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On a finite-dimensional vector space over ${\mathbb{Q}}_p$, the triangle inequality gives an upper bound by a coordinate maximum norm. The coordinate unit sphere is compact. The other norm is continuous and nonzero there, so has a positive minimum. Scaling by powers of $p$ gives the lower bound. The resulting equivalence of norms in finite dimensions proves completeness of finite extensions, and applying the comparison to powers of a field element proves equality of two extending multiplicative absolute values on a field.

On the fixed finite-dimensional space of polynomials of degree at most $d$, uniform convergence of $B_n$ on each monomial and equivalence of norms in finite dimensions imply ${\epsilon }_n=\Vert B_n-I{\Vert }_{\mathrm{op}}\to 0$. A Neumann series then gives $\Vert B_n^{-1}-I{\Vert }_{\mathrm{op}}\le {\epsilon }_n/(1-{\epsilon }_n)$ for sufficiently large $n$. Consequently the displayed convergence holds. A strictly positive polynomial has positive minimum on the compact interval, so its inverse Bernstein approximants are also strictly positive for all sufficiently large $n$.

Let $E=\mathbb{R}[x{]}_{\le d}$ with the supremum norm. By equivalence of norms in finite dimensions, there is $C_d$ such that ${\sum }_{j=0}^d|a_j|\le C_d\Vert p{\Vert }_{\infty }$ whenever $p={\sum }_j{a}_j{x}^j$. Henceso ${\epsilon }_n:=\Vert B_n-I{\Vert }_{\mathrm{op}}\le C_d{\max }_{0\le j\le d}\Vert B_n(x^j)-x^j{\Vert }_{\infty }\to 0$. This upgrades convergence on fixed polynomials to operator norm convergence on the fixed space.

For sufficiently large $n$, ${\epsilon }_n<1$, and a Neumann series givesThus the inverse Bernstein approximation on a fixed-degree polynomial space satisfies $\Vert g_n-f{\Vert }_{\infty }\to 0$. By the extreme value theorem, strict positivity on the compact interval gives $\mu ={\min }_{[0,1]}f>0$. Eventually $\Vert g_n-f{\Vert }_{\infty }<\mu /2$, and consequentlyApplying pointwise convergence to the varying polynomials $g_n$ without this finite-dimensional operator argument would not justify the conclusion.

Choose a basis of the finite-dimensional vector space and let $|x{|}_{\infty }$ be the maximum absolute coordinate. For any norm $N$, the triangle inequality givesso $N$ is continuous in the coordinate topology. On the compact set $\{x:|x{|}_{\infty }=1\}$, the positive continuous function $N$ attains a positive minimum $m$ and a finite maximum $M$. By absolute homogeneity,This proves the equivalence of norms in finite dimensions: every norm is equivalent to the maximum norm, and transitivity of these inequalities proves that any two norms on a finite-dimensional vector space are Lipschitz equivalent.

Choose a $K$-basis of $L$. Each extended absolute value is a norm on the finite-dimensional $K$-vector space $L$, and equivalence of norms in finite dimensions shows that every such norm induces the same topology when $K$ is complete. Consequently any two extended absolute values induce the same topology on $L$. By the result of Question 2a, one is a positive real power of the other. Their restrictions to the nontrivially valued field $K$ are both $|\cdot |$, so that power is one. In particular every extension is equal, and therefore equivalent, to $|\cdot {|}_L$.

For $0<\alpha <2$, write $z=r{e}^{i\theta }$ with $0<\theta <\pi$. The chosen power map sends this to $r^{\alpha }{e}^{i\alpha \theta }$, soAt $\alpha =2$ this is the plane slit along the nonnegative real axis. For every $0<\alpha \le 2$, the map is conformal on the upper half-plane.

To check finiteness of the physical exit time, start a Brownian motion $B$ at $i$ and let $T$ be its real-axis exit. The time change of a continuous process in conformal invariance of planar Brownian motion isWe know $T<\infty$ almost surely. The Cauchy exit law from a Brownian half-plane assigns probability zero to $B_T=0$. On such a path, $B_s\ne 0$ throughout the compact interval $[0,T]$, because it is in the open upper half-plane before exit and has nonzero endpoint. Its modulus therefore has a positive minimum and finite maximum. The displayed integrand is bounded on that interval, proving $A_T<\infty$. The image Brownian motion exits $D$ at this finite changed time.

