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Insert a Fourier mode $U_m^n=G^n{e}^{im\theta }$. With $r=1-2\mu$ and $\eta =e^{i\theta }$, the amplification polynomial isIts two roots have product $-\eta$, of modulus one. Thus if both roots are bounded by one, both must have modulus exactly one. Set $G=e^{i\theta /2}\lambda$. The equation becomesWhen $|r|<1$, both roots have modulus one and their separation obeys $|G_{+}-G_{-}|\ge 2\sqrt{1-r^2}>0$ for every frequency. The companion matrix has bounded entries and can be diagonalized with the two eigenvectors $(G_{+},1{)}^T,(G_{-},1{)}^T$. The uniform separation bounds the inverse eigenvector matrix. Hence every matrix power is bounded uniformly in $n$ and $\theta$, and the Parseval identity proves L2 norm stability for the two starting levels on the whole grid.

If $|r|>1$, frequencies near $\theta =\pi$ give one root outside the unit circle, so the scheme is unstable. If $|r|=1$, the Nyquist polynomial is $(G-r{)}^2$, a repeated unit root. The corresponding companion matrix has a nontrivial Jordan block and solutions containing $n{r}^n$. Thus the endpoints are unstable as well. On the whole line the individual Nyquist plane wave is not square-summable, but Fourier packets localized arbitrarily close to that frequency realize the unbounded power norm by continuity. Therefore the uniform stability of a shifted two-level advection scheme criterion isIn particular $\mu =1/2$ is stably exact with compatible starts, whereas $\mu =1$ is formally exact but unstable under arbitrary perturbations. Checking only the moduli of the roots and retaining both endpoints would miss this essential distinction.

The reproduction hypothesis gives $U1=1$, $U(\cdot )(t)=t$ and $U((\cdot {)}^2)(t)=t^2$. If “quadratic” were read as degree exactly two, the same identities follow by subtracting pairs of such reproduced polynomials. HenceApply the same positive-barrier argument as in part (a), now with no moment errors. It gives $|Uf(t)-f(t)|\le \epsilon$ for every $t$ and every $\epsilon >0$. ThereforeThe identity is itself a positive linear operator on continuous functions reproducing all quadratics, so existence as well as uniqueness is established. In particular, reproducing three small test functions is much stronger for a positive operator than merely fixing one finite-dimensional polynomial space without positivity.

For real continuous $f$, the Chebyshev alternation theorem says that $p\in {\mathcal{P}}_n$ is the unique best uniform approximation to $f$ precisely when there are $n+2$ ordered points $0\le x_0<\cdots <x_{n+1}\le 1$ and a sign $\sigma \in \{-1,1\}$ such thatHere ${\mathcal{P}}_n$ means degree at most $n$. Best approximants exist: a minimizing sequence is bounded in the finite-dimensional polynomial space, and finite-dimensional equivalence of norms gives a convergent coefficient subsequence. When $f=p$, the zero error is treated separately; the derivative-sign hypothesis below excludes that case.

Let $p_n$ be a best degree-at-most-$n$ approximant and suppose for contradiction that $E_n(f)=E_{n+1}(f)$. Since $f^{(n+1)}>0$, $f$ is not a degree-at-most-$n$ polynomial and its error is nonzero. The same $p_n$ would be best in ${\mathcal{P}}_{n+1}$, so the Chebyshev alternation theorem at that degree supplies $n+3$ alternating error extrema. The intermediate value theorem gives at least $n+2$ distinct zeros of $e=f-p_n$, one between each consecutive pair of extrema.

Repeated Rolle's theorem then forces a zero of $e^{(n+1)}$. But $p_n^{(n+1)}=0$, whereas $e^{(n+1)}=f^{(n+1)}>0$ everywhere. This contradiction provesThe argument proves strict decrease of best polynomial approximation under a derivative sign; an everywhere negative $(n+1)$st derivative gives the same strict conclusion after replacing $f$ by $-f$.

The ordinary interpolation formulation is unambiguous for strictly increasing knots; repeated knots require specified confluent interpolation or a limiting convention. First take distinct knots and $k\ge 2$, and for fixed $t$ write $F_t(u)=(u-t{)}_{+}^{k-1}$. The two degree-at-most-$k-1$ Lagrange interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at the $k-1$ shared knots $t_{i+1},\dots ,t_{i+k-1}$. Their difference is thereforeThis remains valid if their difference has lower degree: the coefficient is then zero. The leading coefficient of a degree-at-most-$k-1$ interpolant is its order-$(k-1)$ divided difference. Comparing leading coefficients and using the divided-difference recursion givesConsequently the Lee interpolation identity holds for every real $x,t$:No differentiability of the truncated power in its argument was needed for this distinct-node proof.

