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The first algebra exists. Let $A=L^1(0,1)$ with truncated convolutionThe Tonelli theorem givesCommutativity follows by replacing $s$ with $t-s$; associativity follows from the Fubini theorem applied to the simplex $s,u\ge 0$, $s+u\le t$. Together with completeness of $L^1$, this makes $A$ a commutative Banach algebra, not necessarily unital. Adjoin an identity using unitization:This norm is complete and submultiplicative by the preceding bound, and $e=(1,0)$ is the identity. This is the Volterra convolution algebra on a finite interval with an identity adjoined.

Take $x=(0,1)$, where $1(t)=1$. Direct induction under convolution givesThus no power vanishes. For every $\lambda \ne 0$ the seriesconverges absolutely, since its norms have factorial denominators. Multiplication by $\lambda e-x$ telescopes to $e$, so it is an inverse. On the other hand $x$ cannot be invertible: its scalar coordinate is zero, and scalar coordinates multiply, so $xy$ can never equal $e$. HenceThis explicitly exhibits a nonnilpotent quasinilpotent element without assuming a spectral-radius formula.

The second algebra does not exist. Suppose it did, and choose a nonscalar $y$. Nonemptiness of its spectrum of an element, together with $\rho (y)=0$, gives ${\sigma }_B(y)=\{0\}$. But $y+e$ is still nonscalar: if $y+e=\alpha e$, then $y=(\alpha -1)e$. By translation of the spectrum by a scalar,This contradicts the required zero spectral radius for every nonscalar element. The contradiction already uses the spectral condition alone; the nonvanishing-power condition cannot remedy it.

A complex Banach algebra is an associative complex algebra with a complete norm satisfying submultiplicativity, $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. A field has a nonzero identity $e$, and every nonzero element is invertible. We derive the Gelfand-Mazur theorem from these facts.

If $\Vert z\Vert <1$, the series $e+z+z^2+\cdots$ converges in the complete algebra. Multiplying its partial sums by $e-z$ and passing to the limit proves the Neumann series inverse. In particular, for any $x\in B$ and $|\lambda |>\Vert x\Vert$,This resolvent of an element tends to zero in norm at infinity; the estimate follows from the geometric bound on $\Vert x^n\Vert$.

We next prove that the spectrum of an element cannot be empty. Suppose every $\lambda e-x$ were invertible. At each ${\lambda }_0$, writeThe Neumann series on a small disk about ${\lambda }_0$ proves that $R$ is analytic there. Consequently, for each continuous linear functional $\ell :B\to \mathbb{C}$, the function $\lambda \mapsto \ell (R(\lambda ))$ is entire. It is bounded outside a large disk by the estimate above and bounded on that disk by continuity. The Liouville theorem makes it constant; its zero limit at infinity makes the constant zero.

Continuous complex linear functionals separate points of $B$. Here is the needed consequence of the Hahn-Banach theorem: for $z\ne 0$, define $g(tz)=t\Vert z\Vert$ on the real span of $z$ and extend it as a bounded real linear functional of norm one. Then $\ell (w)=g(w)-ig(iw)$ is complex linear and continuous, and $\mathrm{Re}\ell (z)=\Vert z\Vert \ne 0$. The real extension theorem is proved explicitly in Question 4(i) below. Thus $\ell (R(\lambda ))=0$ for every $\ell$ forces $R(\lambda )=0$, contradicting $(\lambda e-x)R(\lambda )=e\ne 0$. This proves the nonemptiness of the Banach-algebra spectrum.

Choose $\lambda \in {\sigma }_B(x)$. Since $B$ is a field, the noninvertible element $x-\lambda e$ must be zero. Hence every $x\in B$ is a scalar multiple of $e$. The mapis therefore a bijective unital complex-algebra homomorphism. It and its inverse are bounded, since $\Vert \lambda e\Vert =|\lambda |\Vert e\Vert$. Thus it is an isomorphism of Banach algebras. If the usual normalization $\Vert e\Vert =1$ is included in the definition, it is an isometry; without that normalization, it is still the required continuous algebra isomorphism.

For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate continuity and linearity make both maps bounded linear operators. For each fixed $y$,The Uniform boundedness principle therefore gives $C<\infty$ such that $\Vert L_x\Vert \le C$ whenever $\Vert x\Vert \le 1$. Scaling yieldsFor completeness, the relevant Uniform boundedness principle follows from the Baire category theorem as follows. For a pointwise bounded family $\mathcal{T}$ of bounded linear operators on a Banach space, the closed sets $E_m=\{y:{\sup }_{T\in \mathcal{T}}\Vert Ty\Vert \le m\}$ cover the space. One $E_m$ contains an open ball $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\Vert h\Vert <r$, gives ${\sup }_T\Vert Th\Vert \le 2m$. Applying this to $h=(r/2)u$, $\Vert u\Vert \le 1$, bounds every operator norm by $4m/r$.

