A question about $ C‎^{*}$ -‎ algebra‎s

Question

Let ‎‎$‎A‎$ ‎be a‎ ‎Banach ‎algebra ‎with ‎involution‎ ‎$ * : A ‎\to A$, $x ‎‎\mapsto x‎^{*}‎$ ‎‎and ‎for ‎all‎ ‎$ x \in A $ ,‎ ‎‎$‎\|x\|^2 ‎\leq‎ ‎\|x^{*}x\|$‎.‎

How ‎can ‎we ‎show ‎‎that‎ ‎$A$ ‎is‎ a ‎$ C‎^{*}$-‎ ‎algebra‎ ?‎

Answer 1

From your equation it follows $\|x\|^2≤\|xx^*\|≤\|x\|\cdot\|x^*\|$ and $\|x\|≤\|x^*\|$. Doing it again with $\|x^*\|$ gives $\|x^*\|≤\|x^{**}\|=\|x\|$ and $\|x\|=\|x^*\|$.

Now check out: $$\|x\|^2≤\|xx^*\|≤\|x\|\cdot \|x^*\|=\|x\|^2.$$

Answer 2

The answer given by s.harp is already complete. You cannot expect to deduce anti-linearity and the product rule from just your inequality.

For instance consider $A=\mathbb C^2$, with pointwise addition and multiplication, and with the infinity norm $\|(a,b)\|=\max\{|a|,|b|\}$. This is a Banach algebra with $$\|x\|^2=\|(a,b)\|^2=\max\{|a|,|b|\}^2 =\max\{|a^2|,|b^2|\}=\|(a,b)^2\|=\|x^2\|.$$ Consider the "involution", $x^*=x$. Then $\|x\|^2=\|x^2\|$. But $A$ is not a C$^*$-algebra, as can be seen for instance from $$(-1,-1)=(i,i)^*(i,i)\geq0.$$

For an involution to be a "C$^*$-algebra" involution, it has to satisfy $(x+\lambda y)^*=x^*+\overline \lambda\,y^*$, and $(xy)^*=y^*x^*$. As seen above, those do not come from the C$^*$-equality.