TITLE: How to count the numbers $n$ which $n\le p^e$ and $n$ is not coprime to$p^e$
QUESTION [0 upvotes]: To a prime $p$ and a natural number $n$, I want to find $\phi(n)$.
To count how many numbers are there which is less or equal to  $p^e$ and coprime to $p$, I am trying to count all the numbers which are not comprime to $p$.
I divide the cases into consider the numbers $n$ whose $gcd(n,p^e)=p,p^2,...,p^e$.
But then things become messy. Could someone please, if possible, continue on this track to count how many numbers are there are less or equal then $p^e$, but not coprime to it?
Thanks in advance!

REPLY [2 votes]: A natural number $n$ fails to be prime to $p^e$ precisely when it is divisible by $p$.
And there are exactly $p^{e-1}$ integers $1\leq n\leq p^e$ which are divisible by $p$, namely the integers of the form $kp$ for $k=1,2,\dots,p^{e-1}$.


TITLE: The expected edge length of a cube is equal to $e_{D}(f) = f^{1/D}$, what is the definition of `the expected edge length`?
QUESTION [0 upvotes]: section 1.4.3 of the book "Machine Learning - A Probabilistic Perspective" gives an example about KNN:
the input is two dimensional, we have three classes, and K = 10

here is Figure 1.15

here is Figure 1.16

the expected edge length of this cube

$e_{D}(f) = f^{1/D}$
what is the definition of the expected edge length?

REPLY [0 votes]: There the fraction $f$ is ment to give the $D$-dimensional volume of some "expected hypercube", which is designed such that it contains this point subset. If $s$ is the side length of the hypercube, then its volume is given by $s^D$. When equating those 2 values and solving for $s$ you'd just get $s=f^{1/D}$.
--- rk


TITLE: Why does a voltage appears between $A$ and $C$ when this coil is rotated?
QUESTION [0 upvotes]: Find the voltage between the points $A$ and $C$ when a wire is conecting $A$ and $D$.

We know that there is a magnetic field $B$ perpendicular and entering in the papers plane, the semicircular metal has an electrical resistence $\lambda$ by unit of lenght and it is spinning around $C$ with constang angular speed $\omega$. The wire connecting $A$ and $D$ has electrical resistence zero.
What I can't understand in this situation is that, when connecting $A$ and $D$, we get a coil in a closed shape. In order to have some voltage we need some magnetic flux variation, but the coil is only spinning in a plane perpendicular to $B$. So, for me, there is no flux variation and the voltage between $A$ and $C$ is zero. Thinking about this, I also wanted to know if there would be voltage if instead of spinning, the metal were only moving with constant or variable speed in the same plane.

REPLY [1 votes]: $\def\rA{{\rm A}} \def\rC{{\rm C}} \def\rD{{\rm D}} \def\rP{{\rm P}}$
I find it difficult to write an answer without knowing the level I am
supposed to hold. Furthermore, the phrasing of the problem looks
rather strange to me. I don't understand why resistance is given and
which purpose the electrical connection between A and D should serve.
Also the use of "voltage" is liable to misunderstandings. In physical
parlance the right term is "potential difference" (pd), but referring
to a closed circuit "electromotive force" (emf) should be used.
You are right that in the closed circuit ACD no emf is induced and no
current can flow. (This is why I can't see the reason to give the
wire's resistance.) But this does not mean that no pd exists between
different points, whether the circuit is closed or open. Let's see
why.
Consider for instance arc AC of the wire. It is electrically neutral,
holding positive charges bound in the metal lattice and free
electrons. If the wire stands still all these charges stand still too
(for electrons, in the average). When AC rotates around C, every point
P of the wire acquires a speed, higher for points farther from C
and directed perpendicularly to the radius CP.
Now Lorentz' force comes into play: it is perpendicular both to P's velocity and to the field $\vec B$. For a positive charge its direction is towards C, for a negative one, it's opposite. The force's intensity is $qvB$ ($q$ the charge). As to positive charges, Lorentz' force exists but has no effect since these charges are bound. But electrons are free, and are driven towards the extremity A. They would tend to pile up there, leaving free of electrons the other extreme C. But electrons repel each other, so you can see that an equilibrium is reached, with
abundance of electrons near A, scarcity near C.
A more formal description of the process is the following. When
equilibrium is achieved electrons stand still again, which means that
the net force acting on each single electron is zero. This force is
the resultant of the Lorentz force previously described and that of
the electric field generated by all other electrons accumulated near A
and removed from C. Considering that electrons are negative, the
direction of this field is from C towards A, and stronger near A,
as Lorentz force is.
If in a wire an electric field is present, you can speak as well of a
pd between extremes C and A, the potential being higher in C, lower
in A. It would remain to compute the value of this pd. I give only the
result, as I don't know if you could follow the derivation:
$$V(\rC) - V(\rA) = \frac12\,\omega\,B\,l_1^2$$
where $l_1$ is the length of chord CA.
A final look at the other tracts of the wire, with special regard to
the closed circuit. Exactly the same will happen in tract CD:
$$V(\rC) - V(\rD) = \frac12\,\omega\,B\,l_2^2$$
where $l_2$ is the length of chord AD. The same kind of computations
could be done also for segment AD if the circuit is closed, obtaining
$$V(\rD) - V(\rA) = \frac12\,\omega\,B\,(l_1^2 - l_2^2).$$
If you compute the emf as the sum of all these pd's, taken with the
right signs:
$${\rm emf} = [V(\rA) - V(\rC)] + [V(\rD) - V(\rA)] +
[V(\rC) - V(\rD)] = 0.$$


TITLE: Asymptotic of min(#generators times diameter), for a Cayley graph of Sn
QUESTION [2 upvotes]: This post is a sequel of Diameter of symmetric group.
Let $\Sigma$ a generating subset of $S_n$, $\Gamma(S_n, \Sigma)$ the Cayley graph and $d_{\Sigma}$ the diameter of $\Gamma(S_n, \Sigma)$.
Let $s_n = min_{\Sigma}(\vert \Sigma \vert \times d_{\Sigma})$.
Question: What's the asymptotic of $s_n$ (or the best conjecture for that)?

REPLY [5 votes]: For a generating subset $S\subset S_n$ of size $d$, the $n$-ball has size $\le (2d)^n$, which implies that the diameter $D(S)$ satisfies $D\ge\log_{2d}(n!)\simeq n\log n$.
So $|S|D(S)\ge \frac{d}{\log d}\log(n!)\ge \log(n!)\sim n\log n$, and thus $s_n=\min_S |S|D(S)\succeq n\log n$.
This is attained, according to Theorem 7 in this paper by Kassabov, which implies that there exists $\ell$ and generating subsets $F_n\subset S_n$ with $D(F_n)\preceq n\log n$ and $|F_n|\le \ell$, implying $s_n\preceq n\log n$.


TITLE: Equivalence of $P\rightarrow Q$ and $\lnot P\lor Q$
QUESTION [0 upvotes]: How do we explain the logical equivalence
$$(P\rightarrow Q ) \equiv [(\neg P)\; \vee \; Q]$$
and if possible could you please give an example illustrating this equivalence. Thanks alot !!

REPLY [1 votes]: Instead of using a truth table or a series of logical equivalences, I'd like to try a more intuitive method for you.
Try drawing a Euler diagram of $P\to Q$, as in, a small circle $P$ encompassed by a lager circle $Q$.
Now you can see, no matter where you point on this diagram, it is true you will point at $\lnot P$ or you will point within $Q$, or both.


TITLE: Has the conformally field theoretic metric on a Calabi-Yau variety been proved to exist?
QUESTION [20 upvotes]: Let $(X,g)$ be a compact Kähler manifold. Physics allows us to consider a supersymmetric sigma model with target $(X,g)$, which is a N=2 two-dimensional field theory.
From the two-dimensional point of view, the metric $g$ can be thougt as a coupling constant.
At the classical level, this theory is conformally invariant. But it is well known that in a quantum theory, coupling constant are in general renormalized: there is only a effective coupling constant which depends on an energy scale $\mu$. Here, we have an effective metric $g(\mu)$. The dependence on $\mu$ of $g(\mu)$ is controlled by a differential equation of the form $dg/d(log \mu)= \beta(g)$ where the beta function $\beta$ can be computed order by order in perturbation theory. By definition, the quantum theory will be conformal iff the beta function vanishes. At the first order in perturbation theory, $\beta$ is proportional to the Ricci curvature and so one can solve $\beta=0$ if $X$ is Calabi-Yau ($c_{1}(X)=0$) and there is a unique up to scaling such solution in each Kähler class $[\omega]$ (by Yau's proof of the Calabi conjecture). But in general, there are higher order corrections. One can show that the Kähler class remains unchanged but this is not the case for the metric in this class. If one could solve the equation $\beta = 0$ order by order in perturbation theory and prove that the solutions converge, then this should define in each Kähler class of $X$ a unique metric $g_{CFT}$.
All this is rather well known from the mid 80's. At this time, there was an article of Nemeschansky and Sen, "Conformal invariance of supersymmetric $\sigma$-models on Calabi-Yau manifolds", which, I have the impression, has closed the question among physicists. In this article, it is proved that one can solve $\beta =0$ order by order
but there is no result of convergence (as far as I understand).
So my question is:
Has $g_{CFT}$ been mathematically proved to exist?
As the perturbative expansion is only valid for large Kähler class, it is likely that such a restriction is necessary. Maybe a better question is: is the perturbative expansion
expected to converge for a large Kähler class or is it just some asymptotic expansion
(as it is often the case for such question in perturbative quantum field theory)?
In any case, the question of the existence of $g_{CFT}$ is rather well posed mathematically and I would like to know if it has been studied/solved. Any reference will be appreciated.

