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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.

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You might want to express your question better, in particular the introduction. You mention Banach spaces but then ask a question about Banach algebras? – Martin Argerami Sep 17 '10 at 16:29
Editted,thank you for your comment – Jiang Sep 17 '10 at 16:48
Your question might have a much better answer if you restrict to C* algebras, where the norm structure is much more rigid (i.e. there is exactly one norm making the star-algebra into a C* algebra). For a general Banach algebra, perhaps a better question would be "is there an equivalent norm with the following extra properties...?". – Mark Sep 17 '10 at 18:36
Mark, in commutative C*-algebras the question is very easily answered... and I'd be amazed if the original question had a positive answer for general noncommutative algebras – Yemon Choi Sep 17 '10 at 19:29
"General C$^*$-algebra'' is probably too ambitious. But in noncommutative von Neumann algebras (finite or infinite dimensional), it is always possible to find two equivalent projections $p,q$ with orthogonal ranges. Then a partial isometry $v$ with initial projection $p$ and final projection $q$ will satisfy $\|v\|=1$, $\|v^2\|=0$. So the infimum of the norms of the squares of the elements of the unit ball is always zero. – Martin Argerami Sep 17 '10 at 20:23
up vote 2 down vote accepted

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.

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Thank you for your answer ,I modified the statement to add an infinite-dimensional assumption. – Jiang Sep 17 '10 at 16:58
Martin, why can't you just tensor your example with $\ell^\infty$? – Yemon Choi Sep 17 '10 at 18:28
I don't exactly see how to use a tensor product, but I think that a direct sum would work. That is, `$\ell^1(\mathbb{Z}_p)\oplus\ell^\infty(\mathbb{N})$, with $\|a\oplus b\|=\max\{\|a\|_1,\|b\|_\infty\}$ is infinite dimensional, and still has the same property. If you agree this is ok, then maybe I should edit my answer and include this. – Martin Argerami Sep 17 '10 at 19:07
I just meant: "take an infinite direct sum of copies of your finite-dimensional algebra". So if $p=2$ then your original example is 2-dimensional, and we now just take an $\ell^\infty$-direct sum of copies of this 2-dimensional example, with the norm being the sup of the norms on each 2-dimensional block. – Yemon Choi Sep 17 '10 at 19:27
Yeah, that works very nicely. So we have a commutative Banach algebra $A$ where there exists $a\in A$ with $\|a\|=1$ and $\|a^2\|<1$; and this algebra also has the property that $\|a\|=1$ implies $\|a^2\|\geq1/2$. I guess that by choosing weighted $\ell^1$ norms we could also get the lower bound as close to 1 as desired. – Martin Argerami Sep 17 '10 at 19:43

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.

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I fixed up the LaTeX a tiny bit. Click the link after the word "edited" above to see what I did. – Harald Hanche-Olsen Sep 18 '10 at 15:00

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