Sum the identity over $i=1,\dots ,n$. The right side telescopes, giving ${\sum }_i{\omega }_i(x)N_i(t)={\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. When $t_k<t<t_{n+1}$, all the nodes for ${\ell }_1$ lie below $t$, so its interpolation data are zero and ${\ell }_1\equiv 0$. All the nodes for ${\ell }_{n+1}$ lie above $t$, where $F_t(u)=(u-t{)}^{k-1}$ is already a polynomial of the admissible degree. Uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Therefore the Marsden identity isFor order one the empty product is one, the interpolants are constants and a consistent half-open interval convention for $(u-t{)}_{+}^0$ gives the same telescoping proof. Nondecreasing repeated-knot sequences can be treated by the appropriate confluent or limiting convention wherever that definition is valid; merely specifying repeated data values does not uniquely determine the interpolation polynomials. The basic open knot interval in the Marsden identity avoids endpoint convention ambiguity.

Now $B=\overline{\mathbb{C}[1,x]}$. Let $U$ be a bounded component of ${\rho }_A(x)$ and $\lambda \in U$. Its boundary lies in ${\sigma }_A(x)$: an open disk about a boundary point in the algebra resolvent set would meet $U$ and would belong to the same component, contradicting boundary membership. The maximum modulus principle on the bounded domain $U$ therefore gives, for every complex polynomial $p$,Here spectral mapping theorem follows by factoring $p(z)-\mu$: its commuting factors are invertible exactly when every root avoids the Banach algebra spectrum.

The rule $p(x)\mapsto p(\lambda )$ is well-defined by the displayed bound, multiplicative and bounded on the dense polynomial subalgebra. It extends to a unital character ${\phi }_{\lambda }$ on $B$. Since ${\phi }_{\lambda }(\lambda 1-x)=0$, that element cannot have an inverse in $B$. Thus $U\subseteq {\sigma }_B(x)$. Together with part (iii), this proves that the Banach algebra spectrum of a polynomial-generated Banach subalgebra is exactly the original Banach algebra spectrum with all its bounded complementary components filled in.

The hypothesis puts zero in the common unbounded resolvent component from part (iii). Hence $0\in {\rho }_B(x)$ for $B=\overline{\mathbb{C}[1,x]}$, and $(-x{)}^{-1}\in B$, or equivalently $x^{-1}\in B$. By the definition of this norm closure, for each $n$ choose a polynomial $p_n$ with $\Vert p_n(x)-x^{-1}\Vert <1/n$. ThereforeThis is an approximation in the algebra norm, not merely a scalar approximation on the Banach algebra spectrum.

Give $\mathcal{O}(U)$ the compact-open topology. Choose a finite polygonal cycle $\Gamma$ in $U\setminus {\sigma }_A(x)$ whose winding number is one on the Banach algebra spectrum and zero outside $U$. One can take the oriented boundary of a finite union of small squares surrounding the compact Banach algebra spectrum and contained in $U$. DefineThe integral is an algebra-norm integral of a continuous function on finitely many segments. Cauchy's theorem, applied after any bounded linear functional on $A$, makes it independent of the cycle. Moreoverso it is continuous for the compact-open topology.

Here is a direct verification of the algebra properties and uniqueness using the allowed Runge theorem. The version required is that every holomorphic function on an open plane set is a locally uniform limit of rational functions with poles outside that set, allowing a polynomial part. A rational function $r=p/q$ of this kind has the unambiguous value $r(x)=p(x)q(x{)}^{-1}$, since the roots of $q$ avoid the Banach algebra spectrum. These rational values form a unital algebra homomorphism.

They agree with the contour integral. First $\frac{1}{2\pi i}{\int }_{\Gamma }(z1-x{)}^{-1}dz=1$, by replacing the cycle with a large circle and integrating its Neumann expansion. The identity $z^{j}1-x^j=(z1-x){\sum }_{k=0}^{j-1}{z}^{j-1-k}{x}^k$ then proves the assertion for polynomials. For $a\notin U$, the resolvent identity givesThe cycle has winding number zero at $a$, so integration gives $(x-a1{)}^{-1}$. This identity holds on a neighborhood of $a$ disjoint from the cycle and the Banach algebra spectrum. Differentiating with respect to $a$ gives the analogous result for every power $(z-a{)}^{-k}$, and partial fractions give it for all the rational functions under consideration.

