In mathematics, specifically real analysis and functional analysis, the Kirszbraun theorem states that if U is a subset of some Hilbert space H1, and H2 is another Hilbert space, and
f : U → H2is a Lipschitz-continuous map, then there is a Lipschitz-continuous map
F: H1 → H2that extends f and has the same Lipschitz constant as f.
Note that this result in particular applies to Euclidean spaces En and Em, and it was in this form that Kirszbraun originally formulated and proved the theorem. The version for Hilbert spaces can for example be found in (Schwartz 1969, p. 21). If H1 is a separable...
Okay, so the backward implication is true for infinite dimensional vector spaces which aren't complete with respect to the norm induced by the inner-product?
I have a similar question as here.
Lesbesgue-Stieltjes measure $\mu_F$ with respect to the function $F=\lvert x\rvert\lfloor x\rfloor$. i.e. $\mu_F([a,b))=F(b)-F(a)$. I'm trying to find the Lebesgue-Radon-Nikodym decomposition of $\mu_F$ with respect to the Lebesgue measure $m$.
Now, I know thi...
I'm going through Folland's Real Analysis, and has come to an impasse trying to come up with a couple examples in the topic of differentiability.
On $(\mathbb R, \mathcal B_{\mathbb R}, m)$, for locally integrable $f$, define $F(x)=\int_{0}^x f(t)dt$, $A_rf(x)=\frac{1}{2r}\int_{x-r}^{x+r}f(t)...
Motivation
I will briefly elaborate here my motivations for asking the question. If you are not interested in it then please go to the questions.
Recently during trying to understand and prove the Second Hardy-Littlewood Conjecture, I came to notice that the following slightly weak version of ...
@AlessandroCodenotti i showed that the map $t\to (t^2,t^3)$ is homeomorphism. i am asked to explain why it is not isomorphism $\Bbb A ^1 \to Y$ where $Y = Z(y^2-x^3)$
@Astyx yah but we're given that if ab=ca then b=c. so doesn't it hold for any a,b,c? sorry but i do not understand where in my proof i have used specific a,b,c
I'm going through Folland's Real Analysis, and has come to an impasse trying to come up with a couple examples in the topic of differentiability.
On $(\mathbb R, \mathcal B_{\mathbb R}, m)$, for locally integrable $f$, define $F(x)=\int_{0}^x f(t)dt$, $A_rf(x)=\frac{1}{2r}\int_{x-r}^{x+r}f(t)...
if phi is homomorphism from ring R to R' then phi(1) = 1' if R' is an integral domain .How to prove this ? for proving this we probably need to prove that phi is onto
Hi! If the formula is not true, then there's a seq $s$ in $\Sigma$ st $s$ satisfies $(\forall x_i)\mathscr{B}$ and $s$ doesn't satisfy $(\exists x_i)\mathscr{B}$ Here I want to ask if $(\exists x_i)\mathscr{B}$ and $\neg(\forall x_i)\neg\mathscr{B}$ are equivalent, since the lack of parantheses makes it hard to see this for me.
How to prove that $Ф(1) = 1'$ if $R'$ is an integral domain? Since $R'$ is an integral domain , $Ф(b) = Ф(b.1) = Ф(b).Ф(1) = Ф(1).Ф(b)$ .But i can prove this only for those $r'∈R'$ for which there $∃r∈R$ such that $Ф(r)=r'$.
if you assume that $\phi$ does not map all $r$ to $0$, then you can prove it. In that case, you get $\phi(1)=\phi(1 \cdot 1)= \phi(1)^2$, so $\phi(1)$ is idempotent
the only idempotents in an integral domain are $1$ and $0$
It's nice that you can algebraically kind of completely untangle the varying degrees of content in the statement "picking an element fiberwise in a projection map": If $\mathcal{F}$ is a presheaf on $X$, $\widehat{\mathcal{F}}$ be the sheafification and $\prod \mathcal{F}$ be the sheaf which takes value in product of the stalks above a given open set.
Then a section of $\mathcal{F}$ is kind of the most rigid thing, a section of $\widehat{\mathcal{F}}$ is inverse of the local homeomorphism that's the projection from the leafspace of $\mathcal{F}$, a section of $\prod \mathcal{F}$ is choosing a point from each stalk possibly "discontinuously"
There is a sequence of inclusions $\mathcal{F} \to \widehat{\mathcal{F}} \to \prod \mathcal{F}$, of course.
@Mathein: According to my definition, the zero map is not a ring homomorphism. Following Artin, I say it must map identity to identity. I don't remember what Herstein does.
