Why is the following inequality true?

Question

Consider a $C^*$-algebra $\mathfrak{A}$ and let $A,B \in \mathfrak{A}$. Supposedly, the following inequality is true.

$$ \color{blue}{A^*B^*BA \le {\left\lVert B \right\lVert}^2 A^*A }$$

I would like to know where it follows from.

For those of you who do not recall right away, the norm in this context obeys the following rules:

1. $ {\left\lVert A \right\lVert} \ge 0 $ and $ {\left\lVert A \right\lVert} = 0 \iff A = 0 $
2. $ {\left\lVert \alpha A \right\lVert} = |\alpha| {\left\lVert A \right\lVert}$
3. $ {\left\lVert A+B \right\lVert} \le {\left\lVert A \right\lVert} + {\left\lVert B \right\lVert}$
4. $ {\left\lVert AB \right\lVert} \le {\left\lVert A \right\lVert}{\left\lVert B \right\lVert}$
5. $ {\left\lVert A \right\lVert} = {\left\lVert A^* \right\lVert} $
6. $ {\left\lVert A^*A \right\lVert} = {\left\lVert A \right\lVert}^2 $

Playing around with these rules gets me nowhere. Maybe I need some additional information to prove the claim. I cannot seem to figure it out.

Kindly advise.

Answer 1

I know two ways:

1. If you already assume that $\mathfrak A\subset B(H)$, then the inequality is trivial: $$ \langle A^*B^*BAx,x\rangle=\langle B^*BAx,Ax\rangle\leq \|B^*B\|\,\langle Ax,Ax\rangle = \langle \|B\|^2\,A^*Ax,x\rangle, \ \ \ x\in H. $$

The problem with the above is that, to show that a C$^*$-algebra can be represented on a Hilbert space, you might need said inequality.

2. The second way: show that $$\|X\|\leq 1\iff X^*X\leq I.$$ Deduce that $$\tag{1}X^*X\leq\|X\|^2\,I.$$This is not entirely trivial, but it is basic C$^*$-algebra stuff. Also show that $$\tag{2} X\geq 0\implies A^*XA\geq0.$$ Combining $(1)$ and $(2)$, $$ A^*(\|B\|^2\,I-B^*B)A\geq0. $$