REPLY [15 votes]: To the best of my knowledge of the literature on this topic, the answer is: not really (a few exceptions appear below).  Let me first give a rigorous statement of the problem: the physical equation $\beta = 0$ can be expressed as
$\sum_{k=0}^{\infty} \epsilon^k \beta_k = 0$,
where $\epsilon$ is a physical parameter one imagines is "small," and where $\beta_k$
is a symmetric two tensor depending on a Riemannian metric which arises as a certain universal expression in the curvature tensor and its derivatives, and which moreover obeys a natural scaling law and a natural "homogeneity" in terms of the number of derivatives of $g$ which are required to express it.  These terms $\beta_k$ are computable in theory, although my understanding is that this has only been done on a piecemeal basis for the first few terms.  There are many references one can easily find which do this for this and other sigma models.  The first two terms are $\beta_0 = \mbox{Vol}(M) g$, $\beta_1 = \mbox{Rc}$, the term $\beta_2$ is a quadratic expression in the curvature tensor.  Considering the equation up to order $k$ in the $\epsilon$ power expansion is referred to in physics literature as "up to $k$-loops."
So, with this background, there are at least two interesting questions one can ask:
1) Given $N \in \mathbb N \cup \{\infty\}$, can one construct on a given manifold a solution to $\sum_{k=0}^N \epsilon^k \beta_k = 0$?
2) Can one construct on a given manifold a one-parameter family of Riemannian metrics satisfying $\frac{\partial g}{\partial t} + \sum_{k=0}^N \epsilon^k \beta_k = 0$?
The original question was related to 1) above, but 2) is as important in the physics literature (AFAIK).  Of course, the case $N=1$ of question 1) corresponds to solving the usual Riemannian Einstein equation (although one may need to justify ignoring the term $k=0$ in some cases, which physicists have ways of doing).  In the Kähler setting the metric can come from the Calabi-Yau theorem as the poster stated.  For question 2), the case $N=1$ corresponds to the (properly normalized) Ricci flow equation.
Now that I have properly stated the question, let me just say that I focus on studying elliptic/parabolic equations on manifolds, and for a while got interested in exactly this question.  After much literature digging, I found only a few rigorous mathematical results:
http://arxiv.org/abs/0904.1241
http://arxiv.org/abs/1108.0526
http://arxiv.org/abs/1205.6507
The first paper considers the case $N=2$ of question 2).  In this case the operator is still second-order, and so in special settings the equation can be rendered parabolic, and then known techniques can be applied.  This paper is certainly interesting, although the techniques cannot really be extended to $N > 2$.  The last two papers address the homogeneous setting.
Moving beyond $N = 2$ is, IMHO, extremely difficult from a PDE point of view, since one has very little information on the form of the terms $\beta_k$ beyond the general qualitative statements I made above.  For instance, an interesting side question is: do there exist arbitrarily large values of $k$ for which $\beta_k$ corresponds to an elliptic operator of a Riemannian metric?  This is true for $k=1$, but I believe it is very unlikely to be true for higher $k$ since I believe it is "known" in the physics sense that all of the terms arising in $\beta_k$ for $k > 1$ are at least quadratic combinations of curvature and derivatives.  I.e., one cannot expect a term like $\Delta \mbox{Rc}$ to show up, which would be elliptic.
One can imagine considering the case $N > 2$ on homogeneous spaces, where in principle the equations reduce to a system of ODE, but again one does not have much in the way of understanding the form of the general term $\beta_k$.
It is also not inconceivable that if one starts with say a Ricci-flat metric, and is allowed to choose $\epsilon$ very small with respect to this metric, that one can perturb it to a solution to the higher order equations, but again without at least a little more hard data on the form of the $k$-th term this seems tricky.
EDIT: In response to Robert's comments, yes, I meant $\mbox{Vol}(M) g$ (fixed above).  Also, let me give a reference:
http://www.sciencedirect.com/science/article/pii/0550321389904227
This paper computes the terms $\beta_3$ and $\beta_4$ (The formula for $\beta_4$ covers half a page of terms, and each of these is shorthand for a complicated curvature expression already.  Written out fully would take 2 pages at least!).  As far as whether one can expect the $\beta$ functions to be simpler on a Kähler manifold, I can't speak to the physics well enough to really say.  From a mathematical point of view, I don't really see any advantage beyond the fact that of course each term could be expressed in terms of a Kähler potential.
The case of symmetric spaces is probably more tractable.  Looking for solutions within the class $\nabla \mbox{Rm} \equiv 0$ could certainly simplify the terms $\beta_k$ greatly.  My understanding of how these terms are derived is extremely fuzzy, but let me speak anyways: I believe the different summands in $\beta_k$ correspond in some sense to different sized "loop diagrams" (see the above reference for some examples), and moreover the qualitative behavior of such terms (i.e. number of derivatives of $\mbox{Rm}$ which appear is (somehow) related to the topology of the diagram.  One main difficulty in performing these calculations to derive the terms $\beta_k$ is the sheer number of diagrams which must be considered, which grows wildly with $k$.  If, on the other hand, one could a priori rule out a large number of such diagrams (by assuming $\nabla \mbox{Rm} \equiv 0$), perhaps these calculations become more tractable, or at least some stronger qualitative statements could be made.
Lastly, the quantity $\epsilon$ (which, also note is usually $\alpha'$ in physics literature) is, I believe, meant to be small but fixed.  Physically it apparently is meant to represent the "string tension."


TITLE: Solving matrix AX=B matrix for X
QUESTION [1 upvotes]: I'm having the following matrices in the form  A X= B.Where A & B matrix are known values. How can i solve for the matrix X?. Please provide me some formulas or Ideas for solving this.
Note: we need to use the least square method to get the best solution for X $$\left[\begin{array}{ccc}A1&B1&C1\\A2&B2&C2\\...\\A10&B10&C10\end{array}\right]\left[\begin{array}{c}x1&x2&x3\\y1&y2&y3\\z1&z2&z3\end{array}\right]=\left[\begin{array}{c}A11&B11&C11\\A22&B22&C22\\...\\A1010&B1010&C1010\end{array}\right]$$
Can i use this formula for solving this?
$X= A^{-1}B$

REPLY [2 votes]: There are a few options here:

Solve $A\pmatrix{x_1\\y_1\\z_1} = \pmatrix{A_{1,1}\\\vdots\\A_{10,10}}$ et cetera as three LS-problems each with vector unknows
Convert $AX=B$ to a vector unknown system
$$\pmatrix{A&0&0\\0&A&0\\0&0&A} \pmatrix{x_1\\y_1\\z_1\\x_2\\y_2\\z_2\\x_3\\y_3\\z_3} = \pmatrix{A_{1,1}\\\vdots\\A_{10,10}\\B_{1,1}\\\vdots\\B_{10,10}\\C_{1,1}\\\vdots\\C_{10,10}}$$
Use the same formula for the solutions of (1.) to solve the three in parallel
$$X = A^\dagger B = (A^TA)^{-1}A^T B$$


TITLE: In what spaces does the notion of a gradient field make sense?
QUESTION [0 upvotes]: I am doing highschool calculus, and I have been told a gradient field is a field $F(x_1,\cdots,x_n)=f_1(x_1,\cdots,x_n)e_1+\cdots f_n(x_1,\cdots,x_n)e_n$ where we have $$F(x_1,\cdots,x_n)=\nabla q(x_1,\cdots,x_n).$$
(where $\{e_i\}$ are the standard basis vectors for $\Bbb R^n$)
So I guess this definition definitely has $\Bbb R^n$ in mind? Does this notion hold in other spaces? I get that we can define a vectorbundle on really any topological space, and since this notion is apparently equivalent to the notion of a being a conservative field, which seems like it should be very topological, it seems it should (hold in other spaces)?
Related is my previous question: Understanding the conservative property of a vector field

My previous question being related mainly in the sense that it seems that the force field is a section of a bundle, so then we should be able to give a value to each homotopy class of paths, with specific fixed end points, where I feel this is the real definition of a conservative field.

REPLY [0 votes]: Without getting into too much detail on the definitions involved, on a differential manifold the generalization of the gradient is more naturally interpreted as a section of the cotangent bundle, also known as the space of 1-forms.  In this form, the notation would look like:
\begin{equation*}
df = \frac{\partial f}{\partial x_1} d x_1 + \cdots + \frac{\partial f}{\partial x_n} d x_n.
\end{equation*}
In general, though, there is a distinction between the 1-forms which can be expressed in this way, and the generally larger space of 1-forms $\omega$ which satisfy $d \omega = 0$ - I'm not sure which you take as the definition of a "conservative force field".  As an example, the restriction to the unit circle of
\begin{equation*}
\omega := -y\,dx + x\,dy
\end{equation*}
satisfies $d\omega = 0$ but there is no global function $\theta$ on the circle such that $\omega = d \theta$.  (However, there are some functions on subsets of the circle which satisfy this, such as $\theta := \tan^{-1}(y/x)$.)  The study of the difference between the two notions is the start of de Rham cohomology which is closely related to homotopy theory.


TITLE: confidence Interval of multivariate Gaussian Distribution
QUESTION [0 upvotes]: Suppose I have an single variable which is gaussian distributed (with mean value 0 and standard deviation sigma). Then I certainly know that there is 68.2% of the chance that this variable should lie between [-sigma, sigma].
Similarily I have an vector with dimension 1000 and every element of this vector can be seen as independent and gaussian distributed with the same sigma (mean value 0 for all). Then what is the probability that the vector lies within a hyperball with r = sigma?
It would great if somebody can help me with that?
Now make this question more complicated
What if the signal itself is complex valued (the same mean and std for both real and imaginary part. We also assume the real and imaginary part are independent from each other). what is the possibility that the absolute value of the vector |v| lies within an hyperball with radius r?

