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Let $\mathcal{B}$ be a Banach algebra with involution *.

Is it always true that $\forall A \in \mathcal{B}: \| A \|^2 \geq \| A^* A \| $?

(motivation: I read a proof that bounded linear operators on a Hilbert space form a C*-algebra, but for the C*-property they only proved $\| A \|^2 \leq \| A^* A \|$ (after having established that it is a Banach algebra with involution), so I wondered if the other direction is obvious...)

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With the direction established in the motivation above, the other direction follows.

Using the submultiplicativity of the norm, one gets for $0\neq A\in \mathcal{B}$

$$\vert\vert A \vert\vert^2 \leq \vert\vert A^* A \vert\vert \leq \vert\vert A^* \vert\vert \;\vert\vert A\vert\vert$$

and thus

$$\vert\vert A \vert\vert \leq \vert\vert A^* \vert\vert$$

Exchanging the roles of $A^*$ and $A$ one obtains

$$\vert\vert A \vert\vert \geq \vert\vert A^* \vert\vert$$

Thus, the original inequality becomes

$$\vert\vert A \vert\vert^2 \leq \vert\vert A^* A \vert\vert \leq \vert\vert A\vert\vert^2$$

which establishes the C*-property.

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  • $\begingroup$ Yes, C^* property imply B^* property, so one can weaken definition of involutive Banach alebra $\endgroup$ – Norbert Jun 30 '13 at 18:27
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In the definition (that I know) of involutive Banach algebra it is assumed that $\Vert A^*\Vert=\Vert A\Vert$. Now we use submultiplicaive property of the norm to get $$ \Vert A^* A\Vert\leq\Vert A^*\Vert\Vert A\Vert=\Vert A\Vert\Vert A\Vert=\Vert A\Vert^2 $$

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  • $\begingroup$ Did you maybe mean SUBmultiplicative in the first step, so a $\leq$? I actually don't want to require the B*-property, but I think it follows quickly. See following answer. $\endgroup$ – madison54 Jun 30 '13 at 18:04

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