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Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. A character of an algebra is a nonzero multiplicative complex linear functional $\chi :A\to \mathbb{C}$. It satisfies $\chi (1)=1$. Moreover $\chi (a)\in {\sigma }_A(a)$: otherwise $a-\chi (a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the spectrum of an element therefore gives $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.

Every proper maximal ideal $M$ of $A$ is closed. Indeed its closure is an ideal; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann series, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The quotient Banach space $A/M$, with its quotient Banach algebra structure, is a complex normed division algebra. By the Gelfand-Mazur theorem, it is $\mathbb{C}$, so the quotient map gives a character of an algebra with kernel $M$. Conversely, the kernel of every character of an algebra is a maximal ideal, since the character is onto $\mathbb{C}$. The Zorn lemma supplies a maximal ideal containing every proper ideal, so the character space of an algebra $\Delta (A)$ is nonempty.

These facts give the exact relation between the character space and the spectrum of an element:One inclusion was proved above. For the other, if $a-\lambda 1$ is noninvertible, the principal ideal it generates is proper because $A$ is commutative. Contain it in a maximal ideal and use its corresponding character of an algebra to obtain $\chi (a)=\lambda$.

Give $\Delta (A)$ the Gelfand topology, namely its subspace topology from the weak-star topology on $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. This is a continuous function on $\Delta (A)$ by definition of the Gelfand topology. The Gelfand representation theorem gives a contractive unital algebra homomorphism over a fieldMultiplicativity and linearity follow by evaluating at each character of an algebra; the supremum norm equality follows from the preceding spectrum of an element identity. Its kernel isthe Jacobson radical. Equivalently, its elements have spectrum of an element $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a semisimple commutative Banach algebra, and it gives a faithful continuous representation of $A/\mathrm{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta (A)$, because distinct characters of an algebra differ on some $a$. An arbitrary Banach algebra need not have an isometric or surjective Gelfand transform, nor a uniformly dense range: those conclusions require further hypotheses.

For the Banach algebra $C(K)$ on a nonempty compact Hausdorff space $K$, all characters of an algebra are evaluation characters. To see this, let $M$ be a maximal ideal. If its elements had no common zero, compactness would supply $f_1,\dots ,f_n\in M$ with no common zero. The continuous function $h={\sum }_i\overline{f_i}{f}_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq \ker {\delta }_x$ and maximality gives equality. The associated character of an algebra must be ${\delta }_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.

The map $x\mapsto {\delta }_x$ is a continuous bijection $K\to \Delta (C(K))$, using separation of points by continuous functions. Compactness and the Hausdorff property make it a homeomorphism. Under this identification the Gelfand transform is $\boxed{\hat{f}({\delta }_x)=f(x)}$, so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty character space represents the zero function space; it was excluded by the nonzero unital convention above.

Now let $A$ be a commutative unital C-star algebra. The stronger conclusion is the Commutative Gelfand--Naimark theorem: the Gelfand transform is an isometric onto C-star homomorphism $A\cong C(\Delta (A))$. We prove the additional assertions without assuming this conclusion.

First every character of an algebra respects the C-star algebra involution. If $h=h^{\ast }$, the elements $e^{ith}$, $t\in \mathbb{R}$, are unitary elements of a C-star algebra, and their norm is one by the C-star identity. Continuity and multiplicativity give $\chi (e^{ith})=e^{it\chi (h)}$. Thus $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the range of $\Gamma$ is closed under complex conjugation.

Every element of commutative $A$ is a Normal element of a C-star algebra. For a normal $b$, use the C-star identity, and then the same identity for the self-adjoint element $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. The Gelfand transform is therefore an isometry, and its range is complete and closed in the supremum norm. It contains constants, separates points and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence all of $C(\Delta (A))$.

The approximation step in Stone-Weierstrass theorem can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\epsilon >0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.