REPLY [0 votes]: You are not looking for a confidence interval, but for the cumulative probability as a function of the distance from the mean. See this paper for your answer, specifically page 8.


TITLE: What does $R_I$ for a ring $R$ and ideal $I$
QUESTION [1 upvotes]: In this paper, I am confused by the third sentence of the third paragraph, which reads

If $A \subset B$ is a minimal ring extension, it follows from [2 (sic)] that there exists a unique maximal ideal $M$ of $A$ (called the crucial maximal ideal of $A\subset B$) such that the canonical injective ring homomorphism $A_M \to B_M (:= B_{A\setminus M})$ can be viewed as a minimal ring extension while the canonical ring homomorphism $A_P \to B_P$ is an isomorphism for all prime ideals $P$ of $A$ except $M$.

I do not know what the notation $A_M$ denotes. Can someone provide a definition?
For context, two sentences before, they state

Recall (cf. [2]) that a ring extension $A\subset B$ is a minimal ring extension if there does not exist a ring properly contained between $A$ and $B$.

I am assuming that $A\subset B$ is a ring extension if $A$ is isomorphic to a proper subring of $B$.
For completeness, here is their citation for [2]:

D. Ferrand and J.-P. Olivier, “Homomorphismes minimaux d’anneaux,” Journal of Algebra, vol. 16, pp. 461–471, 1970.

REPLY [2 votes]: A subset $S\subseteq R$ (of a commutative unital ring $R$) is called multiplicative if $a,b\in S\implies ab\in S$, $1\in S$ and $0\notin S$. For such a subset we can define the localization (mentioned by Edward Evans in the comments) at $S$:
$$S^{-1}R=\left\{\left.\frac rs\ \right|\ r\in R,s\in S\right\}$$
This construction is related to the field of fractions of an integral domain where we choose $S=R\setminus\{0\}$. The fraction $\frac rs=:(r,s)$ I wrote is formally an equivalence class w.r.t. to the equivalence relation
$$(r_1,s_1)\sim(r_2,s_2)\ \iff \ \exists s\in S:\ s(r_1s_2-r_2s_1)=0\ .$$
The localization naturally admits a ring structure (defined similar to the case of $\mathbb Q$ as field of fractions of $\mathbb Z$) and a universal property. The details may be found elsewhere.
If given a prime ideal $\mathfrak p<R$ we write $S_{\mathfrak p}=R\setminus\mathfrak p$. This subset is multiplicative as $\mathfrak p$ is prime:
$$ a\in S_{\mathfrak p},b\in S_{\mathfrak p}\iff a\notin\mathfrak p,b\notin\mathfrak p\implies ab\notin\mathfrak p\iff ab\in S_{\mathfrak p}$$
So we can define the localization $S_{\mathfrak p}^{-1}R$ which is typically denoted by $R_{\mathfrak p}$. As maximal ideals are prime the same construction can be carried out for maximal ideals $\mathfrak m<R$ and we write $R_{\mathfrak m}$. For an appropriate ideal $I<R$ we also refer tp $R_I$ to as localization at $I$.


TITLE: Creative ways to show that a given matrix is diagonalizable?
QUESTION [1 upvotes]: I know the standard method of calculate the characteristic polynomial, then get the eigenvalues, and look for the dimension of the null space associated to each eigenvalue, then see if their algebraic multiplicity coincides with their geometric multiplicity but I wonder if there are some creative ways instead of the standard.
Do you know any non standard method?
Thanks for your kind attention.

REPLY [1 votes]: This list contains a few basic tools.
1) Use the spectral theorem. ( check if it is normal )
2) If all the eigen values are distinct, then it is diagonalizable.
3) Use the characteristic polynomial as you have mentioned.
4) Check if the matrix belongs to a class of matrices which are already known to be diagonalizable . (Involutions,projections,Normal,..)
5) Check if the matrix belongs to a class of matrices which are already known to be non diagonalizable . ( Nilpotent, some rotation matrices, .. )


TITLE: Norm of function $h \in L^2(\mathbb{R}^3)$
QUESTION [1 upvotes]: Let  $h\colon \mathbb{R}^3 \to \mathbb{C}$, $$h(x)=\frac{1}{|x|(|x|^2+1)^{1/2}}$$
For which $p \text{ does } h \in L^p(\mathbb{R}^3)?$
In case $h \in L^2(\mathbb{R}^3)$ what is the corresponding norm?
Can you can use the residues method?

REPLY [1 votes]: The idea is to exploit radial symmetry (see here for a formal justification):
\begin{align}\|h\|_{L^p(\mathbb{R}^3)}^p&=\int_{\mathbb{R}^3}|h(x)|^pdx=\int_0^{\infty}\int_{|x|=r}|h(x)|^pdx dr=\int_0^{\infty}4\pi r^2\frac{1}{r^p(r^2+1)^{p/2}}dr=\\
&=4\pi\int_0^{\infty}\frac{r^{2-p}}{(r^2+1)^{p/2}}dr\end{align}
and now your problem is reduced to one-dimensional calculus.


TITLE: How can I solve this equality?
QUESTION [0 upvotes]: I need prove this equality
$$ K_O(r/\xi) = \int_{0}^{\infty} \frac{k}{{k^2}+\xi^{-2}}J_o(kr)  dk  $$
Where
$$K_0$$ is a modified bessel function of zero order.

REPLY [0 votes]: Using Laplace Transform-Evaluating integrals over the positive real axis:
$\int_0^{\infty } \frac{k J_0(k r)}{k^2+\frac{1}{t^2}} \, dk=\int_0^{\infty }
   \left(\mathcal{L}_k^{-1}\left[\frac{k}{k^2+\frac{1}{t^2}}\right](s)\right) \left(\mathcal{L}_k[J_0(k r)](s)\right) \,
   ds=\int_0^{\infty } \frac{\cos \left(\frac{s}{t}\right)}{\sqrt{r^2+s^2}} \, ds=K_0\left(\frac{r}{t}\right)$


TITLE: why is $\mathbb{Z}$ an infinite subset of $\mathbb{R}$ with no cluster point.
QUESTION [0 upvotes]: I have read this fact but I can't figure out why.
I know that in order to have a cluster point, we need to find a point $p$ s.t. an open ball centered at $p$ contains infinitely many points. How come $\mathbb{Z}$ has finitely many integers in the ball?

REPLY [3 votes]: It is because all its points are isolated, i.e. a neighbourhood $(n-\varepsilon,n+\varepsilon)$ of an integer $n$, if small enough, contains only that integer. Just talke $\varepsilon<\frac12$.

REPLY [0 votes]: By definition, a ball has finite radius, so it is an interval of the form $(x-\epsilon, x + \epsilon)$ where $x$ is some real number. Any point $p$ in the ball must satisfy $|p-x| < \epsilon$. Since $|p| - |x| \leq |p-x|$ by the reverse triangle inequality, any point $p$ in the interval must satisfy
$$|p| - |x| < \epsilon$$
or equivalently,
$$|p| < |x| + \epsilon$$
Since $|x| + \epsilon$ is a finite number, this inequality can only be satisfied by finitely many integer values of $p$.


text:
\begin{document}

\maketitle

\begin{abstract}
In this paper we study a Schr\"odinger-Bopp-Podolsky system of partial differential equations
in a bounded and smooth
 domain of $\mathbb R^3$  with a non constant coupling factor. Under a compatibility condition
 on the boundary data

we deduce existence and multiplicity of solutions by means of the  Ljusternik-Schnirelmann theory.
\end{abstract}

\section{Introduction}

The Schr\"{o}dinger-Newton equation consists of a nonlinear coupling of the Schr\"{o}dinger equation with a gravitational potential of newtonian form, representing the interaction of a particle with its own gravitational field.

In 1998, Benci and Fortunato \cite{BenciFortunato1998} treated a similar problem,
where the Schr\"odinger equation was coupled with Maxwell's equations. Such coupling represents the interaction of
the particle with its own electromagnetic field. The coupling factor is a constant
$q\neq0$.

In their paper the authors consider  standing waves
solutions in the purely electrostatic field and this  leads to the so-called
Schr\"odinger-Poisson system.
They impose a  Dirichlet boundary condition both on the matter field $u$ and the
electrostatic field $\phi$ and employed variational methods and
critical point theory to develop a
procedure that would become standard to treat other similar problems.

Later, Pisani and Siciliano \cite{PisaniSiciliano2013} treated a Schr\"{o}dinger-Poisson
system with Neumann boundary conditions on the scalar field $\phi$ and considered
the case in which the interaction factor responsible for the coupling of the equations
is non-constant. This gives rise to important and interesting
considerations regarding the geometry of the  manifold on which find the solutions.

\medskip

In this paper we treat a modification of the problem dealt with by Pisani and Siciliano
consisting in the addition of a biharmonic term in the equation of the electrostatic potential and imposing
appropriate boundary conditions. The problem studied  can be interpreted as a
coupling of the
Schr\"{o}dinger equation with the Bopp-Podolsky electrodynamics (for more details
on this  matter, see \cite{dAveniaSiciliano2019} and the
references therein). However here we focus on the mathematical aspects of the
problem.

