Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
"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
show 4 more comments

2 Answers 2

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.

share|improve this answer
    
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
1  
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.