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Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

The minimal polynomial $m_T$ is the monic polynomial of least polynomial degree such that $m_T(T)=0$. Here a polynomial $p(x)={\sum }_j{c}_j{x}^j$ is evaluated at the linear operator by $p(T)={\sum }_j{c}_j{T}^j$, with the constant term acting as $c_{0}I$.

An annihilating polynomial exists because $I,T,T^2,\dots$ lie in the finite-dimensional vector space ${\mathrm{End}}_F(V)$ and are therefore eventually linearly dependent. Uniqueness follows by subtracting two monic annihilators of the same least degree: their difference has smaller degree and still annihilates $T$.

Equivalently,For the forward implication evaluate the factorization at $T$. For the reverse implication divide $p$ by $m_T$ using polynomial division; the remainder annihilates $T$, so minimality makes it zero. This also identifies the annihilating polynomials as the principal ideal generated by $m_T$.

Let $P=b(T)q(T)$ and $Q=a(T)p(T)$. By part (ii), these are complementary projection operators with images $U=\ker (T-\lambda I{)}^{\alpha }$ and $W=\ker (T-\mu I{)}^{\beta }$. Thus $D=\lambda P+\mu Q$ acts as $\lambda I$ on $U$ and $\mu I$ on $W$.

The linear operator $T$ preserves both summands because it commutes with its own polynomials. Therefore $N=T-D$ preserves them too, and its restrictions satisfyPut $k=\max (\alpha ,\beta )$. Both restrictions of $N^k$ vanish, so the direct sum givesHence $T-D$ is a nilpotent linear map. The operator $D$ is diagonalizable, since its two summands are its eigenspaces, and it commutes with $T$. This illustrates the Jordan–Chevalley decomposition in the case of two distinct eigenvalues.

Write $U=\ker p(T)$, $W=\ker q(T)$. The linear operator $E=a(T)p(T)$ vanishes on $U$. On $W$, the identity $E=I-b(T)q(T)$ gives $Ew=w$.

By the direct sum in part (i), every vector has a unique expression $u+w$, andConsequentlyThe complementary projection operator is $I-E=b(T)q(T)$, onto $U$ along $W$. These are polynomial projections for a coprime decomposition. In particular the projection does not depend on which Bézout coefficients are chosen: its action on the two summands uniquely determines it.

Evaluate Bézout's identity for polynomials at $T$:All these linear operators commute because they are polynomials in the same operator. Thus every $v\in V$ can be writtenThe first summand belongs to $\ker p(T)$ and the second to $\ker q(T)$, since applying the corresponding operator introduces the product $p(T)q(T)=0$.

If $v$ lies in both kernels, the same identity gives $v=0+0=0$. Therefore the sum is direct:This is the kernel decomposition for coprime polynomials. It uses both annihilation by $pq$ and the coprimality expressed by the Bézout identity.

To prove the sum is direct, suppose ${\sum }_{j=1}^r{v}_j=0$ with $v_j\in V_{{\lambda }_j}$. Fix $i$ and apply the polynomial in the linear operatorEvery $v_j$ with $j\ne i$ is killed by one factor. On $v_i$, each factor is multiplication by ${\lambda }_i-{\lambda }_j$, givingThe scalar coefficient is nonzero because the eigenvalues are distinct. Thus $v_i=0$ for every $i$, provingIn particular, choosing a basis in each eigenspace yields a linearly independent combined list.

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.

If the real part vanishes identically, the norm-difference identity becomes equality, soThus the same substitution as above gives $\Vert Sz\Vert =\Vert z\Vert$ for every $z$. To prove surjectivity as well, observe that both $T$ and $-T$ satisfy the nonnegative-real-part hypothesis. The preceding part makes both $I+T$ and $I-T$ invertible. Their product $S=(I-T)(I+T{)}^{-1}$ is therefore invertible and onto. It is a surjective norm-preserving linear operator, so the equivalences in (a) prove

Write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$ and $D_z=x{\partial }_y-y{\partial }_x$. The product rule applied to a smooth test function $f$ givesThe mixed second partial derivatives cancel because they commute. Since the orbital angular momentum components are ${\hat{j}}_i=-i\hslash D_i$, their commutator isThis is one of the orbital angular momentum commutation relations; the Planck constant enters through the definition of each linear operator.

For Banach spaces $X,Y$, a nonlinear compact operator $T:X\to Y$ is a continuous map that sends every bounded set to a relatively compact subset of $Y$: the closure of its image is a compact set. For a continuous map, the image condition is equivalent to requiring that every bounded sequence $(x_j)$ has a subsequence along which $T(x_j)$ converges in the norm topology of $Y$. For a linear operator, this is the usual compact operator definition; the continuity requirement is included explicitly here because the subsequent applications are nonlinear.

