@SrivatsanNarayanan I had made my gravatar in response to a comment by t.b. If you look at my profile, there is a link called The Mean Square which links to t.b.'s comment
I just spent a bit of time seeing if it had made it to Google images yet.
I think the book is good, but I(we) do not always get the full picture.
So I do not get the intuition clearly at places.
Well, my question would be this: do you know of any books that takes a similar perspective (i.e., inequalities), but tries to explain the intuition behind these things more?
W.r.t. the background, I recognize that we could be lacking, but at least 2 of us are planning (as of now) to take a functional analysis course next semester (I am not even sure if that's relevant).
The motivation to study these inequalities certainly derives from functional analysis, but I doubt that an ordinary introductory course will cover these things.
In any case, I did not post this as a question in the site, because, as you can perhaps already infer, it's a little vague. I cannot put my finger on something and say: this is what I want.
To be honest, this is a bit far away from my comfort zone. If I understand the motivation for these things correctly, you can encode most of the spaces that arise in practice using some weighted sequence spaces (Besov spaces) and interesting operators between them then have certain summability properties to which you can apply these inequalities, then.
[In any case, while trying to explain it to you, I realize I cannot even state my intention clearly. I guess this is a byproduct of trying to skip pre-requisites and jump to the "interesting" stuff. Maybe, I will wrestle with the book a bit more and see.]
I think of all those, Albiac-Kalton would be the one I'd recommend the most. However, it is hard for me to tell if that's really what you're looking for. Probably I'm just the wrong person to ask. What I can do, however: a friend of mine uses these things in his work on numerical analysis. Maybe he knows some nice down to earth motivations that could give you the motivations that you're looking for, I'll ask him about it when I next see him, and get back to you in the next few days.
I think I addressed this tangentially somewhere above. If there's one thing in that book I want to learn, that would be about hypercontractive inequalities.
No problem, you're welcome of course :) From your background and your motivation, you might be better off asking about applications and motivations on cstheory.SE instead of here. Maybe those people could give some pointers that are closer to your interests.
@robjohn: Yes! I had to scramble to salvage the three upvotes I got for solving the original question. Looks like Sivaram has a slicker answer though. (Though his goes into complex analysis territory.)
@anon: I was at the park, perusing MSE for a bit and I saw that question. I thought, "Oh boy, I can answer that one!" but then I saw that you had already answered it. :-(
@JackSchmidt I have a question regarding ideals; it is well known that if I,J are ideals then the set of all products xy where x \in I and y \in J is not necessarily an ideal
My idea is that it suffices to show that somehow adding two elements in the set will not give us one of the form xy
So if I consider the principal ideals (x-1) = I and J = (x), then x^2(x)(x-1) + 1.x.x = x(x^3 - x^2 +1), I think this is not in the set of all pairs xy, x \in I and y \in J
Sorry for the bad notation I should not have used x earlier to say x \in I
@BenjaminLim: if both I and J are principal and in a commutative ring, then IJ is the set of products ij: Proof if I=Rx and J=Ry, then two typical elements of IJ are (rx)(sy) and (tx)(uy), adding these together we get rxsy+txuy = rsxy+tuxy= ((rs+tu)x)(1y).
@BenjaminLim: however if both I and J are not principal, or if the ring is not commutative, then this breaks and IJ is just sums of various ij.
@BenjaminLim: my grammar is poor (night time here): if I is principal, but J is not, then IJ is still the set of ij, as long as the ring is commutative. same proof works basically.
@Amit: Somewhat luckily for you, the name has no duplicated letters. On the other hand, I'm not terribly fond of manipulating six-variable algebraic expressions by hand myself.
bleeh, I just wasted half an hour writing up an answer for this keyhole contour question, but apparently OP was satisfied with a very complicated solution...
@mix: apart from crypto, what do you think is number theory really good for, practically speaking? (Of course, answer only if you need the practical justification.)
Honestly, I don't think much about the practical implications of what I do. I comfort myself by doing other things on the side that I find more relevant to the world.
@robjohn B is a basis for the set X, It is defined that T is a topology generated by B: A subset U of X is in T if for each x in U there is basis element C such that x is in C and C is a subset of U. How to prove that T is a topology on X?
1.for each x in X there is a basis element which contains x 2.if x is in intersection of two basis elements B_1, B_2 then there is basis element B_3 containing x which belongs to B such that B_3 is a subset of intersection of B_1, B_2
I'm not sure if he (the OP) actually understands what he wants to do. That's why I was hemming and hawing a bit when I answered his question on the SVD...
@J.M. Perhaps someone will actually write a good answer to the question, making the comment bite the dust. It will be like Clarke's First Law coming back to bite it's own tail.
"Since the real numbers are a field, they are trivially a Dedekind domain, aren't they?" <-- this sounds incorrect to me but I don't have the algebraicNT to say why.
@SrivatsanNarayanan there is a joke 'don't ask Russian how is he doing if you don't want to listen how crappy was his breakfast' fair enough I would say
@anon: I think this depends on the author. The usual condition is vacuous for fields, because every non-zero proper ideal... However, I've seen authors explicitly exclude fields.
Well, I don't know, but it seems to me to say that you have unique prime factorization because there are no primes to talk about is a bit of a stretch, isn't it?
I agree with Sri. Additionally, I think an auxillary concern with the comment is that the idealic structure of a ring is not the same as the original ring itself, so s/he's not really addressing real numbers directly but rather an algebraic generalization of sorts...
@JM to be honest, some alcoholic in Russia drink glass-cleaning liquid because it's the cheapest alcohol. some of them go blind or dies because there can be metil in this liquid mixed with etil
@JackSchmidt I realised the fact that I had been trying this out with principal ideals, so could not see why IJ would not be an ideal... The basic idea in my head (though this may seem wild) is like when you have the tensor product of two vector spaces V,W it is not necessarily true that every element in here is of the form v \tensor w where v \in V and w \in W.
@BenjaminLim That is a very good idea indeed. If V has dimension 1, then every tensor is "simple", but if V has dimension 2, then there are v⊗w + v⊗w guys. Same thing happens with polynomials: in k[x,y] and (x,y)*(x,y) you get xx+yy. Replace x by v, y by w, and times by otimes, and it is nearly exactly the same.
@BenjaminLim In fact, the tensors form a polynomial ring, one variable per basis element. In the regular tensor algebra the variables don't commute (non-commutative polynomial rings). In the "symmetric algebra", you set it up so they do commute and you get a regular old polynomial ring.
