Prove that Banach algebra $B(X)$ has a unique (up to equivalence) complete algebra norm.

Question

Let $X$ be a Banach Space, and denote by $∥ · ∥$ the standard norm on $B(X)$, the space of bounded linear functions $T:X\to X$.

(a) Suppose that $||| · |||$ is another algebra norm on $B(X)$. Prove that there exists $C > 0$ such that $∥ · ∥ ≤ C||| · |||$.

(b) Prove that B(X) has a unique (up to equivalence) complete algebra norm.

For (a), Assume not, then $\forall n\in \mathbb{N}, \exists T_n \in \mathcal{B}(X)$, such that $\|T_n\|> n|||T_n|||$, so scale $T_n$ such that $|||T_n||| = 1, \|T_n\|> n$.

$\forall x_0\in X, \lambda\in X^*$, define $S = x_0\lambda, S(x) = \lambda(x)x_0$, then $|S(x)| = |\lambda(x)| \|x_0\| \le \|\lambda\|\|x_0\| \|x\|$, so $\|S\|\le \|\lambda\|\|x_0\|$, and so $S$ is a bounded functional on $X$.

$\forall R \in \mathcal{B}(X), \forall x\in X, (SRS)(x) = S(R(\lambda(x)x_0)) = \lambda(x)\lambda(Rx_0)x_0 = (\lambda(Rx_0)S)(x)$, so $SRS = \lambda(Rx_0)S$. And so $$|||SRS||| = |\lambda(Rx_0)|\cdot|||S||| \le |||S|||\cdot|||R|||\cdot|||S|||.$$ When $S\neq 0, |\lambda(Rx_0)| \le |||S|||\cdot|||R|||$.

Since $|||T_n||| = 1, |\lambda(T_n x_0)| \le |||S|||$. Then let $T'_n = T_n/\|T_n\|$, then $|\lambda(T_n'x_0)| \to 0$,

So $\forall x_0\in X, T_n' x_0 \rightharpoonup 0$ weakly. Thus...???

For (b), assume (a). It remains to show that if $|||\cdot |||$ is a complete norm, then $|||\cdot ||| \le D ||\cdot ||$ for some D. A proof by contradiction starts by finding a sequence $T_n$ such that $|||T_n||| = 1$, $||T_n|| \to 0$, but this does not lead to anything contradictory.

Johnson (1967) THE UNIQUENESS OF THE (COMPLETE) NORM TOPOLOGY shows a related result, but I don't find it helpful.

Answer 1

For part one, continue from "When $S≠0,|λ(Rx_0)|≤|||S|||⋅|||R|||$.".

Let $R = T_n$, then use Uniform Boundedness Principle to find $x_0\in X$ such that $(T_nx_0)_n$ is unbounded, then use Uniform Boundedness Principle again to find $\lambda \in X^*$ such that $(\lambda(T_nx_0))_n$ is unbounded, and that's a contradiction.

This was based on a hint from page 45 of Introduction to Banach Algebras, Operators, and Harmonic Analysis (2003) by George A. Willis, H. G. Dales, Jörg Eschmeier, Kjeld Laursen, and Pietro Aiena.

For part (b), use open mapping theorem on the identity map $id: (X, \|\cdot\|)\to (X, |||\cdot|||)$, which is bounded by part (a).

Answer 2

For part b): if you have two complete norms $\|.\|_1$ and $\|.\|_2$ on a space $Y$ such that $\|.\|_1 \leq C\|.\|_2$ then the two norms are automatically equivalent. This is immediate from Open Mapping Theorem. [Consider the identity map from $(Y,\|.\|_2) \to (Y,\|.\|_1)$.]