The compact-domain version of the Schauder fixed-point theorem saysA useful equivalent version replaces $K$ by a nonempty closed bounded convex set $C$ and assumes $T:C\to C$ is a continuous map with relatively compact image. This latter form will be used for the partial differential equation. Both forms assert existence of a fixed point, without asserting uniqueness.

A linear operator has closed graph when $\{(x,Tx):x\in X\}$ is closed in $X\times Y$. The closed graph theorem states that an everywhere-defined linear operator between Banach spaces is bounded if and only if it has closed graph. Boundedness gives the closed-graph property by continuity. Conversely a closed graph is Banach in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. Projection of the graph onto $X$ is a bounded linear bijection, so the open mapping theorem makes its inverse bounded. Composing that inverse with the second-coordinate projection proves $T$ is bounded.

For the composition assertion, suppose $x_n\to x$ and $T{x}_n\to y$. Boundedness of $S$ gives $ST{x}_n\to Sy$, and boundedness of $ST$ gives $ST{x}_n\to STx$. Injectivity of $S$ forces $y=Tx$. Hence $T$ has closed graph and is bounded by the theorem.

For the derivative bound take $Y=C^1[0,1]$ with norm $\Vert v{\Vert }_{C^1}=\Vert v{\Vert }_{\infty }+\Vert v^{\prime }{\Vert }_{\infty }$, and $Z=C[0,1]$ with its supremum norm; both are Banach. Regard $T(x)=Ux$ as a map into $Y$, and let $S:Y\to Z$ be inclusion. This inclusion is bounded and injective, and $ST=U$ is bounded by hypothesis. The preceding result gives a constant $c$ with $\Vert Ux{\Vert }_{C^1}\le c\Vert x\Vert$. In particular $|(Ux{)}^{\prime }(t)|\le c\Vert x\Vert$ uniformly in $t$.

A linear operator $T:X\to Y$ is a compact operator if the image of the closed unit ball has compact closure in $Y$. Equivalently, every bounded sequence $(x_n)$ in $X$ has a subsequence whose images converge in the norm of $Y$. This implies boundedness of $T$.

The Schauder theorem for compact operators says that for a bounded linear operator between Banach spaces,where the dual map of bounded linear operator is $T^{\ast }f=f\circ T$. The forward implication also holds for normed spaces; completeness of the target gives the usual converse. The compactness definition requires the image subsequence's limit to belong to $Y$, so that completeness qualification matters when stating the theorem beyond Banach spaces.

Here the intended space must consist of bounded linear operators, namely Hilbert-Schmidt operators. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic linear span of $(e_k)$, for example $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$. Extend the linearly independent family $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis. Define a linear functional $\varphi$ on this Hamel basis by $\varphi (v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined linear operator $Rx=\varphi (x)e_1$ satisfiesIts printed column sum is zero although $R\ne 0$. More explicitly, $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined adjoint operator of the bounded kind used in the question. This proves finite column sum does not imply boundedness. We use $T\in \mathcal{B}(X)$ in the intended argument below.

By the Parseval identity for a Hilbertian basis for $(e_j)$, followed by the adjoint operator identity and then the Parseval identity for a Hilbertian basis for $(h_k)$,The interchange is valid by Tonelli's theorem for nonnegative double series, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals ${\sum }_k\Vert T{e}_k{\Vert }^2$. HenceIn particular the Hilbert-Schmidt norm is independent of the Hilbertian basis, and $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

On the complex vector space $\mathrm{CF}(G)$ of class functions, define the linear operatorThis is a complex-linear map and satisfies $T^2=I$. It is well-defined on class functions, because inversion of conjugacy classes takes an entire conjugacy class $K$ to another conjugacy class $K^{-1}$.

Use first the basis consisting of the indicator functions $1_K$ of the conjugacy classes. The linear operator $T$ permutes this basis by $1_K\mapsto 1_{K^{-1}}$. A permutation matrix has trace equal to the number of fixed basis vectors, soThe right-hand side counts the real conjugacy classes.

By character orthogonality and completeness, the irreducible characters are another basis of $\mathrm{CF}(G)$. Part (a) gives $T\chi =\overline{\chi }$. The pointwise complex conjugate of an irreducible character is again an irreducible character: conjugate the representing matrices, and conjugate any hypothetical proper invariant subspace back to contradict irreducibility. Thus $T$ also permutes this basis, andHere $\chi =\overline{\chi }$ means exactly that $\chi$ is a real-valued character. Equality of the trace in these two bases provesThis is the real character and real conjugacy class counting theorem. The linear operator used in the proof is $f\mapsto f\circ \mathrm{inv}$; on the irreducible character basis it agrees with pointwise complex conjugation, although pointwise complex conjugation itself is not a complex-linear map on all class functions.