Every finitely supported sequence $v$ lies in $Y$: if $v_n=0$ for $n>N$, set $x_n={\sum }_{k=n}^N{v}_k$ for $n\le N$ and $x_n=0$ afterwards. Then $x$ is a bounded sequence and $x-Sx=v$. Consequently $L(v)=0$. The finitely supported sequences are dense in the space of sequences converging to zero $c_0$ for the supremum norm, by truncation, so continuity of the bounded linear functional $L$ gives $L{|}_{c_0}=0$. For $x$ in the real convergent sequence space, with ordinary limit $\ell$, we have $x-\ell 1\in c_0$. Therefore $\boxed{L(x)=\ell L(1)=\underset{n\to \infty }{\lim }{x}_n}$.

We prove compactness of zero-boundary Poisson solutions. Here the printed $\partial \Omega$ must mean $\partial B_1$; otherwise the boundary condition refers to an unspecified set. Use this intended boundary condition throughout. Let $f=\Delta u$. A maximum-principle barrier gives a uniform bound on $u$: with $\psi (x)=(1-|x{|}^2)/(2n)$, we have $\Delta \psi =-1$ and $\psi =0$ on $\partial B_1$. Since $|f|\le 1$,The weak maximum principle gives $-\psi \le u\le \psi$, so $\Vert u{\Vert }_{\infty }\le 1/(2n)$.

Standard elliptic regularity on the smooth ball first gives $u\in C^{2,\mu }(\overline{B_1})$. The global Schauder estimate then givesThus the functions and all their derivatives through order two are uniformly bounded and equicontinuous on the closed ball. The Arzela-Ascoli theorem gives a subsequence converging uniformly along with all these derivatives. Integration along line segments identifies the limiting derivatives, so the convergence is in $C^2(\overline{B_1})$.

It remains to keep the limit in the set, rather than merely proving relative compactness. If $u_j\to u$ in $C^2$, then the zero boundary values pass to the limit and $f_j=\Delta u_j\to f=\Delta u$ uniformly. For each pair of distinct points,Taking suprema gives $[f{]}_{\mu }\le {\mathrm{liminf}}_j[f_j{]}_{\mu }$, while $\Vert f_j{\Vert }_{\infty }\to \Vert f{\Vert }_{\infty }$. Hence $\Vert f{\Vert }_{C^{0,\mu }}\le 1$. The limit belongs to the set, proving sequential compactness in $C^2(\overline{B_1})$. The fixed positive Hölder exponent supplies the compactness; a bound on the supremum norm of $\Delta u$ alone would not supply this argument.

The Baire category theorem in the needed version states that a nonempty complete metric space cannot be a countable union of closed sets with empty interior. To prove it, suppose $F_1,F_2,\dots$ are such sets and start with a nonempty open set $U$. Inductively choose closed balls ${\overline{B}}_j$ of positive radius at most $2^{-j}$, with ${\overline{B}}_1\subset U\setminus F_1$ and ${\overline{B}}_{j+1}\subset B_j\setminus F_{j+1}$. This is possible because each complement is open and dense. The centres form a Cauchy sequence; completeness supplies a limit in every closed ball. It lies in $U$ and outside every $F_j$, proving the theorem.

Here the sequence spaces are over the real scalars, as required by the Euclidean-plane assertion. Suppose an isometry $\phi :{\mathbb{R}}^2\to c_0$ existed. On the compact complete space $S^1\times S^1$, the setsare closed. They cover that space: for a nonzero sequence in $c_0$, the supremum norm is attained at a coordinate; for $x=y$ every coordinate works. By the Baire category theorem, one $F_n$ contains a product of open arcs. Shrink the arcs to be disjoint. The real coordinate difference is then continuous and nonzero, so its sign is constant on the product. Thus the chord distance has the separated formIn particular its alternating sum on every four-point rectangle is zero. Parametrize the arcs by angles $s,t$ with $0<t-s<2\pi$. Their chord distance is $D(s,t)=2\sin ((t-s)/2)$, whose mixed derivative isThe zero rectangle identity would force this mixed derivative to vanish, a contradiction. Therefore no such isometry exists.

