Definition: A Banach algebra $A$ is a triple $(A, \Vert \cdot \Vert, m)$ where $\Vert \cdot\Vert$ is a norm on $A$ and $m: A \times A \to A$ is a $\mathbb{C}$-bilinear associative map such that $(A, \Vert\cdot \Vert)$ is a Banach space and such that $(A,m)$ is a $\mathbb{C}$-algebra, such that
$$\Vert xy \Vert = \Vert m(x,y)\Vert \leq \Vert x \Vert \Vert y\Vert$$
for all $x,y \in A$.
My notes say: Assume $A$ is unital with identity $1_A$. Changing to an equivalent norm, we may assume that $\Vert 1_A \Vert=1$.
How exactly is this done? If $(A, \Vert \cdot \Vert)$ is a Banach algebra, then I can define a new norm
$$p: A \to [0, \infty[: a \mapsto \Vert a\Vert/\Vert 1_A\Vert $$
and I see $p$ and $\Vert \cdot \Vert$ are equivalent norms but $(A, p)$ does not satisfy $p(x,y) \leq p(x) p(y)$ so it can't give a Banach algebra.
Any insight is appreciated!
For $a \in A$ define $L_a: A \to A$ by $L_a(x):= ax.$ Then $L_a$ is linear and, since $||L_a(x)||=||ax|| \le ||a|| \cdot ||x||$, $L_a$ is bounded. Now put
$$|||a|||:= ||L_a||.$$
It is your job to show that $||| \cdot|||$ is a norm with the required properties.