The spatial mean is conserved: averaging every term in the partial differential equation over the periodic cell gives ${\partial }_t⟨T⟩=0$. Adding a constant to $T$ also preserves the equation. We therefore fix the mean, conveniently at zero. This qualification is essential: without it the constant Fourier mode contributes an additional neutral direction, so the centre subspace would have five real dimensions rather than four.

On a Fourier mode $e^{i(mx+ny)}$, the linearized linear operator has eigenvalueOn the fixed-mean subspace, the first eigenvalues to reach zero are those with $m^2+n^2=1$. At $\mu =1$ these are the modes $e^{\pm ix},e^{\pm iy}$; every other nonconstant Fourier mode is strictly damped. The reality condition leaves two complex amplitudes, hence a four-real-dimensional centre manifold on the fixed-mean subspace. The trivial state loses linear stability as $\mu$ increases through one.

Use slow time $\tau ={\epsilon }^{2}t$, and write $T=\epsilon W+{\epsilon }^3{T}_3+\cdots$, whereThere is no quadratic nonlinearity, so no forced second-order correction is needed. At order ${\epsilon }^3$, with $\mu =1+{\epsilon }^2\nu$, projection onto the critical Fourier modes givesThe correction $T_3$ does not contribute to the critical projection because the linear operator vanishes on those modes. To find the nonlinear coefficient of $e^{ix}$, split the divergence term intoThe first term contributes $-3|A{|}^{2}A$. In the second, only the zero-$y$ Fourier coefficient of $W_y^2$ contributes; it is $2|B{|}^2$, and differentiating $W_x$ once more gives $-2|B{|}^{2}A$. The third term has zero $y$-average, and the fourth has no $x$-dependent critical component. Interchanging $x,y$ gives the other amplitude equation. Thus the conserved-mean convection amplitude equations areTo establish the selected planform, put $A=r{e}^{i\theta }$, $B=s{e}^{i\psi }$. The phases are constant and the magnitudes obeyFor $\nu >0$, the square state has $r=s=\sqrt{\nu /5}$. Its magnitude Jacobian matrix is $-(2\nu /5)\left(\begin{array}{cc}3& 2\\ 2& 3\end{array}\right)$, with eigenvalues $-2\nu$ and $-2\nu /5$. Its two neutral phases correspond to translation symmetry. The pure roll has $r^2=\nu /3,s=0$; an infinitesimal $B$ disturbance grows at rate $\nu -2\nu /3=\nu /3>0$. The other pure roll is equivalent, while the trivial state is unstable for $\nu >0$. Consequently squares, with $|A{|}^2=|B{|}^2=\nu /5$, are the stable small-amplitude planform, modulo translations. For $\nu <0$ the trivial state is stable.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

Fix a clamped nondecreasing spline knot sequence $t_1,\dots ,t_{n+k}$ on $[a,b]$, with $t_1=\cdots =t_k=a$ and $t_{n+1}=\cdots =t_{n+k}=b$. For continuous splines assume interior multiplicities at most $k-1$, and assume $t_i<t_{i+k}$. An order-$k$ spline has degree at most $k-1$ on each nonempty knot interval; at a knot of multiplicity $r$ it has $k-1-r$ continuous derivatives. Its dimension is $n$, and a normalized B-spline basis $N_1,\dots ,N_n$ has local support, nonnegative values and sum one on $[a,b]$.

These properties are visible in the Cox-de Boor recurrence. Starting with interval indicators at order one, setwith a term of zero denominator defined to be zero. On each contributing support the weights are nonnegative, and summing the recurrence makes the coefficients of each lower-order B-spline add to one. This gives partition of unity and $\mathrm{supp}{N}_{i,k}=[t_i,t_{i+k}]$, with endpoint values defined by limits.

At distinct ordered sites $x_1<\cdots <x_n$, interpolation is the linear systemIt has a unique solution for every data vector exactly when the B-spline collocation matrix is invertible. The Schoenberg–Whitney theorem characterizes this by $N_i(x_i)>0$ for every $i$; away from clamped endpoints this is $t_i<x_i<t_{i+k}$. At the clamped endpoints the first and last B-splines have value one, so the endpoint version uses this positive-value formulation rather than the strict inequalities. Both ordering and the support conditions matter. A spline space with interior knots is generally not a Chebyshev system on the entire interval: a locally supported B-spline vanishes on an interval. Thus arbitrary distinct nodes need not permit interpolation. Merely specifying all data sites as knots also need not fix the remaining freedom; for example, cubic interpolation with simple interior knots at $m-2$ data sites has dimension $m+2$, and two extra boundary conditions are needed to interpolate $m$ values uniquely.

