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The answer is no. There exist analytic functions $f$ of 3 variables such that they cannot be represented as a composition of continuously differentiable functions of two variables. This is old result of Vitushkin. You can find nice story of Hilbert's thirteen problem in Vitushkin, A. G. Hilbert's thirteenth problem and related questions. Russian Math. ...

11

Here is the answer to your specific question (and, in fact, something a little more general):
The hypersurface defined by $f=x_{ij}-x_{kl}$ (assuming that $i$, $j$, $k$, and $l$ are distinct) is smooth if $n\le 5$ and singular for $n>5$. The singular locus consists of the $2$-planes that lie in the codimension $4$ subspace that consists of $0$ plus the ...

answered Aug 31 '11 at 23:15

Robert Bryant

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11

Consider the following:
Take $M$ the plane with the standard (flat) metric, with the origin and the ray $[0,\infty) \times \{2\}$ removed.
Let $S$ be the unit circle centered at the origin. Clearly for every point in $M$ there exists a unique point on $S$ that is closest to it:
When $(x,y)\in M$ is such that either $x < 0$ or $y < 2$, then the ...

11

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen ...

answered Dec 7 '16 at 1:38

Pedro Lauridsen Ribeiro

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11

This is true both over finite and infinite fields.
For infinite fields, see [Jou, Cor. I.6.11(2)]. It works for a general section of any very ample line bundle $\mathscr L$, using that over an infinite field a nonempty open subset of $|\mathscr L| \cong \mathbb P^N$ contains a rational point.
For finite fields, see [Poo, Thm. 1.1]. It requires taking a high ...

answered May 5 '18 at 17:03

R. van Dobben de Bruyn

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The answer is no. In what follows (see OP's comments below) we assume that the arrows of the category of lctvs are Michal-Bastiani smooth maps. Recall that a map $\Phi:E\rightarrow F$ from a lctvs $E$ into another lctvs $F$ is said to be Michal-Bastiani smooth if its directional (Gâteaux) derivatives of order $k$
$$ D^k\Phi[x](y_1,\ldots,y_k) = \left.\frac{\...

answered Jun 15 '15 at 5:36

Pedro Lauridsen Ribeiro

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9

Here is a counter example. Let $E=\ell^2$. Consider the curve $\gamma:\mathbb R \to E$ given by
$$\gamma(t)= \Big(\frac{\sin(2^nt)}{2^n}\Big)_{n\in \mathbb N}$$
In the dual $\ell^2$ consider the dense linear subspace of all sequences $l=(l_n)$ with finite support. For such $l$ with $l_n=0$ for $n\ge N$,
$$(l\circ \gamma)(t) = \sum_{n=1}^N l_n.\frac{\sin(2^...

8

The only restriction I see is that $\hat{X}$ must be normal (because $X$ is): if $\phi$ is a rational function on (some affine open subscheme of) $\hat{X}$ which is integral over $\mathscr{O}_\hat{X}$, then $\phi\circ\pi$ is integral over $\mathscr{O}_{X}$, hence in $\mathscr{O}_{X}$. In other words, $\phi$ lies in $\pi_*\mathscr{O}_{X}=\mathscr{O}_\hat{X}$.

answered Mar 26 '15 at 14:59

Laurent Moret-Bailly

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$\hat{X}$ can be as bad as you want. For example, take your favorite non-Gorenstein variety $\hat{X}$ in $\mathbb{A}^N$. By Noether Lemma there is a finite morphism $\hat{X} \to \mathbb{A}^n =: Y$. Take $X$ to be a resolution of singularities of $\hat{X}$. Then $X \to Y$ is a quasifinite morphism between smooth varieties.