For a real vector $x\in {\ell }_1^n$, list its signed sums over all $2^n$ choices $\epsilon \in \{-1,1{\}}^n$, followed by infinitely many zero coordinates:This is linear, has finite support and lies in $c_0$. Choosing the signs of the components gives ${\max }_{\epsilon }|\epsilon \cdot x|={\sum }_j|x_j|$. Thus $\boxed{\Vert Tx{\Vert }_{\infty }=\Vert x{\Vert }_1}$, the required linear isometry.

Use integer graph vertices in the boxes: ${\Lambda }_n=[-n,n{]}^2\cap {\mathbb{Z}}^2$, and set ${\Lambda }_{-1}=\varnothing$. Thus $\partial {\Lambda }_0=\{0\}$, $g_0=1$, and $|\partial {\Lambda }_m|=8m$ for $m\ge 1$.

If $0\leftrightarrow \partial {\Lambda }_{m+n}$, erase loops from an open graph path to obtain a self-avoiding walk. Let $x$ be its first graph vertex on $\partial {\Lambda }_m$. Split at $x$. The prefix certifies $A_x=\{0\leftrightarrow x\}$, while a segment of the suffix certifies $B_x=\{x\leftrightarrow x+\partial {\Lambda }_n\}$: the final graph vertex is at supremum norm distance at least $n$ from $x$, so the suffix first hits that translated boundary. The two certificates use disjoint edges. Therefore the BK boundary-splitting estimate, the union bound, the van den Berg-Kesten inequality, and translation invariance giveThe cases $m=0$ or $n=0$ follow directly from $g_0=1$, so this proves the bound including the endpoints.

Here is an explicit way to remove the polynomial boundary factor. For $p>0$ put $b_n=32n^2{g}_n$, $n\ge 1$. By symmetry in $m,n$, assume $m\le n$. ThenConsequently $a_n=\log b_n$ is a subadditive sequence. The Fekete lemma states that any real subadditive sequence satisfies $\lim a_n/n={\inf }_{n\ge 1}{a}_n/n$, possibly $-\infty$. In this case the direct horizontal open graph path gives $g_n\ge p^n$, so this limit of a sequence is bounded below by $\log p$. Since $g_n\le 1$, its upper bound is zero. Finally,has the same limit of a sequence. Thus the percolation one-arm decay rate exists andFor $p=0$, every $g_n$ with $n\ge 1$ is zero and $\gamma =0$. No exponential-decay theorem or assumption that $p$ is below the percolation critical probability is needed.

Fix the normalizationThe Lasso penalty uses the L1 norm. All arguments below apply to every minimizer; uniqueness of the coefficient vector is not assumed.

The Karush-Kuhn-Tucker conditions are the subgradient optimality conditionThese conditions are necessary and sufficient because the objective is a convex function. They use the subgradient of the absolute value, coordinate by coordinate.

Put $\delta =\hat{\beta }-{\beta }^0$. Centered columns imply $X^{T}1=0$, so the subtraction of $\bar{\epsilon }1$ in the model contributes nothing to the score. The Karush-Kuhn-Tucker conditions becomeMultiply by ${\delta }^T$. Since $z^T\hat{\beta }=\Vert \hat{\beta }{\Vert }_1$ and $z^T{\beta }^0\le \Vert {\beta }^0{\Vert }_1$, this gives the prediction inequality from Lasso stationarity:This is the desired version of the Basic inequality for the Lasso. Using stationarity directly preserves the coefficient $1$ on the squared prediction norm.

For the Gaussian score event for the Lasso, write $X_j$ for column $j$. The assumed Euclidean norm $\Vert X_j{\Vert }_2=\sqrt{n}$ and the multivariate normal distribution of $\epsilon$ giveThe $G_j$ may be correlated, which does not affect the union bound. Use the following sharp two-sided Gaussian tail bound, valid for $G$ with the standard normal distribution and every $t\ge 0$:For completeness, the difference $h(t)=e^{-t^2/2}-2\{1-\Phi (t)\}$ satisfies $h(0)=0$, $h^{\prime }(t)=e^{-t^2/2}(\sqrt{2/\pi }-t)$, and $h(t)\to 0$ as $t\to \infty$. It first increases and then decreases to zero, so it is nonnegative. This bound avoids introducing an unnecessary factor two.