The positivity governing this existence theorem is total nonnegativity of B-spline collocation matrices: every square minor with increasing row and column indices is nonnegative. In spline terminology this is often called total positivity, but it permits zero minors caused by disjoint supports. It is not the stronger assertion that every minor is strictly positive. A useful proof uses knot insertion. Each refinement step changes coefficients bywith the unchanged endpoint pieces. The refinement matrix is nonnegative bidiagonal, and its minors are nonnegative: any nonzero determinant has its allowed diagonal matching and is a product of nonnegative entries. Products preserve this property by the Cauchy–Binet formula. Insert each sampling site to full multiplicity, so evaluation at that site becomes selection of an ordered refined coefficient. The collocation matrix is then an ordered row submatrix of the refinement product, proving total nonnegativity. The strict support criterion is the strict part of this argument: an ordered path through the consecutive support intervals exists exactly when every diagonal B-spline value is positive, giving the Schoenberg–Whitney theorem determinant criterion.

If $A$ is invertible, its determinant is positive and the adjugate identity yieldsThis checkerboard inverse of a totally nonnegative matrix is the central stability fact. With alternating data $e_i=(-1{)}^i$, the absolute row sums satisfyFor general data, the cardinal splines are ${\ell }_i(x)={\sum }_j{N}_j(x)(A^{-1}{)}_{ji}$. The spline interpolation operator and its exact Lebesgue constant of interpolation areThe equality for the operator norm follows by extending the signs at distinct sites to a continuous unit-norm data function, as in Solution 2. That solution also shows the last inequality can be very strict. Stability concerns the range function as well as its coefficients.

An optimal interpolation set must be specified relative to an objective. For the normalized B-spline coefficients, every admissible sampling set satisfiessince $c(s)=A_{\mathbf{x}}^{-1}(s(x_i))$ and sampling cannot increase the supremum norm. If a unit-norm spline $s_{\ast }$ has $n$ ordered alternating extrema $s_{\ast }(x_i)=(-1{)}^i$ at admissible sites, the checkerboard identity givesTaking the largest row shows $\Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }\le \kappa (\mathcal{S})$, so equality holds. This proves the optimal spline coefficient interpolation sites property and the Chebyshev spline coefficient and dual norm equality. In particular, for the cubic space of Solution 1 the alternating extrema $-1,-2/3,0,2/3,1$ satisfy the support conditions. Its coefficients have maximum magnitude $11/2$, so these sites are optimal for coefficient recovery andThe argument proves optimality whenever this admissible equioscillating spline is available; it does not identify extrema of an arbitrary spline as optimal sites, or transfer coefficient optimality automatically to the Lebesgue constant of interpolation.

For the intrinsic interpolation objective, minimize ${\Lambda }_{\mathbf{x}}$ itself. There is a minimizer for a fixed finite-dimensional continuous spline space. Indeed the sites lie in a compact ordered simplex. On admissible sets, inverse entries and hence ${\Lambda }_{\mathbf{x}}$ vary continuously. As a collocation matrix tends to a singular one, its inverse norm diverges. The uniform-norm stability of a B-spline basis gives the complementary boundchoose sample signs attaining a largest inverse row, extend them to a continuous unit-norm function, and use $\Vert Tc{\Vert }_{\infty }\ge \Vert c{\Vert }_{\infty }/\kappa (\mathcal{S})$. Thus a minimizing sequence with bounded Lebesgue constants of interpolation cannot approach a singular site set. A convergent subsequence has an admissible limit attaining the minimum.

Fekete interpolation sites for a continuous function space offer another useful, constructive criterion: maximize the absolute evaluation determinant. A nonzero maximum exists by compactness and linear independence. Replacing one row by evaluation at $x$ multiplies the determinant by the corresponding cardinal function, so maximality gives $|{\ell }_i(x)|\le 1$, and therefore ${\Lambda }_{\mathbf{x}}\le n$. These sites are determinant-optimal, with a proven interpolation bound; one should not claim without further argument that they minimize ${\Lambda }_{\mathbf{x}}$. Greville abscissae provide inexpensive admissible sites for the usual continuous clamped spline spaces of order at least two. Admissibility alone does not give the best stability bound.