Define the new norm byIt is a norm because $L_{x}e=x$, so $L_x=0$ implies $x=0$. The triangle inequality and scalar homogeneity follow from those of the operator norm. Moreover,so it is equivalent to the original norm and remains complete. By associativity, $L_{xy}=L_x{L}_y$, and the submultiplicativity of the operator norm provesThis is renorming a separately continuous Banach algebra. It even gives $\Vert e{\Vert }_{\ast }=1$. If the algebra is the zero space, the conclusion is immediate separately.

Let $F$ be the convex hull of all terms of the sequence, meaning the set of their finite convex combinations. Suppose $B(0,ϵ)\cap F=\varnothing$. Part (ii) then gives a continuous linear functional $S$ with $Sf\ge 1$ for every $f\in F$. In particular $S{f}_n\ge 1$ for every $n$, contradicting $S{f}_n\to 0$. Therefore the ball meets $F$.

A point of that intersection is a finite combination ${\sum }_{k=1}^m{\alpha }_k{f}_{n_k}$ with nonnegative coefficients summing to one. Set $N={\max }_k{n}_k$, combine repeated indices and insert zero coefficients for omitted indices. This givesThus zero lies in the norm closure of the convex hull. The same argument applies to every tail of the sequence and, by choosing errors tending to zero, also gives the usual Mazur lemma. Completeness of $V$ is not required for this conclusion.

The hypothesis needed is precisely convergence to zero against every continuous linear functional, namely weak convergence. The PDF says only “continuous” for the testing maps. If that is interpreted as including arbitrary nonlinear continuous maps, the constant map $T\equiv 1$ makes the hypothesis impossible. The proof above establishes the substantive conclusion under the intended, weaker linear-testing hypothesis, and therefore also the implication under the literal stronger wording.

Let $p$ be the Minkowski functional of $E$:The contained ball makes $E$ absorbing, so $p$ is finite, nonnegative and satisfies $p(v)\le \Vert v\Vert /ϵ$. It is positively homogeneous. To check subadditivity, if $v\in sE$ and $w\in tE$, the defining property of a convex set gives $(v+w)/(s+t)\in E$, whence $p(v+w)\le s+t$; let $s,t$ decrease to the respective infima. Thus $p$ is a sublinear functional. We have $p(e)\le 1$ for every $e\in E$. Also $p(x)\ge 1$: otherwise $x\in tE$ for some $0<t<1$, which would put $x$ in $E$ because $E$ is a convex set and $0\in E$.

We prove the dominated form of the Hahn-Banach theorem needed here. Suppose $g$ is linear on a subspace $M$ and $g\le p$ there. To extend it to $M+\mathbb{R}v$ with $v\notin M$, choose the value $c$ at $v$ betweenEvery lower candidate is at most every upper candidate, becauseTaking $m=n=0$ among the candidates shows the two endpoints are finite: the lower supremum is at least $-p(-v)$ and at most $p(v)$, and the upper infimum lies between them and $p(v)$. Define $\tilde{g}(m+tv)=g(m)+tc$. For $t>0$, the upper inequality applied to $m/t$ gives domination by $p(m+tv)$. For $t<0$, apply the lower inequality to $m/(-t)$ and multiply by $-t$. At $t=0$ domination already holds. Hence this is a dominated one-dimensional extension.

Order all dominated extensions by inclusion of their domains. Chain unions preserve linearity and domination, so Zorn's lemma gives a maximal extension. The one-dimensional argument shows its domain is the whole vector space. This proves the requisite dominated Hahn-Banach theorem, rather than invoking an unproved separation form.

On the line $\mathbb{R}x$, start with $g(tx)=tp(x)$. For $t\ge 0$ this equals $p(tx)$; for $t<0$ it is nonpositive and therefore at most $p(tx)$. Extend it to $T\le p$ on $V$. ThenApplying domination to $v$ and $-v$ also givesThus $T$ is a continuous linear functional, andThis proves separation from an absorbing convex set with no closedness assumption on $E$. The proof also gives the norm-preserving real extension theorem used earlier: take $p(v)=C\Vert v\Vert$ when the original functional has norm at most $C$.

