Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$ that is not unital as an algebra.

Let $\struct {A_+, \norm {\, \cdot \,}_1 }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.

Let $\norm {\, \cdot \,}_2$ be the direct product norm on $A \times \GF$.

Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent norms.

Proof

Let $u, v \in \R$.

We have:

$\ds \frac 1 2 \paren {\size u + \size v} \le \frac 1 2 \paren {\max \set {\size u, \size v} + \max \set {\size u, \size v} } = \max \set {\size u, \size v}$

and:

$\ds \max \set {\size u, \size v} \le \max \set {\size u, \size v} + \min \set {\size u, \size v} = \size u + \size v$

Let $x \in A$ and $\lambda \in \GF$.

Taking $u = \norm x$ and $v = \cmod \lambda$ in the above we obtain:

$\ds \frac 1 2 \paren {\norm x + \cmod \lambda} \le \max \set {\norm x, \size \lambda} \le \norm x + \cmod \lambda$

That is:

$\ds \frac 1 2 \norm {\tuple {x, \lambda} }_1 \le \norm {\tuple {x, \lambda} }_2 \le \norm {\tuple {x, \lambda} }_1$

So $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent norms.

$\blacksquare$