The endpoint $\alpha =0$ printed in the source is degenerate: $z^0=1$, so its image is the singleton $\{1\}$, not an open domain, and the map is not conformal. If exit from that set is defined for Brownian motion started at one, its exit infimum is zero. It cannot be included in a nondegenerate conformal-domain argument.

Here is the martingale argument for maximal comparisons, including the starting-value issue. For $P=f+ig$, the Cauchy-Riemann equations and Itô's formula show thatare orthogonal continuous local martingales with the same quadratic variation. Localize before large disks and before exit. The Burkholder-Davis-Gundy inequalities then givewith a constant depending only on $p$; Fatou's lemma removes the localization. These are centered martingales. For the uncentered processes this impliessince $|f(i^{\alpha })|\le \sup |f(Z_t)|$.

The source normalizes $P(0)=0$ at a boundary point, rather than the value of $g$ at the starting point. Consequently its displayed constant cannot be uniform in $\alpha$ with only $p$ dependence. To see this, take $P(z)=iz$ and $0<\alpha <\min (1/2,1/p)$. Then $f=-\mathrm{Im}z$ and $g=\mathrm{Re}z$. Inside this narrow wedge,Both $p$th moments are finite and the right one is positive; finiteness follows from the wedge radial-maximum tail of order $R^{-1/\alpha }$, justified below. The source inequality would force $C_p\ge {\cot }^p(\alpha \pi )$, unbounded as $\alpha \downarrow 0$. At $\alpha =0$, the immediate-exit interpretation gives the direct contradiction $1\le C_{p}0$ for the same polynomial.

For a fixed nondegenerate wedge, an uncentered inequality does hold with a constant allowed to depend on $\alpha$ as well as $p$. One rigorous outline for the polynomial class in this question is as follows. Separation of variables in a sector of radius $R$ gives, for exit through its circular arc, the expansionThus at the specified starting point the radial-maximum tail is comparable to $R^{-1/\alpha }$. Hitting any fixed interior subarc has that same power, with positive leading coefficient. If a nonzero polynomial has degree $d$, its real part has magnitude at least $c{R}^d$ on some fixed interior subarc for sufficiently large $R$. Therefore its maximal real-part $p$th moment is infinite whenever $dp\ge 1/\alpha$. In that case the desired extended-real inequality is automatic.

The remaining polynomials have $d<1/(\alpha p)$, so belong to a finite-dimensional vector space space. On that spaceis a genuine norm: if it is zero, Brownian motion's positive probability of reaching any interior neighborhood forces $\mathrm{Re}P=0$ throughout $D$, so analyticity makes $P$ a purely imaginary constant, and $P(0)=0$ makes it zero. The corresponding imaginary-part functional is bounded on this finite-dimensional vector space normed space; its finiteness follows from $dp<1/\alpha$ and the radial tail bound. equivalence of norms in finite dimensions givesThis establishes the fixed-domain interpretation and identifies exactly why the universal centered martingale inequality cannot silently be applied to the printed uncentered boundary normalization. It is the distinction in starting-value obstruction in wedge conjugate maximal inequalities.

Real and imaginary parts of an analytic Brownian image, after subtracting their starting values, have equal quadratic variation. The Burkholder-Davis-Gundy inequalities compare their centered maximal moments with constants depending only on $p$. A boundary-point normalization $P(0)=0$ does not remove the starting imaginary value. For $P(z)=iz$ on a wedge of opening $\alpha \pi <\pi /2$, $\sup |\mathrm{Im}z|\le \tan (\alpha \pi )\sup |\mathrm{Re}z|$. If $\alpha p<1$ both relevant moments are finite, so an uncentered comparison requires a constant at least ${\cot }^p(\alpha \pi )$; no bound depending only on $p$ can hold over all such wedges. For fixed $\alpha >0$, polynomials with finite real-part maximal moment have degree $d<1/(\alpha p)$ by the wedge radial-maximum tail. They form a finite-dimensional vector space, and equivalence of norms in finite dimensions gives a valid constant depending on $p,\alpha$.