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Expand using the bilinearity and symmetry of the real inner product:Addition proves the parallelogram law,For the maximum norm on ${\mathbb{R}}^2$, take $x=(1,0)$ and $y=(0,1)$. All four vectors $x,y,x+y,x-y$ have maximum norm one. The left side of the parallelogram law is then $2$, whereas the right side is $4$. The maximum norm cannot come from an inner product.

A norm on a real vector space $V$ is a map $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly when $x=0$, $p(tx)=|t|p(x)$ for real $t$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d_p(x,y)=p(x-y)$. The three axioms respectively give definiteness, compatibility with scaling, and the triangle inequality.

The Lanford contraction proof of the Feigenbaum fixed point replaces an unstable renormalization operator by a contracting Newton correction. Use a coefficient Banach space for normalized even maps, represented by $p(x)=1-x^{2}h(x^2)$ with an absolutely summable coefficient norm on $h$. Its complex domain and a small ball are chosen so that every map is admissible and the renormalization composition is analytic there.

A high-accuracy polynomial $p_0$ gives an approximate fixed point. Use a bounded invertible approximate inverse $J$ for $D\mathcal{T}(p_0)-I$, and defineRigorous estimates certifyThe finite coefficient computations use interval arithmetic with controlled rounding; analytic estimates bound the infinite tail. Thus the certification applies to the full function space, rather than merely a truncated polynomial system.

Part (b) gives a unique fixed point of $A$ in the certified ball. Invertibility of $J$ makes it a unique fixed point $g$ of $\mathcal{T}$ there. The ball also preserves the quadratic critical maximum and nontrivial shape, excluding the constant formal normalized solution. Consequently an actual locally unique analytic Feigenbaum fixed point exists, with the error bound $\Vert g-p_0\Vert \le \epsilon /(1-\kappa )$. The numerical approximation locates it; rigorous residual, derivative and tail bounds establish its existence and uniqueness. This conclusion is local, not a claim of global uniqueness among all analytic solutions.

The cubic spline is even, so the symmetry of the normalized B-splines gives $N_{j,4}(-x)=N_{6-j,4}(x)$. It therefore suffices to extract three coefficients. The two polynomial pieces join with matching values and first two derivatives at zero, so the function belongs to the stated cubic spline space. At zero its right-hand derivatives areUse the supplied De Boor–Fix spline coefficient functional with the unnormalized knot polynomial ${\psi }_j(x)={\prod }_{r=j+1}^{j+3}(t_r-x)$. In this sign convention the coefficient isFor $j=1$, ${\psi }_1=(-1-x{)}^3$. Taking the right-hand limit at $-1$ kills the first three terms and gives $c_1=f(-1)=1$. For the next two coefficients take the right-hand limit at zero. Their knot polynomials and derivatives there areConsequently $c_2=(-27+6)/6=-7/2$ and $c_3=(27+6)/6=11/2$. Although the third derivative jumps at zero, its multiplier ${\psi }_j(0)$ vanishes for these two functionals; taking either one-sided limit gives the same coefficients. Symmetry gives $c_4=c_2$ and $c_5=c_1$. ThusFor a normalized B-spline basis, define the synthesis linear map $Tc={\sum }_j{c}_j{N}_{j,4}$ from the maximum coefficient norm into the spline space with its supremum norm. Nonnegativity and partition of unity give $\Vert T\Vert =1$, including equality on the all-ones vector. The coefficient condition number of a normalized B-spline basis isOn $[0,1]$, $f^{\prime }(x)=(27/2)x(3x-2)$. The only interior extremum is at $2/3$, where $f=-1$; at zero and one, $f=1$. Reflection therefore gives $\Vert f{\Vert }_{\infty }=1$. Its largest absolute coefficient is $11/2$, proving

Write the spline interpolation operator as $P_{\mathbf{x}}=T{A}_{\mathbf{x}}^{-1}R$, where $R$ samples at the interpolation sites and $T$ synthesizes the normalized B-spline expansion. Sampling has operator norm at most one, and nonnegative B-splines with sum at most one give $\Vert T\Vert \le 1$. HenceHere the matrix norm is the maximum absolute row sum. Existence of the inverse matrix requires the usual ordered distinct interpolation sites and the Schoenberg–Whitney theorem support conditions, with endpoint values interpreted by their one-sided limits. If the displayed diagonal condition is read without the implicit ordering and distinctness, it is insufficient: all three sites equal to $1/2$ in the quadratic example give positive diagonal entries but three identical rows. The estimate applies whenever the interpolating linear map in the question is defined uniquely.

