## real analysis – Application of the mean-value theorem for general intervalls

Is $$Isubseteqmathbb{R}$$ an intervall and $$f: Itomathbb{R}$$ a differentiable function with bounded derivative $$f’: Itomathbb{R}$$, then $$f$$ is Lipschitz-continuous.

This is supposed to be an application of the mean-value theorem.
What gets me is the use of unspecified intervalls. So $$I=(a,b), (a,b), (a,b), (a,b)$$, as the mean-value theorem holds for differentiable functions defined on a compact intervall (a,b).

Every resource I looked it up proofs this result for compact intervalls, but I was unable to give a counterexample for say $$I=(a,b)$$, because of the bounded derivative.

But how does one relax the condition to $$I=(a,b)$$ to apply the mean-value theorem?
I thought that one might can proof that for $$I=(a,b)$$ you are able to continuously extend to $$(a,b)$$.

## 3 manifolds – Parametric general position theorem for foliations

The situation is the following: let $$M$$ be a manifold endowed with a smooth foliation $$mathcal{F}$$ of codimension one (suppose orientable, transversely orientable) and let $$F_t : S rightarrow M$$ be a one parameter family of smooth inmersions. Is it true that there exists an arbitrarily small homotopy from $$F_t$$ to a family of inmersions $$G_t$$ such that $$G_t$$ is in general position with respect to $$mathcal{F}$$? Furthermore can I say something about the number of singularities at each step and how are the transitions? (maybe this is a lot to ask). More concretely I’m thinking in $$M = mathbb{T}^3$$ and the embedded surfaces being $$mathbb{T}^2 times {pt}$$.

Pd: by general position I mean that it is transversal except at finitely many points where the induced foliation on $$S$$ has Morse type singularities.

## multivariable calculus – Corollary of The Immersion Theorem

Here’s what I’m trying to prove;

Let $$D subseteq mathbb{R}^d$$ be a non-empty open set. Let $$d < n$$ and $$k in mathbb{N}_{infty}$$. Let $$phi: D to mathbb{R}^n$$ be a $$C^k$$ embedding. Then, $$phi(D)$$ is a $$C^k$$ submanifold in $$mathbb{R}^n$$ of dimension $$d$$.

The corollary is from Duistermaat and Kolk’s Multidimensional Analysis I so I am using definitions and so on from that book.

Attempted Proof:

Observe that $$phi: D to phi(D)$$ is a homeomorphism and that $$Dphi(y)$$ is injective for every $$y in D$$. Let $$x in phi(D)$$. Then, there exists a unique $$y in D$$ such that $$phi(y) = x$$. By the Immersion Theorem, there exists an open set around $$y$$, call it $$U_y$$, such that $$phi(U_y)$$ is a $$C^k$$ submanifold in $$mathbb{R}^n$$ of dimension $$d$$.

Since $$x in phi(U_y)$$, there exists an open set $$V_x$$ around $$x$$ such that $$phi(U_y) cap V_x$$ is the graph of some $$C^k$$ function $$f: mathbb{R}^d supsetto mathbb{R}^{n-d}$$ defined on an open set. Since $$phi$$ is a homeomorphism, $$phi(U_y)$$ is open and so, $$phi(U_y) cap V_x$$ is open around $$x$$. Then, observe that $$phi(D) cap (phi(U_y) cap V_x) = phi(U_y) cap V_x$$ is equal to the graph of some $$C^k$$ function (the same one as above). This proves that $$phi(D)$$ is a $$C^k$$ submanifold in $$mathbb{R}^n$$ of dimension $$d$$ at $$x$$. Since $$x$$ was arbitrary, we have our result. $$Box$$

Does the argument work? If it doesn’t, then why? How can I fix it?

## complexity theory – Is the multiplicative constant in the Big O notation are ignored because of Linear Speed-Up theorem?

I just want to know if Big O notation was designed to be used as a consequences of the Linear speed up theorem or not. For me I guess the answer is yes.

For example, if we didn’t have a linear speed-up theorem, then does it mean that we would have a different measure of time/space complexity? i.e. multiplicative constants does makes different. For example, $$f(n) = 100 n$$ isn’t the same as $$g(n)=10^{82}n$$. Therefore, in this regard, Big O notation is not useful. So, probably we have another way to measure algorithms.

