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.


Yes, your argument is correct. Here is another version, using the GNS representation theorem:

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 $b^\ast a^\ast a b\leq \lVert a\rVert^2 b^\ast b$.

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