We point out that in the  literature there are few papers
concerning Schr\"odinger-Bopp-Podolski  systems. Beside \cite{dAveniaSiciliano2019}
we cite here \cite{CT,LPT} where the authors study the problem with a critical nonlinearity,
\cite{GG} where solutions with {\sl a priori} given interaction
energy for the Schr\"odinger-Bopp-Podolski  system  are found,
\cite{H} where the problem has been addressed in the context of
closed $ 3-$dimensional manifolds both in the subcritical and  critical case
and
\cite{GK} where the fibering method has been used to prove existence
results depending on a parameter and also nonexistence.

Coming back to our problem, the aim here is to study the following system of
partial differential equations in a connected, bounded, smooth open set $\Omega
\subset \mathbb R^3$:
\begin{align}
- \Delta u + q \phi u  - \kappa |u|^{p-2}u  & = \omega u \quad \text{ in } \ \Omega \label{eq:1}\\
\Delta^2 \phi - \Delta \phi & = q u^2 \quad \text{ in } \ \Omega \label{eq:2}
\end{align}
in the unknowns $u,\phi:\Omega\to \mathbb R$ and $\omega\in \mathbb R$. Here
 $\kappa \in \mathbb R$ and $q:\Omega\to \mathbb R$ are given. We assume the following boundary conditions:
\begin{align}
u & = 0 \quad \text{ on } \ \partial \Omega \label{eq:b1}\\
\dpd{\phi}{\mathbf n} & = h_1 \quad \text{ on } \ \partial \Omega \label{eq:b2} \\
\dpd{\Delta \phi}{\mathbf n} & = h_2 \quad \text{ on } \ \partial \Omega \label{eq:b3}
\end{align}
and for simplicity we assume $h_1, h_2$ continuous.

The symbol $\mathbf n$
denotes the unit
vector normal to $\partial \Omega$ pointing outwards. Since $u$ represents physically the amplitude of the wave function of a particle confined in $\Omega$, we
assume the following normalizing condition:
\begin{equation}
\label{eq:normalization_condition}
\int_{\Omega} u^2 dx = 1.
\end{equation}
We also assume that the coupling factor $q$ is continuous on $\overline \Omega$:
\begin{equation}\label{eq:continuity}
q \in C(\overline{\Omega}).
\end{equation}

Our main result is the following:

\begin{theorem}
\label{thm:main_theorem}
Let $\kappa > 0$, $p \in (2, 10/3)$ and
\begin{equation}\label{eq:alfa}
\alpha := \int_{\partial \Omega} h_2 ds - \int_{\partial \Omega} h_1 ds.
\end{equation}
Assume that $\inf_\Omega q < \alpha < \sup_\Omega q$ and that $|q^{-1}(\alpha)| = 0$.
Then there exist
infinitely many solutions $(u_n, \omega_n, \phi_n) \in H_0^1 (\Omega) \times \mathbb R \times H^2(\Omega)$ to the problem \eqref{eq:1}  and  \eqref{eq:2} under conditions \eqref{eq:b1}-\eqref{eq:continuity}, with
$$
\int_{\Omega} |\nabla u_n|^2 dx \to + \infty.
$$
Moreover the ground state solution can be assumed positive.
\end{theorem}

Our approach is variational, indeed the solutions will be found as critical points of an energy functional
restricted to  a suitable constraint. In this context by a ground state solution we mean the solution with minimal energy.
Moreover as a byproduct of the proof, we obtain that also the energy of these solutions is divergent.

\begin{remark}
It is easy to see that if $\kappa < 0$ the result holds with $p \in (2, 6)$. For $\kappa = 0$, see \cite{Afonso2020a}.
\end{remark}

The paper is organized as follows.

In Section \ref{sec:aux} we introduce an auxiliary problem
which will be useful in order to deal with homogeneous boundary conditions.

In Section \ref{sec:M} we give some properties of the constraint $M$
on which we will find the solution.

In Section \ref{sec:variational} we introduce the energy functional
and show that its critical points on $M$ will give solutions of the problem.

In the final Section \ref{sec:final} by implementing the Ljusternick-Schnirelmann theory
we prove Theorem \ref{thm:main_theorem}.

\medskip

As a matter of notations,
we use the letters $c, c',\ldots $ to denote positive constant whose value can change from line to line.
We use $\|\cdot\|_{p}$ to denote the standard $L^{p}-$norm.

\section{An auxiliary problem}\label{sec:aux}
Our aim is to define a functional whose critical points will be the weak solutions to the problem.
In order to deal with homogeneous boundary conditions, that will permit to write the functional
in a simpler form, we make a change of variable.

Consider the following auxiliary problem (where $\alpha$ is defined in \eqref{eq:alfa})
\begin{align}
\Delta^2 \chi - \Delta \chi &= \alpha/|\Omega| \quad \text{ in } \ \Omega,\label{eq:chi1} \\
\dpd{\chi}{\mathbf n} &= h_1 \quad \text{ on } \ \partial \Omega,  \label{eq:chi2}\\
\dpd{\Delta\chi}{\mathbf n} &= h_2 \quad \text{ on } \ \partial \Omega, \label{eq:chi3}\\
\int_{\Omega} \chi dx &= 0. \label{eq:chi4}
\end{align}
It is easy to see it has a unique solution.

Indeed, let $\theta$ be the unique function satisfying
\begin{align*}
\Delta \theta - \theta &= \alpha/|\Omega| \quad \text{ in } \ \Omega \medskip\\
\dpd{\theta}{\mathbf n} &= h_2 \quad \text{ on } \ \partial \Omega,\medskip\\
\int_{\Omega} \theta dx &= \int_{\partial \Omega} h_1 ds
\end{align*}
and then let $\chi$ the  unique function which satisfies
\begin{align*}
\Delta \chi &= \theta, \quad \text{ in } \ \Omega \medskip\\
\dpd{\chi}{\mathbf n} &= h_1 \quad \text{ in } \ \partial \Omega \\
\int_{\Omega} \chi dx &= 0,
\end{align*}
see e.g. \cite{Taylor}.
Then  it is easy to see that by construction $\chi$ satisfies  \eqref{eq:chi1}-\eqref{eq:chi4}.

The change of variables we make is
$$
\varphi = \phi - \chi - \mu,
$$
where
$$
\mu = \frac{1}{|\Omega|}\int_{\Omega} \phi dx.
$$
With the new variables $(u, \omega, \varphi, \mu)$ our problem becomes
\begin{align}
- \Delta u + q(\chi + \varphi)u - \kappa|u|^{p - 2}u & = \omega u - \mu q u\quad \text{ in } \ \Omega, \label{eq:new_problem_1} \medskip\\
\Delta^2 \varphi - \Delta \varphi & = qu^2 - \alpha/|\Omega| \quad \text{ in } \
\Omega, \label{eq:new_problem_2}  \medskip\\
u & = 0 \quad \text{ on } \ \partial \Omega, \label{eq:new_problem_3} \medskip\\
\int_{\Omega} u^2 dx & = 1, \label{eq:new_problem_4}  \medskip\\
\dpd{\varphi}{\mathbf n} & = 0 \quad \text{ on } \
\partial \Omega, \label{eq:new_problem_5} \medskip \\
\dpd{\Delta \varphi}{\mathbf n} & = 0 \quad \text{ on } \ \partial \Omega, \label{eq:new_problem_6} \medskip \\
\int_{\Omega} \varphi dx &= 0. \label{eq:new_problem_7}
\end{align}
Notice that the compatibility condition between \eqref{eq:new_problem_2}, \eqref{eq:new_problem_5} and \eqref{eq:new_problem_6} now reads as
$$
\int_{\Omega} qu^2 dx = \alpha.
$$
Let us define the sets
\begin{align*}
S & := \cbr{u \in H_0^1(\Omega) \ : \ \int_{\Omega} u^2 dx = 1}, \\
N & := \cbr{u \in H_0^1(\Omega) \ : \ \int_{\Omega} qu^2 dx = \alpha}, \\
M & := S \cap N.
\end{align*}
Recall that $\alpha$ depends on both the boundary conditions to the original problem.

If the problem has a solution, then of course $M \neq \emptyset$. Hence,
\begin{equation}
\label{eq:range_for_alpha}
q_{\min} \leq \alpha \leq q_{\max}
\end{equation}
where
$$
q_{\min} = \inf_\Omega q \quad \text{ and } \quad q_{\max} = \sup_\Omega q.
$$
Indeed, if $\alpha < q_{\min}$, then
$$
\alpha = \int_{\Omega} qu^2 dx \geq  q_{\min} > \alpha,
$$
which is a contradiction. The case $\alpha > q_{\max}$ is analogous.

From \eqref{eq:range_for_alpha} we deduce that $q^{-1}(\alpha)$ is not empty,
and indeed is its measure that will play a major role.

Suppose $\alpha = q_{\min}$ and $|q^{-1}(\alpha)| = 0$. Then
\begin{align*}
\int_{\Omega} qu^2 dx &= \int_{\{ x\in \Omega : q(x)>\alpha\}} q u^2 dx >\alpha,

\end{align*}
so $M=\emptyset$.
If $\alpha = q_{\max}$ and $|q^{-1}(\alpha)| = 0$ we proceed in an analogous manner to conclude that $M$ is empty and so the problem has no solutions. Therefore, we arrive at the following necessary condition for the existence of solutions: either
\begin{equation}
\label{eq:necessary_condition_1}
q_{\min} < \alpha < q_{\max}
\end{equation}
or
\begin{equation}
\label{eq:necessary_condition_2}
|q^{-1}(\alpha)| \neq 0.
\end{equation}

\section{The manifold $M$}\label{sec:M}

We now state some properties of the set $M$, referring the reader to \cite{PisaniSiciliano2013}
for the omitted proofs.

