23
votes

Accepted

### Writing a function on $\mathbb{R}$ as a sum of two injections

The answer is yes. Every function on the reals is the sum of two injective functions, and this can be done in a highly effective manner, constructing the two functions $g,h$ from $f$ without any need ...

13
votes

Accepted

### "Simple" integral equation

That is a rather tough puzzle (took me two full days) with a rather short solution.
The first step is the differential equation Fred already mentioned:
$$
(1-z^2)H'(z)-(1+z)H(z)+2zH(z^2)=0\,.
$$
Now ...

12
votes

### Writing a function on $\mathbb{R}$ as a sum of two injections

It works at least for (locally) absolutely continuous functions.
Such a function is the integral of a locally $L^1$ function.
This weak derivative can be written as a sum of a positive and negative ...

11
votes

### How to show that $\log 2(1/2\log 2\log 4 + 1/3\log 3\log 6 + \dotsb) + 1/2\log 2 - 1/3\log 3 + 1/4\log 4 - \dotsb = 1/\log 2$

Observe that the left-hand side is the sum of two convergent series. Let $N\geq 2$ be an integer tending to infinity. Truncate the first series at the $N$-th term and the second series at the $2N$-th ...

8
votes

Accepted

### Are “most” bounded derivatives not Riemann integrable?

In 1977 Clifford E. Weil showed that $A$ is a first Baire category set (i.e. a meager set) in $X$ (sup norm) -- see The space of bounded derivatives. So the situation, at least with respect to one ...

6
votes

### "Simple" integral equation

Alternative simple proof - integration by parts:
$$
\int_0^{1-a}\frac{H(z)}{1-z}dz=\int_0^{1-a}\frac1{(1-z)^2}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta=
$$
$$
\frac1{1-z}\int_z^1\frac{2\zeta}{1+...

5
votes

Accepted

### $f\in C(B_1)\cap W^{1,2}(B_1\setminus \{f=0\})$ implies $f\in W^{1,2}(B_1)$?

This should follow from the ACL (absolute continuity on lines) characterisation of Sobolev spaces, see for instance, Theorem 4.1.10 here.
Indeed, since $f \in W^{1, 2} (B_1 \setminus \{f = 0\})$, it ...

5
votes

### Possible research directions in analysis?

I find that students often come in with these kinds of ideas, that almost everything is known, it's hard to do anything, the problems are mostly solved. Then they ask how to best position themselves ...

5
votes

Accepted

### A fractional weighted Poincaré inequality

It is not true. Start with a function $u$ which vanishes for $x<0$ and is equal to $1$ for $0 \leq x \leq \frac 12$ and then smooth from $x \geq \frac 12$. The Fourier coefficients behave like $1/n$...

5
votes

### How to get this inequality in Santambrogio's book about optimal transport?

Iosif already pointed out the trivial typo. For the purposes of the argument in that proof (incidentally, it requires that the objects be probability densities on a compact set $\Omega$, not on whole ...

5
votes

Accepted

### How to get this inequality in Santambrogio's book about optimal transport?

$\newcommand\R{\mathbb R}\newcommand\b{\hat\rho(x)}\newcommand\a{\tilde\rho(x)}$This inequality is false in general.
For instance, if $\varepsilon=1/10$, $M=1/10$, and for some $x\in\R^d$ we have $\a=...

5
votes

### Function orthogonal to $|y-x|$ on $[0,1]$ for every $y \in [0,1]$?

No, linear combinations of functions of the form $|x-y|$ are dense in $C([0, 1])$ (because you can approximate any continuous function by the piecewise-linear), so (if $f\in L^1(0,1)$, otherwise the ...

5
votes

Accepted

### On the Riemannian integrability of the bounded derivative

The answer to the question in the body of your post is no, for the reason that if $f' = g$ almost every where and $g$ is Riemann integrable and such that (*) holds, then $f'$ must be Riemann ...

5
votes

Accepted

### Macroscopic sets - a notion of largeness for Lebesgue null sets

By Frostman's lemma, if $E$ is a compact set of positive $\alpha$-Hausdorff content, then there exists a probability Borel measure $\mu$ supported in $E$ such that $\mu(I) \leq c |I|^\alpha $ for ...

4
votes

Accepted

### If all mixed partials of a $C^1$ function exist and are continuous, is the function $C^2$?

No. Let $f(x,y) = x|x|$.
It is easy to check $df = |x| ~dx$ is continuous.
The mixed partial $\partial^2_{xy} f = 0$ exists and is continuous. But the function is not $C^2$.

