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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 $|\cdot {|}_L$.