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.



Are there good inequalities on the norm? 
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.


