C-star unitization by left multiplication ($\Vert a+\lambda 1\Vert =\Vert L_a+\lambda I\Vert$)

The left multiplication representation of a C-star algebra is isometric: test $L_a$ on $a^{\ast }/\Vert a\Vert$ to get $\Vert L_a\Vert =\Vert a\Vert$. For a nonunital algebra, its closed image plus the identity operator is the algebraic unitization. For a formal $z=a+\lambda 1$ and $b\in A$, $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Thus $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$; applying the same estimate to $z^{\ast }$ proves the algebra involution is isometric and the C-star identity holds. This gives a unitization without first assuming an operator representation on a Hilbert space.