If $r_n\to f$ and $s_n\to g$ locally uniformly, the contour estimate makes their evaluated sequences converge. Since $r_n{s}_n\to fg$ locally uniformly and multiplication in $A$ is continuous, their limits satisfy ${\Theta }_x(fg)={\Theta }_x(f){\Theta }_x(g)$. Also ${\Theta }_x(1)=1$ and ${\Theta }_x(Z)=x$. Conversely any continuous unital complex-algebra homomorphism sending $Z$ to $x$ must send $Z-a$ to $x-a1$ and its reciprocal to the inverse. It therefore agrees on rational functions and then, by Runge density and continuity, on all of $\mathcal{O}(U)$. This proves existence and uniqueness of the holomorphic functional calculus.

For its spectral mapping property, first note compatibility on a smaller open spectral neighborhood $W\subseteq U$: composing restriction to $W$ with the calculus there gives the calculus on $U$ by uniqueness. If $\mu \notin g({\sigma }_A(x))$, choose such a $W$ where $\mu -g$ never vanishes. Thenwith the right side evaluated on $W$. Conversely, for $\lambda \in {\sigma }_A(x)$, write $g(z)-g(\lambda )=(z-\lambda )q(z)$ with $q$ holomorphic on $U$. The evaluated factors commute. If their product were invertible, both factors would be invertible, contradicting $\lambda \in {\sigma }_A(x)$. ThusPullback $h\mapsto h\circ g$ is continuous from $\mathcal{O}(V)$ to $\mathcal{O}(U)$ because the image under $g$ of a compact set is compact. The composite $h\mapsto {\Theta }_x(h\circ g)$ is a continuous unital homomorphism taking the coordinate function to $y={\Theta }_x(g)$. Uniqueness of the calculus at $y$ provesFor the exponential conclusion, let $K={\sigma }_A(x)$. There is a simple polygonal arc from zero through its unbounded complementary component to outside a disk containing $K$, continued by a ray to infinity and avoiding $K$. Loops in an initial polygonal path can be removed to make the arc simple. The complement of this slit is a simply connected open neighborhood of $K$ avoiding zero, and hence admits a holomorphic logarithm $\ell$. Set $y={\Theta }_x(\ell )$. The composition rule givesThe calculus of the entire exponential agrees with the usual algebra exponential because its power-series partial sums converge locally uniformly. Finally an invertible matrix over the complex numbers has a finite Banach algebra spectrum avoiding zero. Removing finitely many plane points leaves a connected set, so the hypothesis always holds: every invertible matrix over the complex numbers is an exponential of a matrix over the complex numbers.

A representation of a Banach algebra is a complex-linear multiplicative map $\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$. A nonzero irreducible action of a unital algebra automatically has $\pi (1)=I$: the image of the idempotent $\pi (1)$ is a nonzero invariant subspace and hence is all of $X$. An algebraically irreducible representation of a Banach algebra is a nonzero action if it has no nonzero proper invariant submodule, whether closed or not. A normed representation of a Banach algebra has $\pi (a)$ bounded for each $a$ on a preassigned normed vector space $X$; continuity of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only closed invariant subspaces. Unitizing the algebra extends a nonunital action by $\pi (a+\lambda 1)=\pi (a)+\lambda I$, so it is enough to discuss the unital case.

For $\xi \ne 0$, irreducibility gives $A\xi =X$. The annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal: an intermediate left ideal gives an intermediate submodule of $X$. It is closed by the Neumann-series argument for maximal left ideals. Consequently $A/L_{\xi }$ transports a complete quotient norm to $X$, and left multiplication is contractive in this norm. Every algebraically irreducible representation of a Banach algebra can therefore be realized continuously on a Banach space. This alone does not prove continuity for its preassigned norm.

The kernel of $\pi$ is the intersection of the closed left ideals $L_{\xi }$ and is a closed primitive ideal. The Jacobson radical is the intersection of all primitive ideals, equivalently the intersection of maximal left ideals. Thus these quotient representations detect semisimplicity. Equivalent algebra representations are related by a bijective module homomorphism.

The algebra of module endomorphisms is exactly $\mathbb{C}I$. Indeed, if a module endomorphism $D$ sends a fixed cyclic vector $\xi$ to $a\xi$, then $D(b\xi )=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_{\xi }$, so is bounded for the quotient norm on an irreducible Banach module. Every nonzero module endomorphism is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This commutant of an operator algebra is a closed division subalgebra of the bounded operators on this Banach module; the Gelfand-Mazur theorem makes it $\mathbb{C}I$.

