# Banach Algebra with Unity is Unital Banach Algebra

## Theorem

Let $\mathbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot}$ be a non-trivial Banach algebra over $\mathbb F$.

Suppose that $X$ has an identity element $\mathbf 1_X$.

Then there exists a norm $\norm \cdot '$ on $X$ equivalent to $\norm \cdot$ such that $\struct {X, \norm \cdot '}$ is a unital Banach algebra.

That is:

$(1): \quad\norm \cdot$ and $\norm \cdot '$ are equivalent
$(2): \quad\struct {X, \norm \cdot '}$ is a Banach algebra
$(3): \quad\norm {\mathbf 1_X}' = 1$

## Proof

Define $\norm \cdot' : X \to \closedint 0 \infty$ by:

$\norm a' = \sup \set {\norm {a b} : \norm b \le 1}$

for each $a \in X$.

Note that for each $a, b \in X$ with $\norm b \le 1$, we have:

$\norm {a b} \le \norm a \norm b \le \norm a$

so that:

$\norm a' \in \hointr 0 \infty$

for each $a \in X$.

We now verify that $\norm \cdot'$ is an norm.

We can see that:

$\norm 0' = \sup \set {\norm 0 : \norm b \le 1} = 0$

Conversely, suppose that:

$\norm a' = 0$

for $a \in X$.

Then:

$\sup \set {\norm {a b} : \norm b \le 1} = 0$

while:

$\norm {a b} \ge 0$ for each $b \in X$ with $\norm b \le 1$.

So we have:

$a b = 0$ for all $b \in X$ with $\norm b \le 1$.

In particular, since:

$\ds \norm {\frac {\mathbf 1_X} {\norm {\mathbf 1_X} } } = 1$

we have, setting $b = \paren {\norm {\mathbf 1_X} }^{-1} \mathbf 1_X$:

$\dfrac a {\norm {\mathbf 1_X} } = 0$

so that:

$a = 0$

So Norm Axiom $\text N 1$: Positive Definiteness holds for $\norm \cdot'$.

Now let $a \in X$ and $\lambda \in \Bbb F$.

We then have:

 $\ds \norm {\lambda a}'$ $=$ $\ds \sup \set {\norm {\lambda a b} : \norm b \le 1}$ $\ds$ $=$ $\ds \sup \set {\cmod \lambda \norm {a b} : \norm b \le 1}$ $\ds$ $=$ $\ds \cmod \lambda \sup \set {\norm {a b} : \norm b \le 1}$ Multiple of Supremum $\ds$ $=$ $\ds \cmod \lambda \norm a'$

So Norm Axiom $\text N 2$: Positive Homogeneity holds for $\norm \cdot'$.

We now verify Norm Axiom $\text N 3$: Triangle Inequality.

Let $x, y \in X$.

Then for each $b \in X$ with $\norm b \le 1$, we have:

$\norm {\paren {x + y} b} \le \norm {x b} + \norm {y b}$

by Norm Axiom $\text N 3$: Triangle Inequality for $\norm \cdot$.

Then, we have:

$\norm {\paren {x + y} b} \le \sup \set {\norm {x b} : \norm b \le 1} + \sup \set {\norm {y b} : \norm b \le 1}$

for each $\norm b \le 1$.

Taking the supremum over such $b$ we obtain:

$\norm {x + y}' \le \norm x' + \norm y'$

for each $x, y \in X$.

So Norm Axiom $\text N 3$: Triangle Inequality holds for $\norm \cdot'$.

So, we have that $\norm \cdot'$ is a norm.

We now show that $\norm \cdot'$ is equivalent to $\norm \cdot$.

Note that we have already shown that:

$\norm a' \le \norm a$

for each $a \in X$.

For the other direction, noting that:

$\ds \norm {\frac {\mathbf 1_X} {\norm {\mathbf 1_X} } } = 1$

so that:

$\ds \frac {\norm a} {\norm {\mathbf 1_X} } \in \set {\norm {a b} : \norm b \le 1}$

giving:

$\ds \frac {\norm a} {\norm {\mathbf 1_X} } \le \sup \set {\norm {a b} : \norm b \le 1}$

We therefore have:

$\ds \frac {\norm a} {\norm {\mathbf 1_X} } \le \norm a' \le \norm a$

for each $a \in X$.

So $\norm \cdot'$ is equivalent to $\norm \cdot$.

So we have shown $(1)$.

From Norm Equivalence Preserves Completeness, we have:

$\struct {X, \norm \cdot'}$ is a Banach space.

To show that $\struct {X, \norm \cdot'}$ is a Banach algebra, we now just need to show that $\norm \cdot'$ is an algebra norm.

We clearly have:

$\norm {x y}' \le \norm x' \norm y'$

if $x = 0$ or $y = 0$.

Now take $x, y \in X \setminus \set 0$.

Clearly we have:

$\norm {\paren {x y} b} \le \norm x' \norm y'$

for $b = 0$.

Now let $b \in X \setminus \set 0$.

We have:

 $\ds \norm {\paren {x y} b}$ $=$ $\ds \norm {x \paren {y b} }$ $\ds$ $=$ $\ds \norm {x \paren {\frac {y b} {\norm {y b} } } \norm {y b} }$ $\ds$ $=$ $\ds \norm {y b} \norm {x \paren {\frac {y b} {\norm {y b} } } }$ Norm Axiom $\text N 2$: Positive Homogeneity

Since:

$\ds \norm {\frac {y b} {\norm {y b} } } = 1$

we have:

$\ds \norm {x \paren {\frac {y b} {\norm {y b} } } } \le \norm x'$

Since $\norm b \le 1$, we have:

$\norm {y b} \le \norm y'$

and so:

$\norm {\paren {x y} b} \le \norm x' \norm y'$

for each $b \in X$ with $\norm b \le 1$.

Taking the supremum over $b$ we have:

$\norm {x y}' \le \norm x' \norm y'$

for each $x, y \in X$.

So we have $(2)$.

Now, we show that:

$\norm {\mathbf 1_X}' = 1$

We have:

 $\ds \norm {\mathbf 1_X}'$ $=$ $\ds \sup \set {\norm b : \norm b \le 1}$ $\ds$ $=$ $\ds \sup \closedint 0 1$ $\ds$ $=$ $\ds 1$

and so we have $(3)$, and are done.

$\blacksquare$