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If polynomials with integer monomial coefficients converge uniformly to $f$, their values at zero and one are integers converging to $f(0)$ and $f(1)$. A convergent integer sequence is eventually constant: once all its terms lie in an interval of length less than one they must agree. Hence both endpoint values are integers.

Conversely assume those values are integers. The rounded Bernstein polynomial is an integer-coefficient polynomial, because its displayed scalar coefficients are integers and each $x^k(1-x{)}^{n-k}$ expands with integer coefficients. Part b gives its vanishing distance from $B_{n}f$. For completeness, the remaining uniform convergence $B_{n}f\to f$ follows directly from uniform continuity. Let $X$ have the binomial distribution with parameters $n,x$; then $B_n(f,x)=\mathbb{E}f(X/n)$, $\mathbb{E}(X/n)=x$, and $\mathrm{Var}(X/n)=x(1-x)/n\le 1/(4n)$. Splitting into $|X/n-x|\le \delta$ and its complement, and applying Chebyshev's inequality, giveswhere $\omega$ is the modulus of continuity. First choose $\delta$ to make the first term small uniformly, then take $n$ large. The triangle inequality now gives $\Vert B_n^{\ast }f-f{\Vert }_{\infty }\to 0$. We have proved the integer-coefficient polynomial approximation characterization

The coefficient sum is finite, so the Weierstrass M-test makes the cosine series uniformly convergent and $f$ continuous. Put $Q=5^{m+1}$ and $A_m={\sum }_{k=m+1}^{\infty }{a}_k$. The indicated partial sum belongs to ${\mathcal{T}}_n$ because its largest frequency is $5^m\le n$. Its tail satisfiesand equality holds at $x=0$. Thus its supremum norm error is exactly $A_m$.

At the $2Q$ points $x_j=j\pi /Q$, $0\le j<2Q$, every omitted frequency is an odd integer multiple of $Q$. Indeed for $k\ge m+1$,Uniform convergence permits summing these values, givingThese are cyclically alternating extrema. Since $n<Q$, there are $2Q\ge 2n+2$ of them, so the trigonometric Chebyshev alternation theorem proves that the partial sum is the unique best approximant. Consequently the positive lacunary trigonometric series hasThe positivity of every tail coefficient makes the same tail attain its full coefficient sum with alternating signs; a triangle bound alone would not establish optimality.

Use the standard knot-domain convention in which the spline space contains every polynomial of degree at most $k-1$ on $[a,b]$, with completed or clamped endpoint knots. The De Boor–Fix functional extracts a local spline coefficient, and its Hahn-Banach theorem extension is linear on continuous local data with the stated bound. Only that bound and spline reproduction are needed here.

For $x$ in a knot cell $[t_j,t_{j+1}]$, support and Gram bandwidth of B-splines implies that only indicescan contribute, restricted also to $1\le i\le n$. Every corresponding local support $[t_i,t_{i+k}]$ lies insideNonnegativity and subpartition of unity for B-splines therefore giveTaking the supremum over the cell proves the local stability estimate for spline quasi-interpolation:Endpoint limits follow by continuity for $k\ge 2$; the order-one version uses the usual one-sided interval convention.

Let $h={\max }_i|t_{i+1}-t_i|$ and $f\in C^k[a,b]$. Choose any point $y_j$ in the stencil and form its degree-$k-1$ Taylor polynomial $p_j$. The Taylor theorem remainder givesThe polynomial is itself a spline, so $Q{p}_j=p_j$. By linearity and the local stability estimate,Taking the maximum over cells yieldsThe constant depends only on $k$ and the supplied functional bound, not on mesh ratios or the number of knots.

The endpoint convention is essential to the global assertion. The domain can be the basic knot interval, with any needed exterior data extended smoothly, or the full approximation interval can use clamped endpoint knots. If one instead takes a finite, unclamped collection on the entire outer support interval $[t_1,t_{n+k}]$, every order-$k\ge 2$ basis function vanishes at $t_1$, so $Q1(t_1)=0$ and uniform approximation to the constant one there is impossible. This does not affect the local estimate; it explains the usual boundary completion implicit in the global spline approximation statement.

First take $k\ge 2$, so the simple-knot B-splines are continuous, vanish at both support endpoints, and are strictly positive inside their supports. Thus the diagonal condition is $t_i<x_i<t_{i+k}$. We prove both directions of the Schoenberg–Whitney theorem, deriving the weighted zero-count fact needed for sufficiency rather than relying on an ambiguous coefficient-free hint.