The Jacobson density theorem now says that for finitely many linearly independent ${\xi }_1,\dots ,{\xi }_n$ and arbitrary ${\eta }_1,\dots ,{\eta }_n$, one element of $A$ sends all ${\xi }_j$ to ${\eta }_j$. Here is the induction behind it. The one-vector assertion is $A{\xi }_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a{\xi }_j=0(j<n)\}$. Its image $L{\xi }_n$ is an invariant subspace. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a{\xi }_1,\dots ,a{\xi }_{n-1})\mapsto a{\xi }_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the module endomorphism result, forcing ${\xi }_n$ to be a linear combination of the earlier vectors, a contradiction. This proves the assertion.

We can now give Johnson's continuity argument with its crucial construction. If $X$ is finite dimensional, the closed kernel has finite codimension and $A/\ker \pi$ is finite dimensional, so $\pi$ is continuous. Suppose instead that $X$ is infinite dimensional. LetThis is an invariant submodule, because the orbit map of $\pi (b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.

To rule out $Y=0$, choose linearly independent unit vectors ${\xi }_0,{\xi }_1,\dots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail vectors are independent. In the closed left annihilator of its first vector, density permits arbitrary images of any finite list of the remaining vectors. The condition that finitely many such images are independent is open in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero polynomial in $t$, with only finitely many exceptional values. The Baire category theorem intersects these open dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current vector and retains independence of the entire remaining tail. This is the countable annihilation sequence in an irreducible Banach module.

The gliding-hump continuity principle supplies the other ingredient: if $E,F$ are Banach spaces, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_{n}S{T}_1\cdots T_m$ is continuous whenever $m>n$, then $U_{n}S{T}_1\cdots T_n$ is continuous eventually. The product form of this principle is recorded in Proposition 1.3 of Thomas's paper. Here is the contradiction mechanism. Rescale so $\Vert T_j\Vert ,\Vert U_n\Vert \le 1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ soAlso choose $\Vert x_j\Vert \le 2^{-j}/(1+{\max }_{i<j}\Vert U_{n_i}S{P}_{n_i+1}\Vert )$. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that continuous composite bounds the tail contribution by one. The display forces $\Vert U_{n_j}Sx\Vert >j$, contradicting $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$. Only finite decompositions and the continuous tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.

Apply this principle with $E=A$, $F=\mathcal{B}(\overline{X})$, where $\overline{X}$ is the completion in the original norm, and $S(a)$ the bounded extension of $\pi (a)$. Set $T_j(a)=a{a}_j$ and $U_n(R)=R{\xi }_n$. ThenFor $m>n$ this is zero, hence continuous. For $m=n$ it is the orbit map of a nonzero vector, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.

Finally, apply the Uniform boundedness principle on the Banach space $A$ to the continuous maps $a\mapsto \pi (a)\xi$ indexed by $\Vert \xi \Vert \le 1$. They are pointwise bounded because each individual $\pi (a)$ is bounded in the original norm. It givesThis proves Johnson's continuity theorem for irreducible normed representations, including when the original normed vector space is incomplete. The quotient norm, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired continuity in advance.

It suffices first to treat unital algebras and a unital homomorphism. A surjective homomorphism between unital algebras automatically preserves the identity. If $B=0$, continuity is immediate. In the nonunital case, extend to the forced unitizations by $T^{+}(a+\lambda 1)=T(a)+\lambda 1$. This is still surjective and the unitization of a semisimple algebra is semisimple: its radical projects to zero in its scalar quotient, and its remaining intersection with the original algebra is its original radical. Continuity of the extension implies continuity of the original map.

An algebra homomorphism decreases the spectral radius, even without any continuity assumption, because it takes an inverse of $\lambda 1-a$ to an inverse of $\lambda 1-T(a)$. Suppose $a_n\to 0$ in $A$ and $T(a_n)\to b$ in $B$. Fix any $c\in B$ and choose $u,v\in A$ with $T(u)=b$, $T(v)=c$. PutConsider the polynomial $q_n(z)=T((1-z)w_n+zw)=(1-z)d_n+zd$. Its value at one is exactly $d$. On the outer circle $|z|=R$,On the inner circle $|z|=1/R$, spectral-radius decrease givesApplying part (i) in $B$ and then letting $n\to \infty$, with $R$ fixed, givesThis holds for every $R>1$, so $r_B(cb)=0$ for every $c\in B$.