We first note that $M$ is symmetric with respect to the origin: if $u \in M$ then $-u \in M$. This follows trivially from the definition of $M$. We also note that $M$ is weakly closed in $H_0^1 (\Omega)$.

Now, we want to show that under condition \eqref{eq:necessary_condition_1} the set $M$ is not empty. For this, we introduce the following notation.

Let $A \subset \Omega$ be an open subset and define
$$
S_A := \cbr{u \in H_0^1 (A) \ : \ \int_A u^2 dx = 1}
$$
and
$$
g_A: u \in S_A \mapsto \int_A qu^2 dx \in \mathbb R.
$$

It is immediately seen that
\begin{equation*}
\label{eq:inclusion_1}
g_A(S_A) \subset [\inf_A q, \sup_A q].
\end{equation*}

\begin{lemma}
The following inclusion holds:
\begin{equation*}
\label{eq:inclusion_2}
(\inf_A q, \sup_A q) \subset \overline{g_A(S_A)}.
\end{equation*}
\end{lemma}

We can conclude the following:

\begin{proposition}
\label{prop:M_is_not_empty}
Let $A \subset \Omega$ be an open subset. If $\alpha \in (\inf_A q, \sup_A q)$ then there exists $u \in H_0^1(A)$ such that
$$
\int_A u^2 dx = 1 \quad \text{ and } \quad \int_A qu^2 dx = \alpha.
$$
\end{proposition}

In particular by taking $A=\Omega$ we get
\begin{theorem}
Assume that $\inf_\Omega q < \alpha < \sup_\Omega q$. Then $M$ is not empty.
\end{theorem}

Let us recall the definition of genus of Krasnoselki.
Given $A$ a closed and symmetric subset of some Banach space, with $0\notin A$, the {\sl genus }
of $A$ is denoted as $\gamma(A)$ and defined as the least integer  $k$
for which there exists  a continuous and even map $h:A\to \mathbb R^{k}\setminus\{0\}$.
By definition it is $\gamma(\emptyset) = 0$ and if it is not possible to construct continuous odd maps from $A$
to any $\mathbb R^{k}\setminus\{0\}$, it is set $\gamma(A) = +\infty$.

 It is  known that the genus is a topological invariant (by odd homeomorphism)
and that the genus of the  sphere in $\mathbb R^{N}$ is $N$.

The next result says that  $M$ has subsets of arbitrarily large genus.

\begin{theorem}
\label{thm:existence_of_sets_of_genus_k}
Let $u_1, \ldots, u_k \in M$ be functions with disjoint supports and let
$$
V_k = \langle u_1, \ldots, u_k \rangle
$$
be the space spanned by $u_1, \ldots, u_k$. Then $M\cap V_k$ is the $(k-1)$-dimensional sphere, hence $\gamma(M \cap V_k) = k$.
\end{theorem}

Now, it is natural if one raises the question of whether there exists such functions with disjoint supports for arbitrary $k$. The answer is positive:

\begin{theorem}
If \eqref{eq:necessary_condition_1} holds then for every $k \geq 2$ there exist $k$ functions $u_1, \ldots, u_k \in M$ with disjoint supports. Hence $\gamma(M) = + \infty$.
\end{theorem}

Let
\begin{align*}
& G_1 : u \in H_0^1 (\Omega) \mapsto \int_{\Omega} u^2 dx - 1 \in \mathbb R, \\
& G_2 : u \in H_0^1(\Omega) \mapsto \int_{\Omega} qu^2 dx - \alpha \in \mathbb R
\end{align*}
and
$$
G = (G_1, G_2).
$$
Then
$$
M = \cbr{u \in H_0^1(\Omega) \ : \ G_1(u) = G_2(u) = 0} = G^{-1}(0).
$$

We note that $G$ is of class $C^1$.

Let us show, for the reader convenience, that $G_1'(u)$ and $G_2'(u)$ are linearly independent, so $G$ will be a submersion and $M$ a submanifold of
codimension $2$.

\begin{proposition}
Assume $M$ is not empty. The differentials $G_1'(u)$ and $G_2'(u)$ are linearly independent for every $u \in M$ if and only if
\begin{equation}
\label{eq:measure_is_zero}
|q^{-1}(\alpha)| = 0.
\end{equation}
\end{proposition}

\begin{proof}
First, assume \eqref{eq:measure_is_zero}. We will show that $G_1'(u)$ and $G_2'(u)$ are linearly independent, for all $u \in M$. Suppose that there are $a, b \in \mathbb R$ such that
$$
a G_1'(u) + bG_2'(u) = 0 \quad \text{ in } H^{-1} (\Omega)
$$
for some $u \in M$. Evaluating this expression in $u$, we find that $a + b \alpha = 0$. Then
$$
aG_1'(u)[v] + b G_2'(u)[v] = b \del{-\alpha \int_{\Omega} uv dx + \int_{\Omega} quv dx } = 0 \quad \forall v \in H_0^1(\Omega),
$$
that is,
$$
b \int_{\Omega} (q - \alpha)uv dx = 0 \quad \forall v \in H_0^1 (\Omega).
$$
If $b \neq 0$ then we would have $(q - \alpha)u = 0$ a.e., and hence, in view of \eqref{eq:measure_is_zero}, $u = 0$, a contradiction. Thus $G_1'(u)$ and $G_2'(u)$ are linearly independent for all $u \in M$.

Now, suppose \eqref{eq:measure_is_zero} is not satisfied. Then $q^{-1}(\alpha)$ has not empty interior, hence there is some test function $u$ with support in $q^{-1}(\alpha)$ such that $||u||_2 = 1$. It is immediately seen that $u \in M$ because $qu = \alpha u$ and hence
$$
G_2'(u) = \alpha G_1'(u),
$$
which completes the proof.
\end{proof}

\section{Variational setting}\label{sec:variational}

We now proceed to study the variational framework of the problem. Our aim is to construct a functional whose critical points will be the weak solutions of the problem.

Following \cite{GazzolaGrunauSweers2010}, let
$$
V = \cbr{\xi \in H^2(\Omega) \ : \ \dpd{\xi}{\mathbf n} = 0 \text{ on } \partial \Omega}.
$$
We remark that $V$ is a closed subspace of $H^2(\Omega)$. Indeed, let $\{v_n\} \subset V$ such that $v_n \to v$ in $V$. Then $0 = \gamma_1(v_n) \to \gamma_1(v)$ and hence $\gamma_1(v) = 0$, where $\gamma_1$ denotes the trace operator which for smooth functions gives the directional derivative in the direction of the exterior normal on the boundary. Being a closed subspace, $V$ inherits the Hilbert space structure of $H^2(\Omega)$.

Recall that
$$
\varphi = \phi - \chi - \mu
$$
where
$$
\mu = \frac{1}{|\Omega|} \int_{\Omega} \phi dx.
$$
In this way, we have $\overline \varphi = 0$,
where from now on, given a function $f$, we denote with $\overline f$ its average in $\Omega$.
 Consider then the following natural decomposition of $V$:
\begin{equation}
\label{eq:decomposition_of_V}
V = \widetilde V \oplus \mathbb R
\end{equation}
where
$$
\widetilde V = \cbr{\eta \in V \ : \ \overline \eta = 0}.
$$
On $\widetilde V$ we have the equivalent norm
$$
||\eta||_{\widetilde V} = \del{||\nabla \eta||_2^2 + ||\Delta \eta||_2^2}^{1/2}.
$$

Consider the functional $F:H_0^1(\Omega) \times H^2 (\Omega)$ defined below:
\begin{align*}
F(u, \varphi) = & \frac12 \int_{\Omega} |\nabla u|^2 dx + \frac12 \int_{\Omega} q(\varphi + \chi)u^2 dx - \frac{\kappa}{p} \int_{\Omega} |u|^p dx\\
& - \frac{1}{4} \int_{\Omega} (\Delta \varphi)^2 dx - \frac{1}{4} \int_{\Omega} |\nabla \varphi|^2 dx - \frac{\alpha}{2 |\Omega|} \int_{\Omega} \varphi dx.
\end{align*}
It is easy to see that this functional is of class $C^1$ and that given $u \in H_0^1(\Omega)$ and $\varphi \in H^2 (\Omega)$ we have
\begin{align*}
F'_u(u, \varphi)[v] & = \int_{\Omega} \nabla u  \nabla v dx + \int_{\Omega} q(\varphi + \chi)uv dx  - \kappa \int_{\Omega} |u|^{p - 2}u v dx \\
F'_\varphi(u, \varphi)[\xi] &= \frac12 \int_{\Omega} q \xi u^2 dx - \frac12 \int_{\Omega} \Delta \varphi \Delta \xi dx - \frac12 \int_{\Omega} \nabla \varphi  \nabla \xi dx - \frac{\alpha}{2|\Omega|}\int_{\Omega} \xi dx
\end{align*}
for every $v \in H_0^1(\Omega)$ and $\xi \in H^2(\Omega)$.