For the quadratic Bernstein basis, $N_1=(1-x{)}^2$, $N_2=2x(1-x)$ and $N_3=x^2$. Sampling and inverting giveThe three cardinal Lagrange interpolation polynomials areThe Lebesgue constant of interpolation gives the exact operator norm: the upper bound follows from $|Pg(x)|\le \Vert g{\Vert }_{\infty }{\sum }_i|{\ell }_i(x)|$. To attain it at a maximizing point, prescribe at the three sites the signs of the corresponding cardinal polynomials and extend those values by a continuous piecewise linear function of norm one.

On $0\le x\le 1/2$, the first two cardinal polynomials are nonnegative and the last is nonpositive. Since their sum is one, their absolute sum is $1-2{\ell }_2(x)=1+2x-4x^2$. Its maximum is $5/4$ at $x=1/4$. Reflection gives the same maximum at $3/4$ on the other half. In particular the data $(1,1,-1)$ produce $1+2x-4x^2$, attaining $5/4$. ThusThe coefficient estimate loses cancellation among the B-splines, which explains why it is not sharp for the resulting function norm.

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.

A algebra character $\phi$ is nonzero, linear and multiplicative, with $\phi (1)=1$. Since $\phi (a)\in {\sigma }_A(a)$, it satisfies $|\phi (a)|\le \Vert a\Vert$ and is automatically continuous. If $h=h^{\ast }$ and $t\in \mathbb{R}$, then $(e^{ith}{)}^{\ast }=e^{-ith}$, so $e^{ith}$ is a Unitary element of a C-star algebra and has norm one. HenceUsing both signs of $t$ forces $\mathrm{Im}\phi (h)=0$. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ both self-adjoint C-star elements. Their algebra character values are real, and thereforeThis is the conjugation in the original PDF; the converted TeX loses the bar.

For the final unheaded request, let $B=C^{\ast }(1,x)\subseteq A$. Normality makes this a commutative unital C-star subalgebra. We first justify spectral permanence, so that we do not silently use the possibly larger algebra spectrum in a mere non-star subalgebra. A self-adjoint C-star element $h$ in a C-star algebra has real algebra spectrum: $e^{ith}$ has norm one, and $\lambda \in \sigma (h)$ implies $e^{it\lambda }\in \sigma (e^{ith})$ for each real $t$. The latter assertion follows from the factorization $e^{ith}-e^{it\lambda }1=(h-\lambda 1)g_t(h)$ with commuting factors. An invertible product of commuting factors would make $h-\lambda 1$ invertible. Both signs of $t$ then force $\lambda \in \mathbb{R}$.

A compact subset of the real line has connected planar complement. Apply Question 1 to the self-adjoint C-star element $h\in B$: all of ${\rho }_A(h)$ is its unbounded component and is retained in $B$. Thus ${\sigma }_A(h)={\sigma }_B(h)$. If $a\in B$ is invertible in $A$, both $a^{\ast }a$ and $a{a}^{\ast }$ are self-adjoint C-star elements and invertible in $A$, hence invertible in $B$. The elements $(a^{\ast }a{)}^{-1}{a}^{\ast }$ and $a^{\ast }(a{a}^{\ast }{)}^{-1}$ are respectively a left and a right inverse in $B$, so coincide and give its inverse. This proves spectral permanence for C-star algebras.

In a commutative unital complex Banach algebra, algebra spectra are the ranges of the characters. Indeed a noninvertible $b-\lambda 1$ lies in a maximal ideal. Such an ideal is closed, since its proper closure cannot contain an element sufficiently near $1$. The quotient is a complex Banach division algebra; nonemptiness of the algebra spectrum in that quotient makes each element scalar, producing an algebra character with value $\lambda$ at $b$. Conversely, an algebra character value cannot correspond to an invertible difference.

Apply this to $B$. The preceding involution result gives $\phi (x^{\ast })=\overline{\phi (x)}$, and spectral permanence applies also to $p(x,x^{\ast })$. ConsequentlyThus the second argument is conjugated as in the original PDF, not a second copy of $\lambda$. This is the spectral mapping rule for a star polynomial in one normal element.