Finally, the coefficient condition number of a normalized B-spline basis is bounded by a constant depending only on order. To see the mechanism, for each coefficient choose a longest nonempty knot cell in that B-spline's support. If its length is $h$ and the support width is $W$, then $W/h\le k$. On that cell the spline is a degree-at-most-$k-1$ polynomial; repeated Markov polynomial inequalities bound its $r$th derivative by $C_k{h}^{-r}\Vert s{\Vert }_{\infty }$. In the De Boor–Fix spline coefficient functional, the accompanying derivative of the knot polynomial is bounded by $C_k{W}^r$. Each term is consequently at most $C_k(W/h{)}^r\Vert s{\Vert }_{\infty }$, yielding $\Vert c(s){\Vert }_{\infty }\le C_k\Vert s{\Vert }_{\infty }$ independently of the number and spacing of knots. This is basis stability; it does not bound $\Vert A_{\mathbf{x}}^{-1}\Vert$ at arbitrarily poor sampling sites.

For approximation, reproduction gives the Lebesgue inequality for approximation projectorsAs the maximum knot gap tends to zero with fixed order, spline quasi-interpolation gives an approximant with error at most $\omega (f,kh)$, so the infimum tends to zero. A sequence of uniformly bounded interpolation linear operators therefore converges to every continuous target. Local support and a stable choice of sites make spline interpolation effective; mesh refinement without control of the interpolation operator norm is not by itself a convergence proof.

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.

Proceed by induction on $\dim V$. The zero-dimensional case has the empty orthonormal basis. In positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a unit eigenvector $e$ of $\alpha$, say with eigenvalue $\lambda$. Part (ii) makes it an eigenvector of ${\alpha }^{\ast }$ as well. Thus the line spanned by $e$ and its orthogonal complement are invariant under both operators by the introductory adjoint argument.

On $\{e{\}}^{\perp }$, the restriction of ${\alpha }^{\ast }$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted linear operator is therefore normal. By induction it has an orthonormal basis of eigenvectors. Adjoining $e$ proves an orthonormal eigenbasis exists for $V$, establishing the finite-dimensional spectral theorem by construction.

For the final deduction, use this orthonormal basis $e_j$ and write $\alpha e_j={\lambda }_j{e}_j$. DefineThe real nonnegative diagonal entries make $\beta$ a Hermitian operator, and the modulus-one diagonal entries make $\gamma$ a unitary operator. Both are diagonal in the same orthonormal basis, so they commute and $\beta \gamma e_j={\lambda }_j{e}_j$ even when ${\lambda }_j=0$. This proves the required factorization without assuming invertibility.

Conversely, if $\alpha =\beta \gamma$ with ${\beta }^{\ast }=\beta$, ${\gamma }^{\ast }\gamma =\gamma {\gamma }^{\ast }=I$, and $\beta \gamma =\gamma \beta$, then ${\alpha }^{\ast }={\gamma }^{\ast }\beta$ andThus $\alpha$ is normal. Normality is equivalent to a commuting Hermitian–unitary factorization, with the Hermitian factor even selectable nonnegative. This is the polar decomposition with a unitary extension on the kernel.

Use an inner product linear in its first argument. The adjoint operator ${\alpha }^{\ast }$ is the unique linear operator such thatIn a finite-dimensional inner product space, existence follows by choosing an orthonormal basis: the matrix of the adjoint is the conjugate transpose of the matrix of $\alpha$. Uniqueness follows from nondegeneracy of the inner product.

If $\alpha (W)\subseteq W$, then for $w\in W$ and $v\in W^{\perp }$,Thus ${\alpha }^{\ast }v\in W^{\perp }$. Conversely, if ${\alpha }^{\ast }(W^{\perp })\subseteq W^{\perp }$, the same equality gives $⟨\alpha w,v⟩=0$ for every $v\in W^{\perp }$, so $\alpha w\in (W^{\perp }{)}^{\perp }=W$. This proves the adjoint criterion for an invariant orthogonal complement.

The equality $(W^{\perp }{)}^{\perp }=W$ uses the finite-dimensional setting appropriate here. In a general Hilbert space it holds for closed subspaces, and an adjoint is assured for bounded operators. Without closedness the converse only implies invariance of the closure. For example in ${\ell }^2$, take $W$ to be the finite-support sequences and $\alpha v=⟨v,e_1⟩(2^{-j}{)}_{j\ge 1}$. Then $W^{\perp }=\{0\}$ is automatically adjoint-invariant, but $\alpha e_1\notin W$. Thus the finite-dimensional or closed-subspace hypothesis cannot be omitted from an infinite-dimensional reading.