C* identity

Remarks

The identity implies $\norm{a^*} = \norm{a}$: from $\norm{a}^2 = \norm{a^* a} \le \norm{a^*} \norm{a}$ one gets $\norm{a} \le \norm{a^*}$, and the same with $a^*$ in place of $a$.

Example: let $X$ be a compact Hausdorff space and $A = C(X)$ with the sup norm and $f^* = \overline{f}$. Then $\norm{\overline{f} f}_\infty = \max_x \abs{f(x)}^2 = \norm{f}_\infty^2$. The algebra of bounded operators on a Hilbert space with the adjoint also satisfies it: $\norm{Tx}^2 = \ip{T^* T x}{x} \le \norm{T^* T} \norm{x}^2$ gives $\norm{T}^2 \le \norm{T^* T} \le \norm{T}^2$.

Non-example: $A = \C^2$ with coordinatewise operations, $(x, y)^* = (\overline{x}, \overline{y})$, and the norm $\norm{(x, y)} = \abs{x} + \abs{y}$. It is a Banach algebra with involution, but for $a = (1, 1)$ one has $\norm{a^* a} = \norm{(1, 1)} = 2$, while $\norm{a}^2 = 4$.