Exercise 7, § 4

a) Let $B$ be a complex Banach space and $j$ a point of $B$ such that $\|j\|= 1$. For every $x\in B$, the limit

1

lim $(\|j+\alpha x\| -1)$

$\alpha \alpha >\rightarrow 00\alpha$

exists. Let $\varphi (x)$ be this limit.

b) Let

$\psi (x) =$ sup$_{|\varepsilon\varepsilon\in|=1\mathbf{C}}\varphi (\varepsilon x)$.

The function $\psi$ is a seminorm majorized by the norm of $B$. One has

$\psi (x) =$ sup$_f|f(x)|$,

where $f$ ranges over the set of elements of the dual of $B$ such that $\|f\|=f(j) = 1$. c) We now suppose that $B$ is a Banach algebra and that $j= 1$ is a unit element of $B$. One has:

$\varphi (x) =$ lim$_{\alpha\alpha >\rightarrow 00}\frac{1}{\alpha}$ log($\|$exp($\alpha x$)$\|$) $=$ sup$_{\alpha >0}\frac{1}{\alpha}$ log($\|$exp($\alpha x$)$\|$).

(Observe that log($\|$exp(($\alpha +\alpha '$)$x$)$\|$)$\leqslant$ log($\|$exp($\alpha x$)$\|$) $+$ log($\|$exp($\alpha 'x$)$\|$).) d) Using c), prove that $\|$exp($\lambda x$)$\|\leqslant$ exp($|\lambda |\psi (x)$). By integrating $\lambda \mapsto \lambda^{-2}$ exp($\lambda x$) on the circle determined by $|\lambda |=\varrho$, deduce that

$\|x\|\leqslant \varrho^{-1}$ exp($\varrho \psi (x)$),

whence, by putting $\varrho =\psi (x)^{-1}$, that $\|x\|\leqslant e\psi (x)$, where $e=$ exp(1).

e) For every nonzero $x$ of $B$, there exists a linear form $f\in B'$ such that $f(1) =\|f\|= 1$ and $f(x)\not= 0$.