8

No, it does not. As a counterexample, let $f$ be the inclusion of the closed point $Spec\ k \hookrightarrow Spec\ R$. Equally well, one could also take other irreducible projective smooth $k$-varieties for $X$.

answered Nov 22 '13 at 1:55

Kestutis Cesnavicius

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8

We can say is that $\limsup |a_n|^{1/n}=\infty$, where $a_n=\max_{a\in K}|f^{(n)}(a)|/n!$, for every compact $K\subset\Omega$.

answered Dec 7 '16 at 1:00

Alexandre Eremenko

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8

Theorem 2.8 on p. 30 of Models for smooth infinitesimal analysis by Moerdijk and Reyes. The theorem works with $C^\infty$ rings, which are a bit more general than algebras of functions. The advantage is that the proof is short.

8

Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1.
If $B$ is an algebra with a nilpotent ideal $I$, we need to check that every homomorphism $\varphi:kQ\to B/I$ lifts to a homomorphism $\tilde{\varphi}:kQ\to B$.
...

answered Jun 17 '18 at 10:12

Jeremy Rickard

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7

An algebraic (possibly singular) $K3$ surface is a normal algebraic surface whose minimal resolution is a smooth $K3$ surface. You can obtain these for example by blowing down $(-2)$-curves on a smooth algebraic $K3$. As long as you only do that you can even restrict the kind of singularities you allow. For example blowing down a $(-2)$-curve (and certain ...

7

This desire fails on a much more basic level. A smooth morphism is flat, and you can't even have a flat morphism between two smooth compactifications.
A flat morphism has equidimensional fibers. If it is also proper and birational, then it is finite and a finite birational morphism mapping onto something normal is automatically an isomorphism.
The main ...

answered Apr 14 '16 at 15:34

Sándor Kovács

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6

Edit, 12/2016. The original post was long ago, but I had a reason to revisit this some months back when I was asked a similar question. I am including below the positive answer to this question that I found in a special case. There is a little setup required.
Flatness over $\mathbb{A}^r$. First there is a flatness lemma. Let $k$ be a commutative, ...

6

Let $k$ be a field and $Y=\mathrm{Spec}\,R$ where $R=\bigcup_{n>0}k[[t^{1/n}]]$ is the ring of Puiseux series over $k$. Take $X=\mathrm{Spec}\,(R/tR)$, $f=$ the obvious embedding. The maximal ideal $m$ of $R$ satisfies $m=m^2$, and the same holds in $R/tR$, so both completions are equal to $k$, but of course $f$ is not smooth.

answered Nov 19 '14 at 18:32

Laurent Moret-Bailly

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6

Yes, the spaces are isomorphic as Fréchet spaces. This is often called the exponential law and holds for every compact manifolds $M$ and $N$,
$$C^\infty(M \times N) = C^\infty(M, C^\infty(N))$$
and as a corollary you obtain that both space are tame in the sense of Hamilton.
A proof of this exponential map can be found in the book "The Convenient Setting of ...

6

As noted by Dejan Govc, the set $E$ should also contain those points where the right limit of $F(x)$ equals the left limit of $G(x)$, because any continuous function bounded between $F$ and $G$ will be forced to the unique limit in these points.
Let's first describe a construction for a family of continuous functions, which will later be refined to give a ...

6

For what it's worth, one can say the following sort of thing.
Since $Y$ is log terminal so is $(\hat{X}, -\mathrm{Ram})$. This doesn't mean much since in the pair, the boundary has a negative coefficients (ie, the singularities of $\hat{X}$ can be arbitrarily bad). But it does say things like:
if $\hat{X}$ has really bad singularities at some points, ...

6

EDITED. The following theorem of Bernstein answers the question:
If $f$ is infinitely differentiable on an interval and no derivative changes sign, then $f$ is analytic.
Your condition that all derivatives are monotone of course implies that none
of them changes sign.
Therefore, if such a function is extended on a larger interval with
preservation of the ...

answered Jan 19 '17 at 7:51

Alexandre Eremenko

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6

Take an orthonormal basis in the tangent space of $M$ at some point $p_0$ on your curve $c$. For each vector $u_i$ in that basis, let $p_i=\exp_{p_0}(-(\rho/2)e_i)$, where $\rho$ is the injectivity radius at $p_0$. Then the distance function $f_i(p)=d(p_i,p)$ from $p_i$ has gradient $e_i$ at $p_0$. So $f_1, f_2, \dots, f_n$ is a coordinate system near $p_0$. ...

