Does *-operator be automatically continuous In the C*-algebras, does the * -operator be automatically continuous? I think it is yes, because C*-algebras are semisimple, from the Johnson Theorem, it must be automatically continuous. Am I right?
But for the common Banach*-algebras, I think the *-operator maybe not continuous. Does any counterexample to show this?
 A: You are correct; any involution on a semisimple Banach algebra must be continuous for the reason you give.
The question then remains: Does there exists a Banach algebra with a discontinuous involution? The answer can be found in the following article: 
Does there Exist More than One Banach *-Algebra with Discontinuous Involution?
R. S. Doran, 
The American Mathematical Monthly , Vol. 79, No. 7 (Aug. - Sep., 1972), pp. 762-764. 
Published by: Mathematical Association of America
And, if your institution has access to JSTOR, this is available at:
Article Stable URL: http://www.jstor.org/stable/2316269
Here's the example from that paper:
Let $A$ be an infinite-dimensional Banach space over the complex numbers and make $A$ into a Banach algebra by giving it the trivial multiplication, i.e., 
$$ab = 0$$ 
for $a, b\in A$. 
Let $E$ be a Hamel basis for $A$, chosen so that $\Vert x \Vert = 1$ for each $x\in E$. Let $\{x_n\}$ be a sequence of distinct elements of $E$ and define $x^*$ by
$$
x^*_{2n-1}= nx_{2n}, \quad x^*_{2n} = \frac{1}{n} x_{2n-1} \quad (n =1,2,...).
$$
For all other elements of $E$, let $x^* = x$, and then extend the mapping $x \mapsto x^*$ to all of $A$ by conjugate linearity.
Then $x \mapsto x^*$ is an involution on $A$ which is not continuous since $\Vert x^*_{2n-1} \Vert = n$.
By adjoining an identity, you can obtain an example where the multiplication is not trivial.
In the same paper, he states that it is an open problem whether there exists a more interesting example than the above. (At least it was open in 1972 - perhaps somebody else will know of more recent developments.)
