Well, thanks for that. I was hoping for more meaningful examples. I don't work much (really, at all) with Hardy spaces, but apparently ##H^p## for ##p\in(0,1)## is also a quasi-norm space. Probably some Triebel-Lizorkin and Besov spaces are quasi-norm spaces for appropriate parameters.
Just to be clear, a quasi-norm is like a norm but instead of genuine subadditivity we have
##||x+y||\leq C(||x||+||y||)## where ##C\geq1## is some fixed constant.
To be honest, other than trivial examples the only one that comes to mind is ##L^p## for ##p\in(0,1)##. A quick google search...
Right, it's already been said, but in short, heuristically speaking a vector space is a set equipped with an underlying field and two operations, while a vector field is a vector-valued function.
Thanks, got it!
I actually did something a bit more abstract. Not much though. It goes like this:
Claim: Let ##(X,\tau)## be a locally bounded topological vector space.
If ##U\subseteq X## is such that there exists ##0\neq x\in U## satisfying
for all ##n\in\mathbb{N}## there exists...
I see how taking $(X,\tau)$ to be be Hausdorff can result in some silly things, but I'm still not seeing any contradiction.
In this general setting we have 3 things to work with: our set $U$ being topologically bounded, $+$ being continuous and $\cdot$ being continuous. What would stop there...
Pretty much what the title says.
Suppose we have a topological vector space $(X,\tau)$ and $U\subseteq X$ is topologically bounded. Is it possible for there to be some $x\in X$ such that $cx\in U$ for arbitrarily large $c$? I'm thinking of a real vector space here.
If we try to prove...
Long story short, I've been looking for a program that can produce graphs like mathematica and manipulate 3D objects like google sketchup. Also, I'm going to work these pictures into LaTeX. I'm not sure if a program like this exists (unless maybe mathematica itself can do it).
I have basic...
If the field "defined" on the line actually satisfies the requirements to be a field, yes. Say, for example, if you're in C and choose your line to be R, then you're good. But if you choose the imaginary axis to be your line and do not change the definition of multiplication then you do not...