For necessity, suppose $N_i(x_i)=0$. If $x_i\le t_i$, then for every row $r\le i$ and column $j\ge i$ we have $x_r\le x_i\le t_i\le t_j$, hence $N_j(x_r)=0$. The first $i$ rows have nonzero entries in at most $i-1$ columns and are linearly dependent. If instead $x_i\ge t_{i+k}$, then for $r\ge i$ and $j\le i$ we have $x_r\ge x_i\ge t_{i+k}\ge t_{j+k}$, so those entries vanish. The last $n-i+1$ rows have nonzero entries in at most $n-i$ columns and are again dependent. In either case the B-spline collocation matrix is singular. Thus invertibility implies the diagonal positivity condition.

We establish two elementary spline facts for the converse. For local linear independence of B-splines, extend the strictly increasing knots in both directions. On a cell $(t_j,t_{j+1})$ the $k$ active basis functions have indices $j-k+1,\dots ,j$. Their knot polynomials are ${\psi }_i(y)={\prod }_{\ell =1}^{k-1}(y-t_{i+\ell })$. The Cox-de Boor recurrence proves Marsden's identityHere is the induction behind that identity. For order one the indicator functions sum to one. On passing from order $k-1$ to $k$, collect the coefficient of each lower-order basis function. With $u=t_i$, $v=t_{i+k-1}$, its extra factor isThus the identity at one order multiplies by $y-x$ at the next.

The $k$ polynomials ${\psi }_i$ are independent: evaluate successively at $y=t_j,t_{j-1},\dots ,t_{j-k+1}$. The first evaluation isolates ${\psi }_j$; after its coefficient has been eliminated the next isolates ${\psi }_{j-1}$, and so on, with nonzero diagonal factors because the knots are distinct. Comparing powers of $y$ in the identity then shows that the active $N_i$ span all polynomials of degree at most $k-1$ on the cell. There are $k$ of them, so they are independent. In particular, a spline vanishes on a whole open cell exactly when all its active coefficients vanish. Padding the finite coefficient vector with zeros makes this fact apply also near the outer knots.

Next let $s={\sum }_{i=p}^q{c}_i{N}_i$ be a real spline on $I=(t_p,t_{q+k})$, with no identically zero knot cell there. The simple-knot B-spline regularity gives $s\in C^{k-2}$, and its derivatives through order $k-2$ vanish at both endpoints because it is supported on the closed interval. Suppose it has $Z$ distinct interior zeros. Its zero set initially has $Z+2$ connected components, counting the endpoints; the zeros are isolated because no cell vanishes identically.

A useful form of Rolle's theorem counts zero components, including any zero intervals that arise after differentiation. Between two successive zero components a real differentiable function is nonzero and has an interior extremum, supplying a zero of its derivative. These derivative-zero components are distinct: a zero interval connecting two such extrema would make the original function constant across an intervening zero and nonzero gap, which is impossible. If its derivative also vanishes at both endpoints, those supply two further distinct components. Thus differentiation increases the zero-component count by at least one. Apply this successively through derivative order $k-2$; every required endpoint derivative vanishes. No derivative in this process can vanish identically: that would make $s$ a global polynomial of degree below $k-1$, with all its endpoint derivatives through that degree zero, forcing $s=0$. The final derivative has at least $Z+k$ zero components.

That final derivative is continuous and piecewise affine on $q-p+k$ knot cells. Each closed affine cell can meet at most one zero component: two different zeros would force the whole cell to vanish, joining them into one component. Every zero component meets a cell, so there are at most $q-p+k$ of them. For $k=2$ this argument applies directly without differentiation. We have proved the compact-support spline zero countThis proof also accommodates zero intervals of the derivatives; counting them as infinitely many distinct zeros would invalidate a naive repeated-Rolle argument.

Now assume $N_i(x_i)>0$ for all $i$ and suppose $A_{x}c=0$ for a nonzero coefficient vector. Since $A_x$ is real, a nonzero real null vector can be chosen. Set $s={\sum }_i{c}_i{N}_i$, so $s(x_i)=0$ for all $i$. Consider one connected component of the union of the open supports of the terms with $c_i\ne 0$. If its first and last indices are $p,q$, the component is exactly $I=(t_p,t_{q+k})$. On this interval $s$ equals $s_0={\sum }_{i=p}^q{c}_i{N}_i$. Each knot cell in $I$ has at least one active nonzero coefficient, so local independence ensures $s_0$ does not vanish on any whole cell.