Then,
$(u, \varphi, \omega, \mu)\in H^1_0(\Omega)\times H^2(\Omega)\times \mathbb R\times \mathbb R$ is a weak solution to \eqref{eq:new_problem_1}-\eqref{eq:new_problem_7} if
and only if
\begin{equation}\label{eq:defsol}
\begin{array}{ll}
& (u, \varphi) \in M \times \widetilde V, \\
& \forall v \in H_0^1 (\Omega):  \ F_u'(u, \varphi)[v] =\displaystyle \omega \int_{\Omega} uv dx - \mu \int_{\Omega} q u v dx, \\
& \forall \xi \in V: \ F_\varphi'(u, \varphi)[\xi] = 0.
\end{array}
\end{equation}

\begin{theorem}
Let $(u, \varphi) \in H_0^1(\Omega) \times H^2 (\Omega)$. Then there exist $\omega, \mu \in \mathbb R$ such that $(u, \varphi, \omega, \mu)$ is a solution to \eqref{eq:new_problem_1}-\eqref{eq:new_problem_7} if and only if $(u, \varphi)$ is a critical point of $F$ constrained on $M \times \widetilde V$, in which case the real constants $\omega, \mu$ are the two Lagrange multipliers with respect to $F'_u$.
\end{theorem}

\begin{proof}

Indeed  $(u, \varphi)$ is a critical point of $F$ constrained on $M \times \widetilde V$ if
and only if
\begin{align*}
& \forall v \in T_uM:  \ F_u'(u, \varphi)[v] = 0, \\
& \forall \xi \in \widetilde V: \ F_\varphi'(u, \varphi)[\xi] = 0.
\end{align*}
Note that the tangent space to $\widetilde V$ at $\varphi$ is $\widetilde V$ itself.

Then a weak solution, according to \eqref{eq:defsol} and the Lagrange multipliers rule,
 is a constrained critical point.

Suppose on the contrary that $(u, \varphi)$ is a constrained critical point. Then,
again by the  Lagrange multipliers rule, we have that there exists $\omega, \mu\in \mathbb R$
such that
$$
\forall v\in H^1_0(\Omega): F_u'(u, \varphi)[v] = \omega \int_{\Omega} uv dx - \mu \int_{\Omega} quv dx.
$$
It remains to prove that $F_\varphi'(u, \varphi)[\xi] = 0$ for all $\xi \in V$. But this follows by the
 decomposition \eqref{eq:decomposition_of_V}, noticing that $F_\varphi'(u, \varphi)[r] = 0$ for every constant $r \in \mathbb R$.
 Then \eqref{eq:defsol} is satisfied and this concludes the proof.
\end{proof}

The functional $F$ constrained on $M \times \widetilde V$ is unbounded from above and from below.

 This issue has been addressed by Benci and Fortunato \cite{BenciFortunato1998} and in many subsequent papers. Their standard reduction argument goes as follows:

\begin{enumerate}
\item[(i)] For every fixed $u\in H^{1}_{0}(\Omega)$ there exists a unique $\Phi(u)$ such that $F'_\varphi(u,  \Phi(u)) = 0$.

\item[(ii)] The map $u \mapsto \Phi(u)$ is of class $C^1$.

\item[(iii)] The graph of $\Phi$ is a manifold, and we are reduced to study the functional $J(u) = F(u, \Phi(u)$, possibly constrained.
\end{enumerate}

However the method sketched above fails in our situation, for two reasons. First, we see that $F_\varphi'(u, \varphi) = 0$ with $\varphi \in \widetilde V$ is just
\begin{align*}
\Delta^2 \varphi - \Delta \varphi - qu^2 + \alpha/|\Omega| &= 0 \quad \text{ in } \ \Omega, \\
\dpd{\varphi}{\mathbf n} &= 0 \quad \text{ on } \ \partial \Omega, \\
\dpd{\Delta \varphi}{\mathbf n} &= 0 \quad \text{ on } \ \partial \Omega, \\
\int_{\Omega} \varphi dx &= 0.
\end{align*}
The problem above has not a unique solution  for any fixed $u$: this happens,
due to the compatibility condition,
 if and only if $u \in N$.
 Moreover,
 since $N$ is not a manifold (unless $\alpha \neq 0$) we
 cannot require the map $\Phi: u \mapsto \Phi(u)$ to be of class $C^1$ in $N$. We shall then extend such a map $\Phi$.

\begin{proposition}
\label{prop:auxiiary_problem_to_extend_Phi}
For every $w \in L^{6/5}(\Omega)$ there exists a unique $L(w) \in \widetilde V$ solution of
\begin{align*}
\Delta^2 \varphi - \Delta \varphi - w + \overline w &= 0 \quad \text{ in } \ \Omega, \\
\dpd{\varphi}{\mathbf n} &= 0 \quad \text{ on } \ \partial \Omega, \\
\dpd{\Delta \varphi}{\mathbf n} &= 0 \quad \text{ on } \ \partial \Omega, \\
\int_{\Omega} \varphi dx &= 0.
\end{align*}
The map $L: L^{6/5}(\Omega) \longrightarrow \widetilde V$ is linear and continuous, hence of class $C^{\infty}$.
\end{proposition}
\begin{proof}

The weak solutions to the problem are functions $\varphi$ in the Hilbert space  $\widetilde V$ such that
$$
\int_{\Omega} \Delta \varphi \Delta v dx + \int_{\Omega} \nabla \varphi  \nabla v dx - \int_{\Omega} wv dx = 0 \qquad \forall v \in \widetilde V.
$$
So the result follows by applying the Riesz Theorem since
the bilinear form $b: \widetilde V \times \widetilde V \longrightarrow \mathbb R$ given by
$$
b(\varphi, v) = \int_{\Omega} \Delta \varphi \Delta v dx + \int_{\Omega} \nabla \varphi  \nabla v dx.
$$
is just the  scalar product in $\widetilde V$.
\end{proof}

The following proposition follows from well-known properties of Nemytsky operators.

\begin{proposition}
The map
$$
u \in L^6(\Omega) \mapsto qu^2 \in L^{6/5}(\Omega)
$$
is of class $C^1$.
\end{proposition}

As a consequence of the previous propositions, we can define the following map:

$$
\Phi: u \in H_0^1 (\Omega) \mapsto L(qu^2) \in \widetilde V.
$$

It is clear that
$$
\Phi(u) = \Phi(-u) = \Phi(|u|).
$$
Moreover, for every $(u, \varphi) \in H_0^1(\Omega) \times \widetilde V$ we have that $\varphi = \Phi(u)$ if and only if for every $\eta \in \widetilde V$
\begin{equation*}
\int_{\Omega} \Delta \varphi \Delta \eta dx + \int_{\Omega} \nabla \varphi \nabla \eta dx= \int_{\Omega} qu^2 \eta dx.
\end{equation*}
Taking $\eta = \Phi(u)$ we have  in particular the important relation
\begin{equation}
\label{eq:identity_2}
\int_{\Omega} (\Delta \Phi(u))^2 dx + \int_{\Omega} |\nabla \Phi(u)|^2 dx =  \int_{\Omega} qu^2 \Phi(u) dx.
\end{equation}
The right hand side above is the {\sl interaction energy} term.
Then we infer
\begin{align*}
||\Phi(u)||_{\widetilde V}^2 & \leq ||q||_\infty \int_{\Omega} u^2 \Phi(u) dx \\

& \leq c ||u||_4^2 ||\Phi(u)||_2 \\
& \leq c ||\nabla u||_2^2 ||\nabla \Phi(u)||_2 \\
& \leq c ||\nabla u||_2^2 ||\Phi(u)||_{\widetilde V}
\end{align*}
and hence
\begin{equation}
\label{eq:boundedness_of_Phi}
||\Phi(u)||_{\widetilde V} \leq c ||\nabla u||_2^2,
\end{equation}
that is, $\Phi$ is bounded on bounded sets.
We have

\begin{lemma}
\label{lem:compactness_of_Phi}
If $u_n \rightharpoonup u$ in $H_0^1(\Omega)$ then
$$
\int_{\Omega} q u_n^2 \Phi(u_n) dx \to \int_{\Omega} q u^2 \Phi(u) dx.
$$
Moreover the map $\Phi$ is compact.
\end{lemma}

\begin{proof}
Let $u_n \rightharpoonup u$ in $H_0^1(\Omega)$ and define $B_n, B: \widetilde V \longrightarrow \mathbb R$ by
$$
B_n(\eta) := \int_{\Omega} qu_n^2\eta dx, \qquad B(\eta) := \int_{\Omega} qu^2\eta dx.
$$
Such operators are continuous due to the H\"{o}lder's inequality. For example:
\begin{align*}
\envert{\int_{\Omega} q u^2 \eta dx} &
\leq ||q||_\infty ||u||_4^2 ||\eta||_2 \leq c ||\nabla \eta||_2 \leq c ||\eta||_{\widetilde V}
\end{align*}
(where here $c$ depends on $u$).

Due to the compact embedding of $H^{1}_{0}(\Omega)$ into $L^{p}(\Omega)$ for $p\in[1,6)$,
we get $u_{n}^{2} \to u^{2}$ in $L^{6/5}(\Omega)$

and then
\begin{align*}
|B_n(\eta) - B(\eta)| & \leq ||q||_\infty ||u_n^2-u^2||_{6/5} ||\eta||_6 \\
& \leq c ||q||_\infty ||u_n^2-u^2||_{6/5} ||\eta||_{\widetilde V}.
\end{align*}
Hence
$$
||B_n - B|| \leq \sup_{\eta \neq 0} \frac{c ||u_n^2 - u^2||_{6/5} ||\eta||_{\widetilde V}}{||\eta||_{\widetilde V}} \to 0,
$$
namely $B_n \to B$ as operators in  $\widetilde V$.