Here $\Phi$ is the standard normal distribution function. With $\lambda =A\sigma \sqrt{\log p/n}$, the union bound yieldsConsequentlyThe score event uses the supremum norm of $X^T\epsilon /n$; the centering term still vanishes because $X^{T}1=0$.

We next work deterministically on $\Omega$. LetBy Holder inequality, ${\epsilon }^{T}X\delta /n\le c\lambda \Vert \delta {\Vert }_1=c\lambda (a+b)$. Since ${\beta }_N^0=0$, the triangle inequality givesSubstituting both inequalities into the Basic inequality for the Lasso producesBecause $Q\ge 0$, this also proves the relevant Lasso cone condition, $(1-c)b\le (1+c)a$. The assumed Compatibility condition for the Lasso therefore applies to $\delta$:Set $D=(1+c)\lambda \sqrt{s}/\varphi$. Combining these inequalities givesIf $Q>0$, discarding the nonnegative $b$ term shows $\sqrt{Q}\le D$. If $Q=0$, compatibility gives $a=0$, and the preceding inequality then gives $b=0$. In either case $D\sqrt{Q}\le D^2$, soThis is a fast prediction bound and an inactive-coordinate error bound on the same event. It is the fast-rate Lasso prediction bound under compatibility.

Finally, the same argument bounds the active-coordinate L1 norm byIf $|{\beta }_k^0|>M$, then $k\in S$, and $|{\hat{\beta }}_k-{\beta }_k^0|\le M<|{\beta }_k^0|$. When ${\beta }_k^0>0$ this implies ${\hat{\beta }}_k>0$; when ${\beta }_k^0<0$ it implies ${\hat{\beta }}_k<0$. Thus the sign function obeysThis is Lasso sign recovery under compatibility. It concerns the sufficiently large nonzero coefficients; it does not assert that every estimated inactive coefficient is zero.

For the constant monomial $f_0=1$, the binomial theorem gives $B_n(f_0)(x)=(x+1-x{)}^n=1$. For $f_1$, the binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ givesThus $\boxed{B_n(f_0)=1,B_n(f_1)=x}$, including $x=0,1$ by evaluation of the polynomials.

For $m\ge 1$ and $n\ge 2$, substitute the preceding coefficient identity into the Bernstein polynomial sum and write $j=k-1$. This yields the Bernstein monomial recurrenceThe coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n{)}^{m-1}\to 1$ and the others tend to zero. The weights in any Bernstein polynomial sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le 1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x{x}^{m-1}=x^m$: the $\ell =0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the supremum norm proves uniform convergence on $[0,1]$.

As an independent quantitative bound, take a binomial distribution variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb{E}[(K/n{)}^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the Cauchy-Schwarz inequality and the variance of $K/n$ giveFor $m=0$ the error is exactly zero. Finally the requested sum equals $2^m{B}_{2n}(f_m)(1/2)$, so

The unheaded preliminary requests are addressed before the first example. A norm is a function $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly for $x=0$, $p(ax)=|a|p(x)$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d(x,y)=p(x-y)$. Translation $T_v$ preserves this distance, and scalar multiplication satisfies $d(ax,ay)=|a|d(x,y)$, so both are continuous; multiplication by zero is constant.

The open unit ball determines the norm uniquely through its Minkowski functional:because $x/t\in B$ is equivalent to $p(x)<t$. This also works for $x=0$.