For every $p\le i\le q$, diagonal positivity gives $t_i<x_i<t_{i+k}$, and this places $x_i$ inside $I$. Thus $s_0$ has the $q-p+1$ distinct zeros $x_p,\dots ,x_q$, contradicting the upper bound $q-p$. There is no nonzero null vector, proving sufficiency. Together the two directions giveFor completeness, order $k=1$ consists of disjoint interval indicators, with the usual left-closed, right-open convention. An invertible ordered collocation requires exactly one site in each interval; ordering makes this equivalent to $N_i(x_i)>0$, and the matrix is then the identity. This handles possible sites at left knots without incorrectly replacing positivity by a strict interior inequality for order one.

Write the two formal pseudodifferential operators in normal order aswhere $p={\alpha }_1$ and $q={\alpha }_2$. The formal pseudodifferential composition rule givesThus $\boxed{\mathrm{ord}(P_1{P}_2)=p+q}$ whenever $ab$ is not identically zero, in particular when both leading coefficients are nowhere zero. In general the rigorous statement over the specified coefficient algebra is $\mathrm{ord}(P_1{P}_2)\le p+q$.

The leading scalar coefficients commute. Subtracting the reversed composition also cancels the terms involving $a_1,b_1$, leavingTherefore the order filtration of formal pseudodifferential operators giveswith equality precisely when the displayed coefficient is not identically zero. The bound can be strict: constant-coefficient powers of $\partial$ commute, giving the zero operator, whose order can be assigned $-\infty$.

There is also a genuine qualification to product-order addition in $C^{\infty }(\mathbb{R})$. Choose nonzero smooth functions $a,b$ with disjoint compact supports and take $P_1=m_a,P_2=m_b$. Each has order zero, but their product is zero. Nonzero coefficients as elements of this algebra need not have a nonzero product. Thus an unqualified assertion of exact product order requires an additional leading-symbol hypothesis.

Use the formal pseudodifferential composition rule on each coefficient $w_i(x)$ and then the exponential eigenfunction convention. For every integer $n$,ConsequentlyCollecting all terms of a fixed power of the spectral parameter gives the useful answerThe sum defining each $v_s$ is finite. For example,andThis is the action on a formal pseudodifferential wave function. It is a formal expansion near $k=\infty$, with $k^{\beta }$ carried as a formal prefactor. It does not assert convergence for finite $k$, and negative powers are not evaluated at $k=0$.

First let $n\ge 0$. The ordinary Leibniz rule, applied to a test function $h$, givesEquivalently, the composition of differential operators isFor completeness, this formula follows by induction: applying $\partial$ differentiates either the coefficient or the remaining derivative of $h$, and the two contributions combine by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$. The initial case is multiplication by $f$.

For all integers $n$, define $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. The formal pseudodifferential composition rule isFor $n=-s<0$, $\left(\genfrac{}{}{0ex}{}{-s}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$, so this is usually an infinite series. In particularApplying $\partial$ on the left gives $m_f$: all later coefficient derivatives cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in $n$, gives every negative power. More explicitly, applying $\partial$ to the proposed $n$-series yields coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ at $f^{(r)}{\partial }^{n+1-r}$, so it is consistent with the already determined $(n+1)$-series. Negative powers here are formal; differentiation on all of $C^{\infty }(\mathbb{R})$ has a nontrivial kernel and has no genuine two-sided inverse there.

Use the canonical representatives of functionals of $u$: $P_f=f{\partial }^{-1}$, $P_g=g{\partial }^{-1}$. The formal pseudodifferential composition rule givesand the reversed product has $f,g$ exchanged. ThereforeMultiplying by $L={\partial }^2+u$ on the left, only the leading displayed commutator term can produce a residue. Differentiation of its coefficient produces powers at most ${\partial }^{-2}$, and multiplication by $u$ produces powers at most ${\partial }^{-3}$. Thusand, with the trace convention of part (i),There is no $u$ dependence in this bracket.