On the other hand, we have that $\Phi(u_n) \rightharpoonup \Phi(u)$ in $\widetilde V$. Indeed, let $g \in
\widetilde{ V}^{'}$. Then there is some $v_g \in \widetilde V$ such that
$$
g(\Phi(u_n)) = \int_{\Omega} \nabla \Phi(u_n)  \nabla v_g dx + \int_{\Omega} \Delta \Phi(u) \Delta v_g dx = \int_{\Omega} q u_n^2 v_g dx.
$$
But then
\begin{align*}
g(\Phi(u_n)) - g(\Phi(u)) & = \int_{\Omega} q (u_n^2 - u^2) v_g dx \\
& \leq ||q||_\infty ||u_n^2 - u^2||_2 ||v_g||_2 \to 0
\end{align*}
since $u_n^2 \to u^2$ in $L^2(\Omega)$ as well.

We then conclude that
$$
\int_{\Omega} q u_n^2 \Phi(u_n) dx \to \int_{\Omega} q u^2 \Phi(u) dx
$$
and by  \eqref{eq:identity_2} that $\| \Phi(u_{n})\|_{\widetilde V} \to \|\Phi(u)\|_{\widetilde V}$.
Consequently  $\Phi(u_n) \to \Phi(u)$ in $\widetilde V$.
\end{proof}

Note that for every $u \in N$ we have that $F_\varphi'(u, \Phi(u)) = 0$. Indeed, $\Phi(u)$ is the unique solution to the problem in Proposition \ref{prop:auxiiary_problem_to_extend_Phi} with $w = qu^2$.

We now define the reduced functional of a single variable:
$$
\fullfunction{J}{H_0^1(\Omega)}{\mathbb R}{u}{F(u, \Phi(u))}
$$
With the notation $\varphi_u := \Phi(u)$ the functionl $J$ is explicitly given by (recall \eqref{eq:identity_2})
\begin{align*}
J(u) & = \frac12 \int_{\Omega} |\nabla u|^2 dx + \frac12 \int_{\Omega} q \varphi_u u^2 dx + \frac12 \int_{\Omega} q \chi u^2 dx - \frac{\kappa}{p} \int_{\Omega} |u|^p dx \\
& \quad - \frac{1}{4} \int_{\Omega} (\Delta \varphi_u)^2 dx - \frac{1}{4} \int_{\Omega} |\nabla \varphi_u|^2 dx - \frac{\alpha}{2 |\Omega|}\int_{\Omega} \varphi_u dx \\
& = \frac12 \int_{\Omega} |\nabla u|^2 dx + \frac14 \int_{\Omega} (\Delta \varphi_u)^2 dx + \frac14 \int_{\Omega} |\nabla \varphi_u|^2 dx + \int_{\Omega} q \chi u^2 dx \\
& \quad - \frac{\kappa}{p} \int_{\Omega} |u|^p dx.
\end{align*}

We note that $J$ is of class $C^1$ on $H_0^1 (\Omega)$ and even. Moreover, for every $u \in M$ we have that
$$
J'(u)[v] = F_u'(u, \varphi_u)[v] + F_\varphi'(u, \varphi_u)[\Phi'(u)[v]] = F_u'(u, \varphi_u)[v] \quad \forall v \in H_0^1(\Omega)
$$
and hence we deduce the following

\begin{theorem}
The pair $(u, \varphi) \in M \times \widetilde V$ is a critical point of $F$ constrained on $M \times \widetilde V$ if and only if $u$ is a critical point of $J|_M$ and $\varphi = \Phi(u)$.
\end{theorem}

\section{Proof of the main result} \label{sec:final}

The next lemma will be useful.

\begin{lemma}\label{lem:ps}
Let $D$ be a regular domain of $\mathbb R^N$ and
\begin{equation*}
\label{eq:lemma_sobolev_spaces_1}
1 \leq s \leq N,
\end{equation*}
\begin{equation*}
\label{eq:lemma_sobolev_spaces_2}
s < p < s^* = \frac{Ns}{N - s}
\end{equation*}
and
\begin{equation*}
\label{eq:lemma_sobolev_spaces_3}
0 < r \leq N \left(1 - \frac{p}{s^*} \right).
\end{equation*}
Then there exists a constant $C > 0$ such that for every $u \in W^{1, s}(D)$ it holds that
\begin{equation*}
\label{eq:lemma_sobolev_spaces_conclusion}
||u||_p^p \leq C ||u||_{W^{1, s}}^{p - r}||u||_s^r
\end{equation*}
\end{lemma}

\begin{proof}
See \cite[Lemma 3.1]{PisaniSiciliano2007}.
\end{proof}

\begin{remark}
If $D$ is bounded, then the conclusion of the lemma is true also in the case $p \in [1, s]$ with $r < p$. Also, if $D$ is bounded and $u \in W_0^{1, p} (D)$, then, by Poincar\'e inequality,
$$
||u||_p^p \leq C ||\nabla u||_s^{p - r} ||u||_s^r.
$$
\end{remark}

The following lemma gives the existence of solutions to our modified problem.

\begin{lemma}\label{lem:minimo}
\label{lem:coercivity_and_weak_lower_semicontinuity}
The functional $J$ on $M$ is weakly lower semicontinuous and coercive. In particular, it has a minimum $u \in M$,
and it can be assumed positive.
\end{lemma}

\begin{proof}
We have
\begin{align*}
J(u) & = \frac12 \int_{\Omega} |\nabla u|^2dx + \frac14 \int_{\Omega} (\Delta \varphi_u)^2dx + \frac14 \int_{\Omega} |\nabla \varphi_u|^2dx
 + \int_{\Omega} q \chi u^2 dx - \frac{\kappa}{p} \int_{\Omega} |u|^pdx\\
& \geq  \frac12 \int_{\Omega} |\nabla u|^2 dx - \|q\|_{\infty}\|\chi\|_{\infty}
- \frac{\kappa}{p} \int_{\Omega} |u|^pdx.
\end{align*}

Finally, we apply Lemma \ref{lem:ps} with $s = 2$ and $N = 3$. Since $p \in (2, 10/3)$ it holds that
$$
p - 2 < 3 \left(1 - \frac p6 \right) < 2
$$
and we can choose
$$
p - 2 < r < 3 \left(1 - \frac p6 \right),
$$
so that by the Lemma it follows that
$$
\frac{\kappa}{p} \int_{\Omega}|u|^p dx \leq c||\nabla u||_2^{p - r}.
$$
Hence,
$$
J(u) \geq \frac12 \int_{\Omega} |\nabla u|^2 dx - \|q\|_{\infty}\|\chi\|_{\infty}- c'||\nabla u||_2^{p - r}
$$
and thus $J$ is coercive and bounded from below on $M$.

Now, let $\{u_n\} \subset M$ such that $u \rightharpoonup u$. Since $M$ is weakly closed, $u \in M$. By Lemma \ref{lem:compactness_of_Phi} we know that
$$
\frac14 \int_{\Omega} (\Delta \varphi_{u_n})^2 dx + \frac14 \int_{\Omega} |\nabla \varphi_{u_n}|^2 dx \to \frac14 \int_{\Omega} (\Delta \varphi_u)^2 dx + \frac14 \int_{\Omega} |\nabla \varphi_u|^2 dx.
$$
We also know that $u_n^2 \to u^2$ in $L^{6/5}(\Omega)$ so
\begin{align*}
\int_{\Omega} q \chi (u_n^2 - u^2) dx & \leq c \int_{\Omega} u_n^2 - u^2 dx \\
& \leq c ||u_n - u||_{6/5} \to 0.
\end{align*}
Finally, the first and last terms are the norms of $u$ in $H_0^1(\Omega)$ and $L^p (\Omega)$ (up to constants), so they are weakly lower semicontinuous.

Thus $J$ is weakly lower semicontinuous and the  existence of the minimum follows by standard results.

Note that $J(u) = J(|u|)$ so the minimum may be assumed to be positive.

\end{proof}

We will use a deformation argument to show that there are infinitely many solutions. A crucial point is that the functional satisfies the Palais-Smale condition.
We recall that in general, it is said that the $C^{1}$ functional $\mathcal I$
satisfies the Palais-Smale condition on the manifold $\mathcal M$, if
any sequence $\{u_{n}\}\subset \mathcal M$ such that
$\left\{\mathcal I(u_{n})\right\}$ is bounded and   $\mathcal I(u_{n}) \to0$
in the tangent bundle, admits a convergent subsequence to an element $u\in \mathcal M$.

\begin{proposition}
\label{prop:Palais_Smale}
The functional $J$ satisfies the Palais-Smale condition on $M$.
\end{proposition}

\begin{proof}
Let $\{u_n\} \subset M$ be

such that

$$
\{J(u_n)\} \ \text{ is bounded }
$$
and
\begin{equation}
\label{eq:derivative_goes_to_zero}
J|_M'(u_n) \to 0.
\end{equation}
By \eqref{eq:derivative_goes_to_zero}  there exists two sequences of real numbers $\{\lambda_n)\}, \{\beta_n \} $
and a sequence $\{v_{n}\} \subset H^{-1}$ such that $v_n \to 0$ and
\begin{equation}
\label{eq:derivative_goes_to_zero_meaning}
- \Delta u_n + q(\varphi_n + \chi)u_n - \kappa |u_n|^{p - 2}u_n = \lambda_n u_n + \beta_n q u_n + v_n
\end{equation}
where $\varphi_n := \varphi_{u_n}$.

Since $J$ is coercive and $\{J(u_n)\}$ is bounded, we know that $\{u_n\}$ is bounded in $H_0^1 (\Omega)$. Hence there exists $u \in H_0^1(\Omega)$ such that $u_n \rightharpoonup u$, up to a subsequence. By the compact embeddings and Lemma \ref{lem:compactness_of_Phi} we know that
\begin{equation}\label{eq:conv}
u_{n}\to u \quad\text{in } \  L^{p}(\Omega), \quad
\varphi_n \to \varphi_{u}\quad  \text{ in } \  H^{2}(\Omega).
\end{equation}
 Also, since $M$ is weakly closed, we know that $u \in M$. It only remains to show that $u_n \to u$ in $H_0^1 (\Omega)$.

