# $b^* a^* ab \leq \Vert a\Vert^2 b^* b$ in a $C^*$-algebra.

Let $$A$$ be a $$C^*$$-algebra and $$a,b \in A$$. In a proof I'm reading the following is claimed: $$b^* a^* ab \leq \Vert a\Vert^2 b^* b$$. I want to understand this:

Here is my reasoning: we view $$A \subseteq \tilde{A}$$ with $$\tilde{A}$$ the unitisation of $$A$$. Then we know that $$a^* a \leq \Vert a^* a \Vert 1$$ since this holds in every unital $$C^*$$-algebra (by a Gelfand-representation argument). Then $$b^* a^*a b \leq b^* \Vert a^* a \Vert 1 b = \Vert a \Vert ^2 b^* b$$

Is the above correct? I find arguments with unitisations always a bit tricky.

If $$a,b$$ are bounded linear operators on the Hilbert space $$H$$ and $$\xi\in H$$, then $$\langle b^\ast a^\ast a b \xi,\xi\rangle=\lVert ab \xi\rVert^2\leq \lVert a\rVert^2\lVert b\xi\rVert^2=\lVert a\rVert^2\langle b^\ast b\xi,\xi\rangle.$$ Thus $$b^\ast a^\ast a b\leq \lVert a\rVert^2 b^\ast b$$.