Moreover $\mathrm{res}(P_{f}L)=fu$, so $l_{P_f}=\int fudx$ and $\delta l_{P_f}/\delta u=f$. If $K(x,y)=\{u(x),u(y)\}$ is the distributional field bracket, it is characterized bySince $\int {\partial }_x\delta (x-y)g(y)dy=g^{\prime }(x)$, the answer isThe Dirac delta is periodic in the periodic setting, or the ordinary Dirac delta on the real line with admissible test functions. This is the first Hamiltonian structure of the KdV equation, with Poisson operator $-2{\partial }_x$. Its skewness follows by integration by parts; its coefficients are constant, so the functional Jacobi identity has no coefficient-variation terms.

A qualification is needed if the displayed trace formula is interpreted as a bracket on the reduced space for arbitrary operator representatives. Multiplication by a function $h$ gives $l_{m_h}({\partial }^2+u)=0$, yet direct expansion givesWith $h=g=\cos x$ on a period $[0,2\pi ]$, the latter is $\pi$. Hence the formula does not descend through every possible representative of a functional on this reduced affine space. The requested $f{\partial }^{-1},g{\partial }^{-1}$ representatives, used consistently for variational gradients, give the well-defined constant field bracket just computed; a general restriction from the ambient operator algebra needs an appropriate reduction.

Write $P={\sum }_{j\le \alpha }{p}_j(x){\partial }^j$, setting unused coefficients to zero. Split the Adler trace into its fixed leading part and variable parts:Cyclicity from part (i) givesThus this Hamiltonian trace functional on monic differential operators has the required formTo make each coefficient explicit, expand ${\partial }^l{p}_j$ by the Leibniz rule. A term with $k$ derivatives of $p_j$ contributes to the residue exactly when $l-k+j=-1$. Consequently,These are smooth functions depending on $P$, not on the variable coefficients of $L$. In particular $a_0=p_{-1}$ and $a_1=p_{-2}+p_{-1}^{\prime }$. The coefficient $a_0$ must be included: it appears in the displayed sum even though the printed coefficient list starts at $a_1$.

The monic operators form an affine space, not a vector space. Accordingly $l_P$ is affine in the coordinates $u_l$, with constant $c$, and its differential is linear in variations. Its variational derivatives are $\delta l_P/\delta u_l=a_l$. It is the restriction of a linear trace pairing on the full operator space, which explains the terminology without discarding the constant term.

Put every formal pseudodifferential operator in normal order, with all coefficient functions on the left. Its formal pseudodifferential residue is the coefficient of ${\partial }^{-1}$. The Adler trace isHere and below, use periodic coefficients and integrate over a period, or impose convergence and vanishing boundary terms on the real line. Equivalently, the algebraic integration functional must annihilate total derivatives. This convention is needed for the requested trace property; arbitrary elements of $C^{\infty }(\mathbb{R})$ do not supply it automatically.

It suffices to compute with monomials $A=a{\partial }^p$, $B=b{\partial }^q$. Write $r=p+q+1$. If $r<0$, no term in either product has power ${\partial }^{-1}$, so both residues are zero. If $r\ge 0$, the formal pseudodifferential composition rule givesSince $q=r-1-p$, reversal of the factors in the numerator provesAfter $r$ integrations by parts, $\int a{b}^{(r)}dx=(-1{)}^r\int b{a}^{(r)}dx$. Hence the two product traces agree. More explicitly, for $r\ge 1$ their residue difference is the total derivativeFor $r=0$ the residue difference is $ab-ba=0$. In products of operators bounded above in order, only finitely many monomial pairs can contribute to a residue: the required $r$ is nonnegative, so the two downward coefficient indices have a bounded sum. Summing the monomial result therefore proves

The boundary convention is substantive. Without it, take $A={\partial }^2$, $B=(\tanh x){\partial }^{-2}$. Both individual residues vanish, but $\mathrm{res}[A,B]=2(\tanh x{)}^{\prime }=2{\mathrm{sech}}^{2}x$ and its integral over $\mathbb{R}$ is $4$, not zero. The primitive has unequal limits at the two ends. Also a generic smooth residue need not have a convergent integral at all. Thus the proof supplies the intended cyclic trace under its necessary analytic or formal integration convention.

The coefficients of the formal series must be smooth functions $H_n\in C^{\infty }(M)$, rather than points of $M$. Interpret complex parameter values in the complexified function algebra. The Casimir function condition for the Poisson pencil is a coefficientwise formal identity:Equality of the formal coefficients provesThese are the Lenard-Magri recursion relations, valid for every smooth function $f$.