Under the given radial and convexity hypotheses, set $p_B(0)=0$ and $p_B(v)=1/\lambda (v)$ for $v\ne 0$. The finite symmetric segment along each line gives positivity, absolute homogeneity and $v\in B⟺p_B(v)<1$. If $s>p_B(x)$ and $t>p_B(y)$, then $x/s,y/t\in B$, and the stated convexity condition yields $(x+y)/(s+t)\in B$. Hence $p_B(x+y)<s+t$; taking $s\downarrow p_B(x)$ and $t\downarrow p_B(y)$ proves the triangle inequality. Thus this is a norm determined by a convex radial unit ball. The case $V=\{0\}$ is immediate.

For the coordinate box, the Minkowski functional is the supremum norm:Indeed $x/t\in B$ exactly when $t>{\max }_i|x_i|$.

Use the paper's classical norm convention: $|u{|}_{2;U}$ is the derivative supremum norm through order two, not an $L^2$ norm. Write $H(u;U)=[D^{2}u{]}_{\alpha ;U}$ and $Q(u,f)=|u{|}_{2;B_1}+|f{|}_{0,\alpha ;B_1}$, where the latter is the full Hölder norm of the forcing.

Fix $0<\delta <1$. If the asserted estimate failed, rescaling the functions by their global Hessian matrix Hölder seminorm would produce $u_j,f_j$ withIn particular $Q(u_j,f_j)\to 0$. Choose $x_j,y_j\in B_{1/2}$ for whichThe supremum norm of $D^2{u}_j$ tends to zero, so $r_j=|y_j-x_j|\to 0$. Let $P_j$ be the quadratic Taylor polynomial of $u_j$ at $x_j$, and defineThe domains contain $B_{1/(2r_j)}$, and $U_j,D{U}_j,D^2{U}_j$ vanish at zero. MoreoverThe right side is bounded by $[f_j{]}_{\alpha }|z{|}^{\alpha }$ and tends to zero locally uniformly. The normalized Hessian matrix bound controls $|D^2{U}_j(z)|\le |z{|}^{\alpha }$; integration along line segments then bounds $D{U}_j,U_j$ on every compact Euclidean ball. Arzela-Ascoli theorem and a diagonal subsequence argument give convergence in $C^2$ on compact subsets to an entire harmonic function $U$, with $[D^{2}U{]}_{\alpha ;{\mathbb{R}}^n}\le 1$ and $D^{2}U(0)=0$.

After a further subsequence $z_j=(y_j-x_j)/r_j\to z_{\ast }$ with $|z_{\ast }|=1$. The normalization ensures $|D^{2}U(z_{\ast })|\ge \delta /2$. But each second derivative of $U$ is harmonic and has finite global Hölder seminorm with exponent below one. The supplied Liouville theorem makes each derivative constant; it is the globally Hölder case of the polynomial-growth Liouville theorem for harmonic functions. Its value at zero makes it zero, a contradiction. This blow-up compactness proof of an interior Schauder estimate establishes

To absorb the term on the larger Euclidean ball, rescale this estimate to Euclidean balls contained in $B_s$. For $1/2\le r<s<1$, pairs separated by less than $(s-r)/2$ are controlled in a Euclidean ball centred at their first point; farther pairs are controlled directly by $\Vert D^{2}u{\Vert }_{\infty }$. ThusThis is the covering step in the Simon absorption lemma. Take $r_j=1-2^{-j-1}$ and fix $\delta <2^{-2-\alpha }$. Iterating makes the remainder ${\delta }^{j}H(u;B_{r_j})$ tend to zero, since the global seminorm is finite; the error terms form a convergent geometric series. Consequently

For the requested failure of the weaker estimate, take $u_m=m^{-2}\sin (m{x}_1)$ and $f_m=-\sin (m{x}_1)$. The derivative supremum norm $|u_m{|}_2$ and $\Vert f_m{\Vert }_{\infty }$ remain bounded, but comparison at $0$ and $\pi e_1/(2m)$, both in $B_{1/2}$ for large $m$, givesThis proves the necessity of Hölder forcing for Schauder estimates. Every member of the sequence is smooth; failure comes from the increasing frequency, not from any lack of individual regularity.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.