This forces $b$ into the Jacobson radical. To see the criterion directly, if $b$ were outside a maximal left ideal $L$, then $L+Bb=B$, so $1=l+cb$ for some $l\in L$, $c\in B$. The equality $r_B(cb)=0$ makes $1-cb$ invertible, but this element equals $l\in L$, impossible in a proper left ideal. Hence $b$ lies in every maximal left ideal. Since $B$ is semisimple, $b=0$.

We have shown that the separating space of $T$ is zero. If $a_n\to a$ and $T(a_n)\to y$, apply the result to $a_n-a$ to obtain $y=T(a)$. Its graph is closed, and the closed graph theorem now provesThe crucial estimate uses part (i) with an exactly fixed central value; it does not assume the false general assertion that the spectral radius is continuous under norm limits.

We first extend the scalar Hadamard three-circle theorem to a holomorphic Banach-space-valued function $F$. Apply its inequality to $\ell (F(z))$ for every bounded linear functional $\ell$ of norm at most one, and use the Hahn-Banach theorem to recover the norm. For the radii $1/R,1,R$ this givesNow take $F(z)=p(z{)}^{2^n}$. Even for noncommuting coefficients this is an algebra-valued holomorphic polynomial. Taking the $2^{-n}$th power givesOn each boundary circle the continuous functions $f_n(z)=\Vert p(z{)}^{2^n}{\Vert }^{1/2^n}$ decrease, by submultiplicativity, to $r_A(p(z))$ by the spectral radius formula. Their maxima decrease to the supremum of this pointwise limit. To justify that passage without assuming continuity of the spectral radius, take maximizing points along a convergent subsequence on the compact circle. For any fixed $k$, all later $f_n$ are bounded by $f_k$; continuity of $f_k$ therefore bounds the limiting maxima by $f_k$ at the limiting point. Taking the infimum over $k$ gives the desired upper bound, while the reverse bound is immediate.

Passing to the limit proves the spectral-radius three-circle inequality at the geometric mean:For a nonunital algebra, use its unitization of an algebra; the spectral radius of an original algebra element is unchanged. Zero boundary suprema cause no difficulty in the decreasing-limit argument.

First $N(A)$ is finite: by Cauchy-Schwarz, $\Vert Ax{\Vert }_2\le ({\sum }_{ij}{a}_{ij}^2{)}^{1/2}\Vert x{\Vert }_2$. It is nonnegative and absolutely homogeneous. If $N(A)=0$, every $A{e}_j=0$, so every column is zero and $A=0$; the converse is immediate. The triangle inequality for the Euclidean norm gives $N(A+B)\le N(A)+N(B)$ after taking suprema. Thus $N$ is a norm.

For any vector $x$, rescaling gives $\Vert Ax{\Vert }_2\le N(A)\Vert x{\Vert }_2$. Applying this twice givesFurthermore $|a_{ij}|\le \Vert A{e}_j{\Vert }_2\le N(A)$, proving the requested strict entry bound when $N(A)<\epsilon$.

For a polynomial in matrix entries, expand each monomial at $A+H$. The terms involving exactly one entry of $H$ give a linear map $Df(A)[H]$, with coefficients polynomial in the entries of $A$. Every remaining term has at least two entries of $H$ and is $O(N(H{)}^2)$ for $N(H)\le 1$, because $|h_{ij}|\le N(H)$. This proves differentiability. Its derivative is continuous as an operator-valued function: each coefficient is continuous, and for $N(H)\le 1$ every $h_{ij}$ is bounded by one, so the operator-norm difference is bounded by the sum of the coefficient differences. Coordinate continuity follows from the same entry bound. Hence every such polynomial function is continuously differentiable.

In the Leibniz expansion of $\det (I+H)$, the identity permutation supplies $1+{\sum }_i{h}_{ii}$ plus terms of degree at least two. Any nonidentity permutation moves at least two indices, so each of its terms has at least two $H$ entries. ConsequentlyIf $A$ is invertible, multiplicativity of the determinant and the preceding expansion giveThe adjugate identity gives $\mathrm{adj}A=\det (A)A^{-1}$. Thus $d^{\prime }(A)[H]=\mathrm{tr}((\mathrm{adj}A)H)$ for invertible $A$.

For singular $A$, take invertible $A_r\to A$, using the permitted density result. Both $d^{\prime }$ and the adjugate entries are continuous polynomial expressions, so taking limits for every fixed $H$ proves the derivative of the determinant formula on all matrices:Here the standard adjugate is the transpose of the cofactor array, as required by $(\mathrm{adj}A)A=(\det A)I$. In components this derivative is ${\sum }_{ij}{\mathrm{cof}}_{ij}(A)h_{ij}$, also valid at singular matrices.