For $i\ge 1$ and $j\ge 0$, use the recursion in the first argument, then antisymmetry and the recursion in the second:Each step lowers the first index while raising the second. Iterating $i$ times givesbecause $H_0$ is a Casimir function of the first Poisson structure. This includes $i=0$ directly. The recursion one step higher then givesThus the commuting coefficients of a Poisson pencil Casimir satisfyThey are therefore Poisson-commuting functions for every member of the Poisson pencil. The argument proves involution, not functional independence; an infinite sequence can contain repetitions or zero coefficients.

Since $|T_j(x)|\le 1$ on $[-1,1]$ and the coefficients are summable, the Weierstrass M-test gives uniform convergence of this positive lacunary Chebyshev series and makes its sum continuous.

Let $r$ be the largest integer with $3^r\le n$, and set $r=-1$ when $n=0$. The candidate and its tail size areThe empty sum for $n=0$ is zero. The candidate has degree at most $n$. The triangle inequality bounds its error by the proposed tail sum, and equality holds at $x=1$, since every Chebyshev polynomial has $T_j(1)=1$ and all tail coefficients are positive. It remains to prove that another polynomial cannot improve this error.

Put $L=3^{r+1}$, the first omitted degree, so $L>n$. At the points $y_j=\cos (j\pi /L)$ for $j=0,\dots ,L$, every omitted term hasbecause $3^{k-r-1}$ is odd. Uniform convergence therefore givesThese $L+1$ distinct points run in decreasing order; reversing their order still gives alternating signs. Since $L+1\ge n+2$, select $n+2$ consecutive ones and apply the Chebyshev alternation theorem. It proves both optimality and uniqueness of the displayed candidate. Thus the best polynomial and the error stay unchanged between successive powers of three; this exact truncation property depends on the synchronized tail signs, not on a general rule that truncating a Chebyshev expansion is optimal.

This addresses the general interpolation bound in part (I); the subsequent (a)–(c) address the cubic case in part (II). Write the interpolant in the B-spline basis asThe stipulated uniquely defined spline interpolation operator presupposes invertibility of the B-spline collocation matrix. For the standard increasing distinct sites, the conditions $N_i(x_i)>0$ give it by the Schoenberg–Whitney theorem. Thus $c=A_{\mathbf{x}}^{-1}Rf$.

Use the standard partition-normalized B-splines of the question: they are nonnegative and satisfy the subpartition of unity for B-splines, ${\sum }_j{N}_j(t)\le 1$. This also holds outside the basic knot interval; one may extend the knots and regard the finite family as a subfamily of a full partition of unity. The recurrence proof of the full partition is given in Question 6(c). Repeated knots retain the same nonnegativity and subpartition properties, with zero-denominator recurrence terms interpreted as zero or by knot limits. ConsequentlySampling has norm at most one, $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, and soHere the induced matrix norm is the maximum absolute row sum. The label $L^{\infty }$-normalized means the usual partition normalization, not rescaling each spline to have maximum value one. Such a rescaling would invalidate the specific coefficient bound and the cardinal values used below.

We first derive the recurrence needed for the partition of unity. Denote an order-$r$ partition-normalized spline by $N_{i,r}$. For $r\ge 2$, set $h_t(u)=(u-t{)}_{+}^{r-2}$, so $(u-t)h_t(u)=(u-t{)}_{+}^{r-1}$. The elementary product identities for divided differences areThey follow from the Leibniz rule for divided differences because the divided differences of the linear factor vanish above first order. Use the first identity on $t_i,\dots ,t_{i+r-1}$ and the second on $t_{i+1},\dots ,t_{i+r}$. Subtracting cancels the common middle-block term. The defining divided-difference recurrence for $N_{i,r}$ then givesThis is the Cox-de Boor recurrence, derived here from the specified normalization rather than assumed as a partition theorem.

Extend the distinct knot sequence in both directions to a strictly increasing sequence $(t_i{)}_{i\in \mathbb{Z}}$ tending to $\pm \infty$. For order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, so the full sum equals one. All higher-order splines are supported on $[t_i,t_{i+r}]$, making the following sums locally finite. Reindexing the second term of the recurrence yieldsInduction proves that the full sum is one for every order. The same recurrence gives nonnegativity: each coefficient is nonnegative wherever its associated lower-order spline is nonzero. Removing splines therefore gives the subpartition of unity for B-splines on the whole line.