For downvoters, thank you for reading, just try to put your comments below in order to improve the question or in worst case I will delete the question.

## Proof of Theorem 12.2.7 from Function Theory of One Complex Variable by Greene and Krantz

I am reading the book $$textit{Function Theory of One Complex Variable}$$ by Greene and Krantz. In Chapter 12 Rational Approximation Theory, Theorem 12.2.7 on p. 378 says that if $$K$$ is a compact set such that $$widehat{mathbb{C}}-K$$ has only finitely many connected components, then for function $$f$$ continuous on $$K$$ and analytic in the interior of $$K$$, it can be approximated by rational functions. They haven’t written the proof. Instead, they only said that “only the final argument in the proof of Theorem 12.2.1 (Mergelyan’s Theorem) needs to be modified so that $$textbf{sets E_j are selected in each component of mathbb{C} -K.}$$” I don’t understand why the bold part is true. The construction of $$E_j$$ is given in the following:

I have found many books on complex analysis and no book explains in details why Mergelyan’s Theorem can be extended to this general case. In fact, I am really frustrated with this. Can anyone help me?

## combinatorics – Looking for formula for variation of binomial theorem

Is there variation of the binomial theorem as follows?

$$sum_{i=0}^n {m choose i} a^{n-i} b^i$$

I am trying to find a formula for it that is a function of $$n$$ without summation notation. I think you would call this is a solution in closed form. Any help would be great. Thanks.

## fa.functional analysis – lifting theorem for n case

I am aware of the following statement of the lifting theorem. For i∈{1,2} let Bi be a contraction on a Hilbert space Hi and let Ai, acting on the Hilbert space Ki, be the minimal unitary dilation of Bi. Let Pi be the orthogonal projection of Ki onto Hi. Then an operator X from H1 to H2 satisfies B2X=XB1 if and only if there exist an operator Y from K1 to K2 such that

A2Y=YA1,
∥X∥=∥Y∥,
P2YP1=XP1.
I am looking for a similar theorem but for i∈{1,2,…n}. So now X satisfies n operator equations and I want a lift Y of X which will further satisfy n equations. Any reference /suggestion are most welcome for the above question.

## at.algebraic topology – Reference request for widely used theorem

Thanks for contributing an answer to MathOverflow!

But avoid

• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

## solution verification – All triangles are isosceles Theorem

Let ABC be any triangle. Bisect BC at D, and from D draw DE at right angles to
BC. Bisect the angle BAC.

(1) If the bisector does not meet DE, they are parallel. Therefore the bisector is at
right angles to BC. Therefore AB = AC, i.e., ABC is isosceles.

(2) If the bisector meets DE, let them meet at F. JOin FB, FC, and from F draw
FG, FH, at right angles to AC, AB.

Then the triangles AFC, AFH are equal, because they have the side AF common,
and the angles FAG, AGF equal to the angles FAH, AHF. Therefore AH = AG, and FH
= FG.

Again, the triangles BDF, CDF are equal, because BD = DC, DF is common, and
the angles at D are equal. Therefore FB = FC.

Again, the triangles FHB, FGC are right-angled. Therefore the square on FB = the
squares on FH, HB; and the square on FC = the squares on FG, GC. But FB = FC,
and FH = FG. Therefore the square on HB = the square on GC. Therefore HB = GC.
Also, AH has been proved = to AG. Therefore AB = AC; i.e., ABC is isosceles.
Therefore the triangle ABC is always isosceles.

Questions :

1. Draw pictures for all cases of the proof described above.
2. Explain the flaws in the above reasoning.

## dg.differential geometry – The Swan–Serre theorem as a monoidal equivalence

Let $$X$$ be a compact Hausdorff The well-known Swan–Serre theorem gives an equivalence between the continuous vector bundles over a compact Hausdorff space $$X$$, and finitely-generated projective $$C(X)$$-modules. Both the category of vector bundles and the category of projective modules have evident monoidal structures. With respect to these structures, is the Swan–Serre equivalence a monoidal equivalence? I would guess that this is the case but I cannot find a reference.