almost unital Banach algebra's Let $A$ be an "almost unital Banach algebra", in the sense that it satisfies all the usual axioms but not necessarily that $\|1\|=1$. From the product inequality $\forall x,y \in A$
\begin{equation}
\|xy\| \leq \|x\|\|y\|
\end{equation}
we can deduce, by applying this inequality to $x=y=1\neq0$, that necessarily
$\|1\|\geq 1$. But there's nothing more which can be said in general (that is, for all $\lambda>1$ there exists actually such an "almost unital Banach algebra" with $\|1\|=\lambda$. For example consider the set of continuous functions $f:\mathbb{C} \rightarrow \mathbb{C}$ which converge at infinity. Equip this set with the pointwise addition and multiplication operations and scalar multiplication. And take as a norm
\begin{equation}
\|f\| = \|f-1\lim_{x \rightarrow \infty } f\|_{\infty} + \lambda |\lim_{x \rightarrow \infty } f|
\end{equation}
where $1:\mathbb{C} \rightarrow \mathbb{C}:x\mapsto1$
(More examples can be constructed via the standard "adding a unit"-procedure for non-unital algebra's, where you can freely choose the norm of the added unit to be some number $\geq 1$)
)
In general, can we equip $A$ with a modified norm $\|\|'$ which turns it into a unital Banach algebra?
 A: The answer is yes, and $\|\|'$ is defined as follows:
\begin{equation}
\|x\|'=\sup_{y \in A_0} \frac{\|xy\|}{\|y\|}.
\end{equation}
This is well defined because of the original product inequality $\|xy\| \leq \|x\|\|y\|$ from which we can immediately establish that $\|\|' \leq \|\|$. This inequality also transfers the completeness of $A$ for the old norm to completeness for the new norm: to see this, suppose $(x_n)_n$ is a sequence in $A$ which is Cauchy for the $\|\|'$-norm. Then we can estimate
\begin{equation}
\|x_n-x_m\|=\|(x_n-x_m)1\|\leq\|(x_n-x_m)\|'\|1\|,
\end{equation}
implying that the sequence is also Cauchy for the $\|\|$-norm, hence convergent to a certain $x \in A$ for this norm. The inequality $\|\|'\leq\|\|$ implies then also convergence to $x$ for the $\|\|'$-norm.
The norm properties of $\|\|'$ are easily verified.
The multiplication inequality is also satisfied: suppose for this purpose that for $y$ there exists a $z\in A$ such that $yz\neq0$. (Things are trivial if this is not the case)
\begin{eqnarray}
\|xy\|'&=& \sup_{z \in A_0, yz \neq 0} \frac{\|xyz\|}{\|z\|} \\
&=& \sup_{z \in A_0, yz \neq 0} \frac{\|xyz\|}{\|yz\|}\frac{\|yz\|}{\|z\|}\\
&\leq& \sup_{\alpha \in A_0} \frac{\|x\alpha\|}{\|\alpha\|}\sup_{z \in A_0}\frac{\|yz\|}{\|z\|}\\
&=&\|x\|'\|y\|'
\end{eqnarray}
It is evident that for the (algebraic) unit, we have $\|1\|'=1$.
