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Special functions, orthogonal polynomials, harmonic analysis, ordinary differential equations (ODE's), differential relations, calculus of variations, approximations, expansions, asymptotics.
23
votes
Accepted
Microwaving Cubes
You can uniformly cook the cube if $f$ is harmonic, i.e. $\Delta f=0$. Note that, if e.g. $\Delta f>0$ everywhere, then the center of the cube will always receive less heat than the average over a sp …
13
votes
Locally complete space is topologically equivalent to a complete space
This is not a reference but a short direct proof.
Let $\bar X$ be the completion of $X$. Define $f:X\to\mathbb R$ by $f(x)=dist(x,\bar X\setminus X)$. Obviously $f$ is continuous, and the local compl …
5
votes
Hidden convexity
It seems that the following interpretation is not covered yet: the domain is allowed to change but $f$ is not required to be proper. In this case the answer is no. Take any connected domain on the pla …
8
votes
Accepted
Vertical Diameter of Convex Domains
Yes. Incidentally, just recently I had to write down a proof of a similar fact in one of my papers. It is quite technical.
Let us work at the endpoint $x=0$. We have to prove that the function $x\map …
3
votes
Is there a non-trivial example for a 1-homogeneous function satisfying a specific inequality...
You can define $A(x_1,x_2,\dots,x_n)=B(x_1,x_2)$ where $B$ is a solution for $n=2$. As for $B$, one can e.g. define it on the unit circle by $B(\sin t,\cos t)=\cos 3t$ and extend by homogeneity.
13
votes
Polynomial positive on an interval
The following extension of Keivan Karai's comment proves the result.
Consider $q(x):=p(x)-c$ where $c=\frac12 \min p([0,1])$.
Approximate $q(x)$ as Keivan suggested: let
$$
q_n(x) = \sum_{i=0}^n q( …
69
votes
How to solve $f(f(x)) = \cos(x)$?
There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of $f$. And f should be injective and hence monotone (increasi …
29
votes
Accepted
Counterexample to Sard's theorem for a non-C1 map
No, such functions do not exist. More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ …
15
votes
Accepted
A generalization of intermediate value theorem on R^k
The statement is true. It is almost precisely Lemma 2 in the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210. The proof is short but not easy to invent. The paper can be read …
8
votes
Accepted
A plausible positivity
The sum $\sum a_n/n$ can be negative. Below I construct a finite sequence; one can always add a negligibly small tail to get infinitely many non-zeroes.
Begin with $a_1=1$ and $a_2=-1$.
This gives $A …
3
votes
Uncountable preimage of every point
Here is a formalization of André Henriques' answer to the Hausdorff dimension variant of the question.
Let $K=\{0,1\}^\infty$ be the standard Cantor set. Define a map $f:K\to[0,1]$ as follows:
for a …
18
votes
5
answers
3k
views
Smoothness of $f(\sqrt x)$
I found that I need to use the following facts in a paper that I am writing.
Let $f\in C^\infty(\mathbb R)$, then
If $f(0)=0$, then $f(x)=x g(x)$ for some $g\in C^\infty(\mathbb R)$.
If $f$ is even …
10
votes
Accepted
Explicit extension of Lipschitz function (Kirszbraun theorem)
I like a recent proof by Akopyan and Tarasov:
A. V. Akopyan, A. S. Tarasov, "A constructive proof of Kirszbraun's
theorem"(Russian), Mat. Zametki 84 (2008), no. 5, 781--784;
translation in Math. Note …
9
votes
Accepted
Growth of the "cube of square root" function
Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.
If $f(n)=n^{3/2}$ without the floor …
4
votes
Accepted
Analogue of an orthogonal subspace in a noneuclidian normed space
Here is a simple proof that the property holds only for Euclidean norms, at least if the norm in question is $C^1$ smooth and strictly convex. Surely it was known way before Gromov was born.
Let $S$ …