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

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.

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 $$(Y,\|.\|_2) \to (Y,\|.\|_1)$$.]
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.
For part (b), use open mapping theorem on the identity map $$id: (X, \|\cdot\|)\to (X, |||\cdot|||)$$, which is bounded by part (a).