5

The inner functions $\phi_{qp}$ are independent in Kolmogorov's superposition theorem and they are very "bad" functions like cantor function. The smoother $f$ is, the more singular $\Phi_q$ has to be to cancel out the singularities of $\phi_{pq}$ except in some special cases, such as $f$ is a constant. There is no linear relation between the smoothness of $...

5

I don't know the general story, but I think the correspondence between Weil divisors and line bundles breaks down already for $\mathbf{A}^\infty$, which is pretty much the smoothest ind-variety out there. Using the standard ind-structure on $\mathbf{A}^\infty$, the example I have in mind is the union of all hyperplanes $x_n-x_1=0$. This is a codimension ...

5

For the second question only:
For $f(x,t) = (x-t)^2$, you have that if $t > 1$
$$ \int_0^1 \frac{1}{f(x,t)} ~dx = \frac{1}{t-x} \Big|^1_0 = \frac{1}{t - 1} - \frac{1}{t} = \frac{1}{t(t-1)} $$
So you functional
$$ G(f)(t) = \begin{cases}
0 & t \leq 1 \\
t^2(t-1)^2 & t > 1\end{cases} $$
and is not a smooth function of $t$ (not even $C^2$). ...

5

Faa di Bruno's formula for derivatives of compositions of functions says
$$
(f\circ g)^{(n)}(t)=n!\sum_{k\ge 0}\ \sum_{n_1,\ldots,n_k\ge 1}
\mathbf{1}\left\{\sum_{i=1}^{k} n_i=n\right\}
\ \frac{f^{(k)}(g(t))\ g^{(n_1)}(t)\cdots g^{(n_k)}(t)}{k!\ n_1!\cdots n_k!}
$$
where $\mathbf{1}\{\cdots\}$ stands for the indicator function of the condition within braces.
...

answered Apr 28 '17 at 19:57

Abdelmalek Abdesselam

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5

Notice that $e^x$ does not have an inverse on the whole real line.
Extension of iterates is possible if $f$ has a fixed point $x_0$. Suppose for example, that this fixed point is repelling that is $f(x_0)=x_0$ and $\lambda=f'(x_0)>1.$
I assume that $f$ is analytic, strictly increasing on $R$ and maps $R$ onto
itself. The Poincare equation
$$F(\lambda y)=...

answered Nov 19 '17 at 20:46

Alexandre Eremenko

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4

It is possible, indeed.
Assume that you have smooth functions $f:\mathbb{R} \longrightarrow \mathbb{R}^2$ and $g: \mathbb{R} \longrightarrow \mathbb{R}^2$ such that $f(0)=g(0)$. Define $h:\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ by $h(x,y) = f(x) + g(y) - f(0) $. Then, $h$ is clearly smooth and satisfies $h(x,0) = f(x) + g(0) - g(0) = f(x)$ and $h(0,y) = ...

4

Yes, the Hochschild-Konstant-Rosenberg theorem has a converse. More generally you have vanishing characterizations of smoothness in terms of Hochschild homology (one of them is e.g. Avramov, Luchezar L.; Vigué-Poirrier, Micheline, Hochschild homology criteria for smoothness, Internat. Math. Res. Notices 1992, no. 1, 17–25: If $HH_i(R,R)=0=HH_j(R,R)$ for some ...

4

Under your assumptions, "smooth" is equivalent to "flat with smooth fibers". So the only problem is flatness. Use the flatness criterion by fibers (EGA IV, 11.3.10).

answered Sep 1 '13 at 14:00

Laurent Moret-Bailly

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