By \eqref{eq:derivative_goes_to_zero_meaning} we have that
\begin{equation}
\label{eq:derivative_goes_to_zero_applied_to_un}
\frac12 \int_{\Omega} |\nabla u_n|^2 dx + \frac12 \int_{\Omega} q(\varphi_n + \chi)u_n^2 dx - \frac{\kappa}{p} \int_{\Omega} |u_n|^p dx - \langle v_n, u_n \rangle = \lambda_n + \alpha \beta_n.
\end{equation}
By \eqref{eq:conv} we infer
\begin{eqnarray*}\label{eq:}
\left| \int_{\Omega} \left(q(\varphi_n + \chi)u_n^2 -  q(\varphi_u + \chi)u^2  \right)dx\right| &\leq&
c \int_{\Omega} \left| \varphi_{n}+\chi \right| |u_{n}^{2} - u^{2}| dx +
\int_{\Omega} u^{2} \left| \varphi_{n} - \varphi \right| dx \\
&=&o_{n}(1)
\end{eqnarray*}
where we are denoting with $o_{n}(1)$ a vanishing sequence.

Then the right-hand side of \eqref{eq:derivative_goes_to_zero_applied_to_un} is bounded

and we can assume that
$$
\lambda_n + \alpha \beta_n =\xi + o_{n}(1)
$$
with $\xi \in \mathbb R$.

Then \eqref{eq:derivative_goes_to_zero_meaning} becomes
\begin{equation}
\label{eq:derivative_goes_to_zero_becomes}
-\Delta u_n + q(\varphi_n + \chi)u_n - \kappa |u_n|^{p - 2}u_n - v_n =(\xi + o(1))u_n - \beta_n(q - \alpha)u_n.
\end{equation}

Now, since $u \in M$ we know that $||u||_2 = 1$. This, together with the assumption $|q^{-1}(\alpha)| = 0$ implies that $(q - \alpha)u$ is not identically zero. Then there exists a test function $w \in C^{\infty}_{0}(\Omega)$ such that
$$
\int_{\Omega} (q - \alpha)u w dx \neq 0.
$$
Evaluating \eqref{eq:derivative_goes_to_zero_becomes} on this $w$ we get
\begin{multline}\label{eq:beta}
\int_{\Omega} \nabla u_n  \nabla w dx + \int_{\Omega} q(\varphi_n + \chi)u_n w dx -\kappa \int_{\Omega}|u_n|^{p - 2}u_n w dx \\ - \langle v_n, w \rangle - (\lambda + o_{n}(1))\int_{\Omega} u_n w dx
= \beta_n \int_{\Omega} (q - \alpha)u_n w dx
\end{multline}
and using again \eqref{eq:conv} we see that

every term in the left-hand side converges.

Also, by the weak convergence of $\{u_n\}$,
$$
\int_{\Omega} (q - \alpha)u_n w dx \to \int_{\Omega} (q - \alpha)u w dx.
$$
This implies, coming back to \eqref{eq:beta}, that $\{\beta_n\}$ is bounded, which in turn implies that $\{\lambda_n\}$ is bounded.

Applying \eqref{eq:derivative_goes_to_zero_becomes} to $u_n - u$ we get
\begin{multline}\label{eq:ultimaconv}
\int_{\Omega} \nabla u_n \nabla(u_n - u) dx + \int_{\Omega} q(\varphi_n + \chi)u_n(u_n - u) dx - \kappa \int_{\Omega} |u_n|^{p - 2}u_n (u_n - u)\\
- \langle v_n, u_n - u \rangle = (\lambda + o(1))\int_{\Omega} u_n (u_n - u) dx + \beta_n \int_{\Omega} (q - \alpha)u_n(u_n - u) dx.
\end{multline}
Since (again by \eqref{eq:conv}) we have
\begin{align*}
& \int_{\Omega} q(\varphi_n + \chi)u_n(u_n - u) dx \to 0, \quad \langle v_n, u_n - u \rangle \to 0, \\
& \int_{\Omega}|u_n|^{p - 2} u_n(u_n - u) dx \to 0, \quad (\lambda + o(1))\int_{\Omega} u_n (u_n - u) dx \to 0
\end{align*}
and
$$
\beta_n \int_{\Omega} (q - \alpha)u_n(u_n - u) dx \to 0,
$$
we conclude by \eqref{eq:ultimaconv}
 that $||\nabla u_n||_2 \to ||\nabla u||_2$ and so $u_n \to u$ in $H_0^1(\Omega)$.

\end{proof}

\medskip

Now we can give the proof of Theorem \ref{thm:main_theorem}.

By Theorem \ref{thm:existence_of_sets_of_genus_k}, $M$ has compact, symmetric subsets of genus $k$ for every $k \in \mathbb N$.

Let us recall now a classical result in critical point theory. We give the proof for the reader convenience.
\begin{lemma}
\label{lem:sublevels_have_finite_genus}
For any $b \in \mathbb R$ the sublevel
$$
J^b = \cbr{u \in M \ : \ J(u) \leq b}
$$
has finite genus.
\end{lemma}

\begin{proof}
We argue by contradiction. Suppose that
$$
D = \cbr{b \in \mathbb R \ : \ \gamma(J^b) = \infty} \neq \emptyset.
$$
Since $J|_M$ is bounded from below, then $D$ is bounded from below. Then
$$
- \infty < \overline b = \inf D < \infty.
$$
Moreover, since $J|_M$ satisfies the Palais-Smale condition, the set
$$
Z = \cbr{u \in M \ : \ J(u) = \overline b, \ J|_M'(u) = 0}
$$
is compact. Hence there exists a closed symmetric neighborhood $U_Z$ of $Z$ such that $\gamma(U_Z) < \infty$. By the Deformation Lemma, there exists an $\varepsilon > 0$ such that $J^{\overline b -  \varepsilon}$ includes a deformation retract of $J^{\overline b + \varepsilon} \setminus U_Z$. Then, by the properties of the genus,
$$
\gamma(J^{\overline b + \varepsilon}) \leq \gamma(J^{\overline b + \varepsilon} \setminus U_Z) + \gamma(U_Z) \leq \gamma(J^{\overline b - \varepsilon}) + \gamma(U_Z) < \infty,
$$
a contradiction.
\end{proof}

Let $n \in \mathbb N$. By Lemma \ref{lem:sublevels_have_finite_genus} there exists some $k \in \mathbb N$ depending on $n$ such that
$$
\gamma(J^n) = k.
$$
Let
$$
A_{k+1} = \cbr{A \subset M \ : \ A = -A, \overline A = A, \gamma(A) = k + 1}
$$
that we know is not empty by   Theorem \ref{thm:existence_of_sets_of_genus_k}.

 By the monotonicity property of the genus, any $A \in A_{k+1}$ is not contained in $J^n$, then $\sup_A J > n$ and therefore
$$
c_n = \inf_{A \in A_{k+1}} \sup_{u \in A} J(u) \geq n.
$$
Well known results (see e.g. \cite{Szulkin1988})
say that $c_{n}$ are critical levels for $J|_{M}$ and then there is a
sequence $\{u_{n}\}$ of critical points such that
$$J(u_{n})= c_{n}\to +\infty$$

The critical points give rise to Lagrange multipliers $\omega_n, \mu_n$ and then, recalling the decomposition $\varphi = \phi - \chi - \mu$, to solutions $(u_n, \omega_n, \phi_n) \in H_0^1(\Omega) \times \mathbb R \times H^2 (\Omega)$ of the original problem.

\medskip

We show that $||\nabla u_n||_2 \to + \infty$.

Since
$$
\int_{\Omega} q \chi u_n^2 dx \leq ||q \chi||_\infty,
$$
and by \eqref{eq:boundedness_of_Phi} it is
$$
||\varphi_n||_{\widetilde V} = \int_{\Omega} (\Delta \varphi_n)^2 dx + \int_{\Omega} |\nabla \varphi_n|^2 dx \leq c ||\nabla u_{n}||_2^2,
$$
we see that
$$
|J(u_n)| \leq (1+c) ||\nabla u_n||^2 + c' ||\nabla u_n||_2^{p } + ||q \chi||_\infty
$$
and then $\{u_{n}\}$ can not be  bounded.

This concludes the proof of Theorem \ref{thm:main_theorem}.

config: arxiv


TITLE: "Converse" of Taylor's theorem
QUESTION [26 upvotes]: Let $f:(a,b)\to\mathbb{R}$. We are given $(k+1)$ continuous functions $a_0,a_1,\ldots,a_k:(a,b)\to\mathbb{R}$ such that for every $c\in(a,b)$ we can write $f(c+t)=\sum_{i=0}^k a_i(c)t^i+o(t^k)$ (for any $t$ in a neighbourhood of $0$). Can we conclude that $f$ is of class $C^k$?

REPLY [3 votes]: Another answer: You are prescribing a Whitney jet, and the Whitney extension theorem gives the answer:

Whitney, Hassler: Analytic extensions of differentiable functions defined in closed sets, Trans. AMS 36 (1934), 63--89.

Also, another reference building on the paper of Marcinkiewicz and Zygmund, is

G. Glaeser: Etude de quelques Algebres Tayloriennes, J. Anal. Math. 11 (1958), 1-118.