For $t_k<t<t_{n+1}$, a spline with index $i\le 0$ ends no later than $t_k$, and one with $i\ge n+1$ starts no earlier than $t_{n+1}$. They all vanish, leaving exactly the indices $1,\dots ,n$. For $k\ge 2$ the splines are continuous and vanish at their support endpoints, so the equality extends to the closed basic knot interval:If order one is allowed, the standard half-open indicator convention gives equality on $[t_1,t_{n+1})$; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or $L^{\infty }$ norms. For the continuous splines of order at least two no such convention is necessary.

Two absolute values on a field are equivalent absolute values when they induce the same topology. For nontrivial absolute values on a field this is equivalent tofor a fixed real number $c>0$. The trivial absolute value is equivalent only to itself.

Restriction of a Non-Archimedean absolute value plainly preserves the ultrametric inequality. Conversely, if the restriction is non-Archimedean, every integer, interpreted in either field, has absolute value on a field at most one. In particular every binomial coefficient has absolute value on a field at most one. The triangle inequality applied to the binomial expansion gives, for $x,y\in L$ and every positive integer $N$,Raise both sides to the power $1/N$ and let $N\to \infty$. Since $(N+1{)}^{1/N}\to 1$, this yieldsThis is the bounded-integer criterion for a non-Archimedean absolute value. The proof works for arbitrary field extensions, not just algebraic ones, and in positive characteristic as well.

An extending absolute value on a field is non-Archimedean by Question 1(a). We prove finite-dimensional non-Archimedean norm equivalence over a complete field, using completeness rather than local compactness.

Fix a basis $e_1,\dots ,e_d$ of a finite-dimensional vector space over $K$, and let $\Vert \cdot \Vert$ be a norm homogeneous for the given absolute value on a field and satisfying the ultrametric inequality. Write $\Vert \sum a_i{e}_i{\Vert }_0={\max }_i|a_i|$. The upper boundis immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\dots ,e_{d-1}$ is complete for its restricted norm by the induction comparison with the coordinate norm and completeness of $K$. A complete vector subspace of a metric space is closed, soFor $v=w+a{e}_d$ with $a\ne 0$, scaling by $a$ gives $\Vert v\Vert \ge |a|\delta$; the same coefficient bound is trivial when $a=0$. AlsoThe induction bound controls every coordinate of $w$ by a constant times $\Vert w\Vert$. Together with the bound on $a$, this proves $\Vert v{\Vert }_0\le C^{\prime }\Vert v\Vert$.

Two extending absolute values on a field on $L$ are norms of this kind on its finite-dimensional vector space over $K$. Hence there is $B>0$ with $|x{|}_2\le B|x{|}_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:Letting $N\to \infty$ gives $|x{|}_2\le |x{|}_1$. Interchanging the two norms gives equality. ThusThe proof covers the trivial base absolute value on a field too. It proves uniqueness, without assuming an existence theorem or compactness of the coordinate sphere $\{v:\Vert v{\Vert }_0=1\}$.

A local field is nondiscrete and locally compact. In the non-Archimedean case its valuation ring is compact by the scaling argument in part (b). Any Cauchy sequence has a tail in a translate of a compact ball, so has a convergent subsequence; the Cauchy property forces the whole sequence to converge. Thus the field is complete. Part (b) gives a discrete valuation, a uniformizer $\pi$, and a finite residue field $k={\mathbb{F}}_q$, $q=p^f$.

We construct the coefficient field rather than presuming it exists. The polynomial $T^q-T$ has derivative $-1$ in characteristic $p$. Every residue element therefore has a unique polynomial root lift by Hensel's lemma. The set $S$ of these $q$ lifts is a field: qth powers preserve addition and multiplication, so sums and products remain polynomial root lifts; a nonzero lift has inverse $a^{q-2}$. Reduction identifies $S$ with ${\mathbb{F}}_q$. This is the finite coefficient field in positive-characteristic local fields.

For $x\in {\mathcal{O}}_K$, choose $a_0\in S$ with the same residue, write $x=a_0+\pi x_1$, and repeat with $x_1\in {\mathcal{O}}_K$. Completeness givesThe first nonzero coefficient determines the valuation, so the expansion is unique. Multiplying an arbitrary field element by a suitable power of $\pi$ reduces it to this case. Consequently substitution $T\mapsto \pi$ gives a valued-field isomorphismAddition and multiplication agree with those of Laurent series by convergence, and a nonzero series has a nonzero leading coefficient, establishing injectivity as well as surjectivity.