Take $n\ge 1$ and $1<p<\infty$. The Lp space is strictly convex because $z\mapsto |z{|}^p$ is strictly convex: for distinct unit functions its pointwise midpoint inequality is strict on a set of positive measure, hence integration gives norm strictly below one. The finite-dimensional linear subspace ${\mathcal{T}}_{n-1}$ has a best approximant: any minimizing sequence is bounded by $\Vert t{\Vert }_p\le \Vert f-t{\Vert }_p+\Vert f{\Vert }_p$, and finite-dimensional compactness gives a convergent subsequence whose limit minimizes the continuous error. Uniqueness follows from part 1.

Write that unique minimizer as $t^{\ast }$ and its residual as $F=f-t^{\ast }$. Translation on the circle preserves the Lp norm. Since $f(x+2\pi /n)=f(x)$, the translated polynomial $t^{\ast }(x+2\pi /n)$ is another minimizer. Uniqueness forcesAlso $f(x+\pi /n)=-f(x)$. Therefore the polynomial $-t^{\ast }(x+\pi /n)$ has residual $-F(x+\pi /n)$, with the same norm as $F$, and is again a minimizer. HenceTo extract the consequence, write the trigonometric polynomial as $t^{\ast }(x)={\sum }_{|\ell |\le n-1}{c}_{\ell }{e}^{i\ell x}$. The first identity gives $(e^{2\pi i\ell /n}-1)c_{\ell }=0$. Only $\ell =0$ in the indicated frequency range is a multiple of $n$, so $t^{\ast }$ is constant. The second identity makes this constant its own negative and therefore zero. Thus the lower-frequency Lp approximation to a single harmonic isFor $n=1$ the space already consists only of constants, and the negative-translation argument gives the same result.

Let $y=(f(x_1),\dots ,f(x_n){)}^T$ and write the interpolating spline as $P_{x}f={\sum }_{j=1}^n{c}_j{N}_j$. The interpolation equations are $A_{x}c=y$, so $c=A_x^{-1}y$. Since $P_x$ is defined as a unique interpolant for arbitrary data, its B-spline collocation matrix is invertible. Ordered sites satisfying diagonal positivity ensure this by question 7. Positivity alone without admissible ordering does not: two equal sites in overlapping basis supports could give two identical rows. The norm argument below uses the existence of the operator stipulated here.

Standard partition-normalized B-splines satisfy $N_j\ge 0$ and ${\sum }_{j=1}^n{N}_j\le 1$. The latter is a subpartition of unity for B-splines: extending the knot sequence gives a full partition of unity, and discarding basis functions cannot increase its sum. HenceTaking the supremum over $t$ and then over unit-norm $f$ gives the B-spline interpolation operator norm estimateHere the matrix norm is the induced maximum-row-sum norm, not the maximum of its individual entries. The proof only needs nonnegative basis functions, their sum at most one, and invertible interpolation.

The centered quadratic spline interpolation matrix is the symmetric tridiagonal Toeplitz matrixFor $n=1$, this means simply $A_x=(3/4)$. To evaluate the inverse norm exactly, let $D=\mathrm{diag}((-1{)}^i{)}_{i=1}^n$ and $M=8D{A}_{x}D=\mathrm{tridiag}(-1,6,-1)$. If $J$ is the nonnegative adjacency matrix of the path, then $M=6I-J$ and $\Vert J/6{\Vert }_{\infty }\le 1/3$. The convergent Neumann seriestherefore has nonnegative entries. Since $A_x^{-1}=8D{M}^{-1}D$, its entrywise absolute values are $8M^{-1}$. Let $r_i={\sum }_j|(A_x^{-1}{)}_{ij}|$. Then $r=8M^{-1}1$, orThe homogeneous characteristic roots are $3\pm 2\sqrt{2}$. Put $\tau =3-2\sqrt{2}\in (0,1)$. The particular solution $r_i=2$ and the two boundary conditions giveThese row sums are symmetric and increase toward the centre: the exponential sum in the numerator is minimized at the middle integer or pair of integers. With $m=\lfloor (n+1)/2\rfloor$, the exact finite-size inverse norm is thusFor example it equals $4/3$ at $n=1$ and $8/5$ at $n=2$, tending to $2$ as $n\to \infty$. Stating only the bound two would not evaluate the finite matrix norm requested here.

The general estimate from part 1 and the exact inverse norm from part 2b implyand in particular $\Vert P_x{\Vert }_{L^{\infty }}\le 2$, uniformly in the number of knots. This bound concerns the support-centred sites $x_i=i+3/2$; shifting the sampling sites changes the B-spline collocation matrix and can destroy uniform stability.