Right. It seems to me that problem is right if Herstein rules out the 0 map. Go back to the definition. Neraj, what's the definition of a ring homomorphism?
@MatheinBoulomenos I'm confused. On the Spec level there's only one map $f:\operatorname{Spec}(X)\to\operatorname{Spec}K$ so we only need a morphism of sheaves $\mathcal O_X\to f_\ast\mathcal O_K$, right?
A mapping¢ from the ring R into the ring R' is said to be a homomorphism if I. ¢(a + b) = ¢(a) + ¢(b), 2. ¢(ab) = ¢(a)¢(b), for all a, b belonging to R. @TedShifrin which point contradicts zero mapping ?
Nope. If that's the definition, then the zero map is a homomorphism and the exercise is in fact wrong. This is a different Herstein book than I used as a college student, but I don't remember anything wrong back then.
@AlessandroCodenotti I think a scheme in general is so general that without more adjectives it's natural for them to be not too intuitive - after all even if you only cared about affine schemes - that's already the opposite category of commutative rings - and you can imagine one can cook up some pretty crazy examples of rings
@Alessandro: It always gets confusing with permutation cycle notation, too.
There are times I use the right action and row vectors. Certain things in differential geometry. Chern did it, and I blindly followed him instead of doing what other people do.
@Pig yeah what I was thinking of is similar to the $L^1(G)$ example. If $G$ is a locally profinite group, then the Hecke algebra $\mathcal{H}(G)$ are locally constant compactly supported functions $G \to \Bbb{C}$ (also denoted $C^{\infty}_c(G)$), then this is a algebra without identity under convolution. It turns out that smooth representations of $G$ are equivalent to certain modules over $\mathcal{H}(G)$ (the condition is that $\mathcal{H}(G)M=M$ (which isn't automatic due to the lacking identity)
@AlessandroCodenotti probably it's easier to understand locally ringed spaces first, after all morphisms of schemes = morphisms of locally ringed spaces
@MatheinBoulomenos I feel like this isn't a "true" example of lacking identity, in that we can probably enlarge $H(G)$ (p-adic or archimedean) to include delta functions
so it isn't "real" per se, but i can imagine in other cases the corresponding group algebra wouldn't have identity
My naive attempt would have been to try to define a morphism of schemes as a morphism of ringed spaces, but we saw in class that this cannot work, basically because the induced map on stalks can be a mess, right?
Oh, yeah, the issue is that not every morphism $\operatorname{Spec}(R)\to\operatorname{Spec}(S)$ as ringed spaces is induced by a morphism of rings $S\to R$
While it holds for morphisms as locally ringed spaces
So anyway I guess the counterintuive part for me is how pullbacks are additional data one needs to specify in a morphism of schemes, while a map of the underlying spaces isn't enough
My understanding is that schemes have smoothness inbuilt in them as the stalk, which is a local ring, and the tangent space is m/m^2 where m is the maximal ideal of the stalk. You want the map to preserve the maximal ideals stalk-wise
morally your stalk at p remembers the functions defined around p, and your maximal ideal consists of functions vanishing at p
and so asking for the induced map on stalks means that "you had a function on Y vanishing at f(p) and you want the function obtained via your map of sheaves O_Y -> f_*O_X to also vanish at p
I guess the point is that morphism of ringed spaces is too general to force the above (very natural!) condition
Do diffeomorphisms allow self-intersections? For example, is a line segment in $\mathbb{R}^2$ diffeomorphic to some smooth curve with a self-intersection?
Actually if you take the bigger picture, you find that the middle illustrations are diffeomorphic to the left one (ie circle and sphere) and the rightmost are examples that are not diffeomorphic to the rightmost one
@Ted I believe has a list of errata for G-P somewhere. (This detail isn't in it IIRC, but there are a lot of other things that would otherwise make problems difficult and so on.)
They give an example of non-diffeomorphic things with a V and a loopy curve. I don't like that too much, my diffeomorphisms always have domain/range manifolds.
I don't know about "clearly". Notice that they define diffeomorphisms for subsets of R^n, as restrictions of diffeomorphisms from open subsets containing them.
Oh by the way @Balarka a question you might find interesting: I was trying to construct a connected topological space $X$ with at least two points such that there exist an $x\in X$ such that $x$ has a nbhd basis $(B_i)_{i\in I}$ and $B_i$ is homeomorphic to $X$ for every $i$
Kinda like an hawaiian earring but a little worse
But then I realized $\Bbb R$ works so I'm not sure what's the right question to ask to get interesting spaces :/