit of bounded finite-rank operators. If the target is a Banach space, every approximable operator is a compact operator. For a given tolerance, approximate the operator in norm by one finite-rank operator, cover that operator's bounded unit-ball image by finitely many small balls in its finite-dimensional range, and transfer the cover across the operator-norm error. This makes the original image totally bounded; its closure is complete in the Banach target and therefore compact. The converse on arbitrary Banach spaces would require an additional approximation hypothesis and is not part of this assertion.

Asymptotic regularization stops the gradient flow $x^{\prime }=-A^{\ast }(Ax-f)$ at $t=1/\alpha$, starting from zero. With the singular system of a compact operator convention $A{v}_j={\sigma }_j{u}_j$, its spectral filter givesThe scalar coefficient solves $c_j^{\prime }=-{\sigma }_j^2{c}_j+{\sigma }_j⟨f,u_j⟩$ with zero initial value. Since $1-e^{-s}\le \min (s,1)$ for $s\ge 0$, the operator norm is at most ${\alpha }^{-1/2}$. On the domain of the Moore–Penrose inverse of an operator, dominated convergence theorem of the squared spectral coefficients proves $R_{\alpha }f\to A^{†}f$. The noise-bias decomposition for linear regularization then gives noisy-data convergence when $\delta /\sqrt{\alpha }\to 0$.

For a $C^1$ flux $F$ and a weak solution of $\mathrm{div}F(Du)=0$, discrete integration by parts and the fundamental theorem of calculus along a line segment show that $w_h={\delta }_{\ell ,-h}u$ solves $\mathrm{div}(A_{h}D{w}_h)=0$, with $p_{-}=Du(x-h{e}_{\ell })$ and $p_{+}=Du(x)$. Coercive quadratic-form bounds and operator norm bounds for $DF$ on all segments between these gradients pass to $A_h$. This produces a divergence-form elliptic operator without initially assuming second partial derivatives of $u$.

A Banach limit is a positive linear functional on the real l-infinity sequence space that extends the ordinary limit on the convergent sequence space, has operator norm one, and is invariant under the left shift on bounded sequences. The Hahn-Banach theorem constructs one by separating the constant sequence $1$ from the subspace $(I-S){\ell }^{\infty }$, whose distance from $1$ is one. The functional is generally not unique.

If $X$ is a normed vector space and $Y$ is a Banach space, then the bounded linear maps $\mathcal{B}(X,Y)$ form a Banach space under the operator norm. Completeness of the domain is unnecessary.

The Born series is the Neumann series expansion of the Lippmann-Schwinger equation. Its terms describe successive orders of wave scattering; convergence is ensured, for example, by a scattering operator norm smaller than one.

A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

For distinct interpolation sites and an invertible B-spline collocation matrix, the interpolant is $P=B{A}^{-1}R$, where $R$ samples data and $Ba={\sum }_i{a}_i{N}_i$. Sampling and basis synthesis have operator norms at most one in the supremum norm. For the reverse inequality, realize the signs of a maximal absolute row sum of $A^{-1}$ as values of a continuous function of norm one, and apply uniform-norm stability of a B-spline basis. This yields both displayed bounds.

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

For local outcomes $A_0,A_1,B_0,B_1\in \{-1,1\}$, the Clauser–Horne–Shimony–Holt inequality isAn arbitrary no-signalling correlation can reach the algebraic maximum $4$, while quantum correlations are bounded by $2\sqrt{2}$.

For a short proof of the local bound, fix a deterministic local hidden-variable model assignment. The combination is $A_0(B_0+B_1)+A_1(B_0-B_1)$. Exactly one bracket vanishes and the other is $\pm 2$, so every assignment contributes $\pm 2$. Averaging with a single setting-independent probability distribution preserves the interval $[-2,2]$. Mixtures also cover stochastic local hidden-variable theories. Rejecting a different sample at each setting would invalidate precisely that common-average step.

The quantum bound follows just as economically. For Hermitian operators with $A_i^2=B_j^2=I$, the CHSH inequality operator $C=A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)$ obeys $C^2=4I-[A_0,A_1]\otimes [B_0,B_1]$. Each commutator has operator norm at most two, giving $\Vert C\Vert \le 2\sqrt{2}$. The Polarization CHSH count model attains this bound without allowing either station to control the distant marginal probability distribution.

For dichotomic observables with $A_i^2=B_j^2=I$, let $\mathcal{B}=A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)$. Expanding gives ${\mathcal{B}}^2=4I-[A_0,A_1]\otimes [B_0,B_1]$. Each commutator has operator norm at most two, so $\Vert \mathcal{B}{\Vert }^2\le 8$ and $|⟨\mathcal{B}⟩|\le 2\sqrt{2}$. Appropriate spin measurements of a Bell state attain equality. The local hidden-variable bound of the corresponding CHSH inequality is two.