For a self-adjoint C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $x$ is a normal C-star element, thenAll powers are normal C-star elements, so induction gives $\Vert x^{2^j}\Vert =\Vert x{\Vert }^{2^j}$. The general spectral radius formula now gives $r(x)=\Vert x\Vert$.

One can also see the needed growth estimate directly. For $R>r(x)$ the resolvent Cauchy coefficient formula givesThe formula follows by the Neumann series on a larger circle and contour deformation through the resolvent annulus. Put $m=2^j$, take $m$th roots and let $j\to \infty$ to get $\Vert x\Vert \le R$. Let $R\downarrow r(x)$; the reverse inequality already follows from the norm bound on the algebra spectrum. ThusThis proves spectral radius norm equality for normal elements from the norm identity.

An algebra involution is a conjugate-linear map $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $(a^{\ast }{)}^{\ast }=a$; on a unital algebra it fixes $1$. In the Banach-star convention it is continuous, although the following spectral identity uses only its algebraic properties. An element $a$ is invertible precisely when $a^{\ast }$ is: its inverse is $(a^{-1}{)}^{\ast }$. Applying this to $a=\lambda 1-x$ provesFor a nonunital algebra, extend the involution to its unitization of an algebra by $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$ and use the spectrum in that unitization; the same argument proves the identity.

A unital C-star algebra is a complex Banach algebra with an involution satisfying the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A Normal element of a C-star algebra satisfies $x^{\ast }x=x{x}^{\ast }$. The norm identity for the nonzero identity implies $\Vert 1\Vert =1$.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

Since every eigenvalue has modulus less than one, $1$ is not an eigenvalue and $I-H$ is invertible. Hence the iteration has a unique fixed point $\hat{x}=(I-H{)}^{-1}v$. The stated consistency implication gives $A\hat{x}=b$, and nonsingularity of $A$ identifies $\hat{x}=x^{\ast }$. This justifies using the fixed-point relation even though consistency was stated in only one direction.

Expand the initial error in the eigenvector basis, $x^0-x^{\ast }={\sum }_i{c}_i{w}_i$. Subtract the fixed-point equation from the stationary iterative method for a linear system and iterate:With $q={\max }_i|{\lambda }_i|<1$, any norm obeys $\Vert x^k-x^{\ast }\Vert \le q^k{\sum }_i|c_i|\Vert w_i\Vert \to 0$. ThusThe argument uses the given diagonalizable eigenbasis; the more general convergence criterion is spectral radius less than one.

For $t\ne 0$, ordinary differentiation of the two smooth components givesAt zero, use the definition of the derivative in the Euclidean norm:Thus $f$ has differentiability everywhere and $\boxed{f^{\prime }(0)=(0,0)}$. The two vectors $(\sin (1/t),\cos (1/t))$ and $(-\cos (1/t),\sin (1/t))$ are orthogonal unit vectors, so the squared norm of the displayed derivative is $4t^2+1$. ConsequentlyThis discontinuous derivative of a differentiable plane curve also shows why componentwise scalar mean value theorem conclusions do not supply a single intermediate point reproducing an entire vector-valued difference quotient.

By equivalent norms, choose $c,C>0$ with $c\Vert h{\Vert }_1\le \Vert h{\Vert }_2\le C\Vert h{\Vert }_1$. Differentiability in the first norm means that there is a bounded linear map $L:V\to \mathbb{R}$ with $f(a+h)-f(a)-Lh=r(h)$ and $|r(h)|/\Vert h{\Vert }_1\to 0$. The same $L$ is bounded in the second norm, since $|Lh|\le K\Vert h{\Vert }_1\le K{c}^{-1}\Vert h{\Vert }_2$. Moreover,The two norms define the same approach $h\to 0$, so this proves differentiability with the same Fréchet derivative in the second norm. Reversing the norms proves the converse.

For completeness, the derivative is unique: if $L_1,L_2$ both work, fix $v$ and take $h=tv$. The two remainder estimates imply $|(L_1-L_2)v|=o(1)$ as $t\to 0$, so $(L_1-L_2)v=0$ for every $v$. Thus differentiability and its derivative are independent of the chosen norm. This is differentiability under equivalent norms, not merely invariance of continuity.