Hopefully I’ll have time to look at this later. Besides being counterinuitive, it seems potentially useful: consider a vector of composite length , you could partition this vector into blocks of length then apply this inequality to get a family of bounds on the euclidean length of the vector. Exploit the fact that and you could derive some very weird bounds indeed.

Speaking of my research, I fell in love with the subject all over again. Remember my problem is to find bounds on the error in approximating a given matrix with a sparse matrix ; ideally, these bounds should be a function of only: the whole point of sparsification is to avoid making convoluted computations involving the original matrix ; if we had to make such computations to bound the error, we might as well just calculate it exactly. Anyhow, I saw anew how awesome it is that you *can* get useful bounds that are just a function of . I think it helped a lot that I got nice results

I thought about this a little while last night: it’s equivalent to asking how many hollow (the diagonal is zero) symmetric binary matrices satisfy for all . I haven’t seen a way to exploit this equivalence, so maybe there is another approach that would be more fruitful.

BUT, it makes for a nice problem in the one-dimensional case. Show that if (i.e. its second derivative exists and both it and its first and second derivatives are square integrable), then , and find .

I haven’t a clue how to proceed, but it looks like mighty fun.

Update: the proof is, in retrospect (always in retrospect, damn it), obvious.

Recall that a (complex) Banach algebra is a (complex) Banach space with a multiplication operation which is continuous w.r.t. the norm. The Gelfand representation theorem says that we can represent an abelian Banach algebra as the algebra of continuous functions on a locally compact Hausdorff space . The important case to consider is that of a unital Banach algebra, one with an identity element; in this case, we can take the space to be compact. For the remainder, will be a unital abelian complex Banach algebra.

The obvious question is, what is this space ? Can it be intrinsically connected to , or is the proof nonconstructive? It turns out that is a very natural space; in order to construct it, we need the concept of the spectrum of an algebra.

Let . We call the set the spectrum of . One can think of the spectrum as a generalization of the eigenvalues of a square matrix or the range of a bounded function. Recall that a (continuous) homomorphism between two algebras is a linear map which preserves the multiplication operation. A character on is a non-zero homomorphism between and . Let denote the set of characters on ; it turns out, magically, that for all .

One can show that is a weak* closed subset of the unit ball of A*, and in fact if we give the relative weak* topology, it is compact. From here it’s easy to guess that , the character space or spectrum of , is the appropriate . Given , define its Gelfand transform by . By the choice of topology on the character space, each such functional is continuous (and in fact, vanishes at infinity).

The Gelfand representation theorem specialized to unital abelian complex Banach algebras then says

The map is a norm-decreasing homomorphism, and for all . Furthermore, .

Anything?

Quick recap: a Banach algebra is a Banach space with a multiplication operation, such that the norm is submultiplicative (). A prototypical (non-abelian) finite dimensional Banach space is that of all matrices with the spectral norm, and a prototypical (abelian) infinite dimensional Banach space is that of all continuous functions on with the sup norm. An ideal is a vector subspace that ‘absorbs’ under multiplication: if is in the space and is in the ideal, both and are in the ideal. A proper ideal is one strictly contained in the ambient space.

This statement caught my eye because it gives a way in which ideals topologically differ from vector subspaces. It is easy to come up with examples of proper subspaces whose closures are the entire space; e.g. take the set of differentiable functions in the above infinite dimensional Banach space: every continuous function on is the uniform limit of differentiable functions, so the closure of this subspace is the space itself. It’s harder for me to come up with ideals nontrivially exemplifying the above statement.

The question is: exactly how does one use the multiplicative closure property of ideals to prove the above?