For a nonnegative partition-normalized B-spline basis, the synthesis linear map $Tc={\sum }_j{c}_j{N}_j$ has operator norm one. Its condition number is consequently $\Vert T^{-1}\Vert$, with the supremum norm on the spline space and the maximum norm on its coefficients. The uniform-norm stability of a B-spline basis bounds this number independently of the spline knot sequence when the order is fixed. This is different from the condition number of a particular sampled B-spline collocation matrix.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

If $H={\sum }_j{h}_j$ and all local terms commute, thenexactly. Approximating each fixed-size factor to error at most $\epsilon /m$ and applying the telescoping bound for products of operators gives total operator norm error at most $\epsilon$.

A bounded complex-linear linear functional on a complex linear subspace of a normed space extends to the whole space with the same operator norm. To deduce this from the real Hahn-Banach theorem, extend the real part $u$ under the bound $\Vert \ell \Vert \Vert x\Vert$ on the underlying real vector space, then put $L(x)=u(x)-iu(ix)$. This is complex-linear and agrees with $\ell$ on its original domain. Given $x$, choose a complex number $a$ of modulus one with $aL(x)=|L(x)|$. Then $|L(x)|=\mathrm{Re}L(ax)=u(ax)\le \Vert \ell \Vert \Vert x\Vert$. Restriction gives the reverse operator norm inequality.

The continuous dual space $X^{\ast }$ of a topological vector space is the vector space of its continuous linear functionals. Unlike the algebraic dual, it requires that a small change in an input produce a small change in its scalar measurement. For a normed vector space, this is equivalent to a finite boundfor all vectors $x$ and some constant $C\ge 0$. The least such constant is the operator norm of $\phi$.

This is the continuity-restricted subspace of the algebraic dual space; in infinite dimension the two can differ.

For the linear functional $\phi (x,y)=2x-y$ on ${\mathbb{R}}^2$ with its Euclidean norm, the Cauchy-Schwarz inequality givesEquality holds on the unit vector $(2,-1)/\sqrt{5}$, so its operator norm is exactly $\sqrt{5}$. The same estimate controls measurement error: $|\phi (v+h)-\phi (v)|\le \sqrt{5}\Vert h{\Vert }_2$. The linear functional diagram shows which changes are entirely invisible to this measurement; its kernel of a linear map consists of the changes with zero response. All linear functionals on a finite-dimensional real or complex normed vector space are continuous, but discontinuous functional detected by vanishing sequence tails shows why the restriction matters in infinite dimension.

A stationary iterative method for a linear system converges to its specified fixed point for every initial vector if and only if its iteration matrix satisfies $\rho (H)<1$, where the spectral radius is the largest modulus of an eigenvalue.^[ref]

For a diagonalizable matrix, write $H=V\Lambda V^{-1}$. Then $H^k=V{\Lambda }^k{V}^{-1}$, so each eigenvector coefficient tends to zero exactly when its eigenvalue has modulus less than one. For a general matrix, the Jordan normal form gives the same result: a Jordan block $J=\lambda I+K$ with a nilpotent matrix $K$ of size $m$ satisfiesIf $0<|\lambda |<1$, every coefficient tends to zero because exponential decay beats any fixed polynomial in $k$; if $\lambda =0$, the block vanishes once $k\ge m$. Conversely, an eigenvector with $|\lambda |\ge 1$ provides an initial error that cannot tend to zero. For a real matrix with complex eigenvalues, the same obstruction appears in the real invariant subspace spanned by the real and imaginary parts of an eigenvector.

The words "every initial vector" matter. For example, $H=\mathrm{diag}(3/2,1/2)$ fails the criterion, but $e^{(0)}=(0,1{)}^T$ still tends to zero. Here a generic initial error has a nonzero first coordinate and diverges; the exceptional errors lie on the second coordinate axis. With $H=\mathrm{diag}(-1,1/2)$, generic errors oscillate rather than diverge in norm, while the same exceptional axis still converges. Both are valid matrix splittings with $M=I$ and $A=I-H$. In general the convergent errors form the direct sum of the generalized eigenspaces for $|\lambda |<1$; if $\rho (H)\ge 1$, this is a proper linear subspace.

A small spectral radius describes eventual convergence, but does not imply that the Euclidean norm of the error decreases on every step for a non-normal matrix. An operator norm bound $\Vert H\Vert <1$ does give the stepwise estimate $\Vert e^{(k)}\Vert \le \Vert H{\Vert }^k\Vert e^{(0)}\Vert$. For a real symmetric matrix, the matrix 2-norm equals its spectral radius, making this particularly direct.

For proper lower semicontinuous convex functions on finite-dimensional Hilbert spaces, a nonempty saddle point set, and constant positive steps satisfying the displayed inequality, the primal-dual hybrid gradient method with extrapolation parameter one converges to a saddle point. The strict condition controls the bilinear coupling by the operator norm. It does not remove the need for existence of a saddle point. The fixed-step theorem and assumptions are recalled in Malitsky and Pock, A first-order primal-dual algorithm with linesearch, Section 1.