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 holds for $\norm \cdot'$.

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

We then have:

So holds for $\norm \cdot'$.

We now verify.

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 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 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:

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:

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