Are there good inequalities on the norm?  It's well known that in a Hilbert space, good inequalities exist concerning the norm due to the existence of inner product.Now let X be a general Banach algebra, are there good inequalities concerning the norm? To be precise, let's consider an example, let X be a commutative Banach algebra with identity I,is the following claim ture or not(especially when X is infinite dimension)?
   Either for every element b in X with norm 1, we have the norm of b^2 is also 1, or inf ||b^2||=0, with b running over all elements in X with norm 1.
P.S.This problem is derived from a question concerning the existence of a nilpotent element in X, in other words, the linear span of all the multiplicative linear functionals may not equal to the dual space of X.
 A: The answer is no. Take the space $B$ of $2\times2$ matrices of the form
$$\begin{matrix} a & b \\ 0 & a+ b \end{matrix}$$
This is an algebra, in which $A^2=0$ implies $A=0$ (because they are diagonalizable). 
Now take a norm over ${\mathbb R}$, and endow $B$ with the induced norm. There are so many of them that you will find that in general $\|M^2\|$ is not identically equal to $\|M\|^2$. Thus there exist matrices of norm one, whose square is not of norm one. But because $B$ is finite dimensional, the ratio $\|M\|^2/\|M^2\|$  remains bounded, which is the same as saying that the infimum of $\|b^2\|$ over the unit sphere is strictly positive.
A: The way it's formulated, the claim can fail in the finite-dimensional case. For example, consider $\ell^1(\mathbb{Z}_p)$. Then if we take an element $a$ of norm 1, $\sum_{k=1}^p|a_k|=1$. This implies that there is $k$ with $|a_k|\geq1/p$. Then $\|a^2\|\geq1/p^2$ (it's likely that a sharper inequality can be found, but that's not necessary to answer your question). 
Edit: on the suggestion of Yemon, we now know how to provide an infinite dimensional counterexample. So let $A_0$ be the algebra $\mathbb{C}^2$ with the norm $\|(\lambda,\mu)\|_1=|\lambda|+|\mu|$. As mentioned in the first paragraph, this algebra has the property that if $\|a\|=1$, then $\|a^2\|\geq1/2$, and this bound is achieved. And now construct $A=\ell^\infty(\mathbb{N},A_0)$ with the supremum norm. This one is infinite-dimensional, and it still has the same lower-bound-for-the-square property.