4
votes

Accepted

### A Gaussian measure $\mu$ on $\mathcal{E}'(S^1)$ by Minlos theorem and its value for Sobolev spaces $H^{\alpha}(S^1)$

$\newcommand\al\alpha\newcommand\EE{\mathcal E}\newcommand\ip[2]{\langle #1,#2\rangle}$The answer is
$$\mu(\EE(S^1))=\mu(H^\al(S^1))=0$$
for all real $\al\ge0$.
Indeed, since $\EE(S^1))\subseteq H^\al(...

4
votes

Accepted

### Can we approximate a Hölder pdf by higher-order Hölder pdf's?

No, take any $p$ such that $p(x)=|x|^\alpha$ in some neighbourhood of the origin.

3
votes

### "Simple" integral equation

This is an incomplete answer, the last step is missing (yet).
We can differentiate the OPs equation to get
\begin{align}\tag{1}\label{eq:1}
(1-z^2) H'(z)-(z+1) H(z)+2 z H(z^2)=0.
\end{align}
The ...

3
votes

### Does convergence in probability implies L^1 convergence in probability density function, for bounded random variables?

No. Let $Y$ be uniform on $[0,1]$ and $X_n$ have density $f_n=1+\sin (2\pi nx)$. Then $X_n\to Y$ in distribution. You can represent them on the same probability space $(0,1)$ (with Lebesgue measure) ...

3
votes

Accepted

### Extending a $C^1$ function on $\mathbb R^n$ to a set of finite $\mathcal H^{n-2}$ measure

The claim holds under the much weaker assumption that the exceptional set satisfies $\mathcal H^{n-1}(E)=0$.
Since the limits of $f$ and $\nabla f$ exist everywhere we get two continuous functions $F\...

3
votes

Accepted

### Does there exist a continuous choice of maximizing balls for the Hardy Littlewood maximal function?

Not in general. You already can find an example in $d = 1$.
Let $f$ be given by the function
$$ f(x) = \begin{cases}
1 & x\in [-1,0] \\
2 & x \in [4,5] \\
0 & \text{otherwise} \end{cases} $...

3
votes

Accepted

### Sufficient conditions for ensuring that a monic polynomial in $\mathbf{Z}[x]$ possesses exclusively simple roots

Discriminant is the resultant
of $f$ and $f'$. In your case, it is simply an integer.
If this integer is $\neq 0$ then $f$ has no multiple root
(and vice versa).

3
votes

### Oscillation functions and similar constructs

Oscillation and related quantities arise in harmonic analysis among other places e.g., probability and ergodic theory. The buzz words to look for are variational and jump inequalities; see this paper. ...

3
votes

Accepted

### Does every real number $r\in [0,1]$ have a rational sequence $q_n\to r$ s.t. $q_n$ has (simplified) denominator $n$?

For each $n$ we set $q_n = \frac{\text{smallest number more than $nr$ coprime to $n$}}n$. Note that the next prime following $nr$ or the next prime following it is coprime to $n$ for large enough $n$ (...

2
votes

Accepted

### Matrices and vectors of intervals

$\newcommand\R{\mathbb R}$Any operation you can define on intervals on the real line, you can define (entry-wise) on any arrays of such intervals.
For any function $f\colon\R^n\to\R$, you can define ...

2
votes

### If a continuous function is differentiable at a point, is it differentiable in some neighborhood around that point?

The answer is negative. Take a bounded nowhere differentiable continuous map $f : \mathbb{R} \to \mathbb{R}$ and consider the map $g(x) = f(x) \cdot x^2$, which is differentiable at $x = 0$, but not ...

2
votes

Accepted

### Equivalent characterization of weak derivative in Bochner space

You can argue by density of $C_0^\infty(0,T)\otimes H$ in $C_0^\infty(0,T;H)$. A reference for that (probably there are more "natural" references but I do not know them) could be Herbert ...

2
votes

Accepted

### Is the Boltzmann entropy continuous in the supremum norm?

The answer is no. For instance, let $d=1$, $\rho(x):=e^{-x}\,1(x>0)$, $$\rho_n(x):=c_n\big(e^{-x}\,1(0<x\le n)+p_n\,1(n<x\le2n)\big),$$
where $c_n:=1/(1-e^{-n}+np_n)$, $p_n\in(0,\infty)$ for ...

2
votes

### Nature of $ \sum_{n \geq 1} \frac{ \cos(n) \sin(n+1) }{n} $

You might try to regularize the sum,
$$S(\alpha)=\sum_{n=1}^\infty \frac{ \cos(\alpha n) \sin(n+1) }{n}=
-\tfrac{1}{4} i \left[e^{-i} \ln \left(1-e^{i (\alpha-1)}\right)+e^{-i} \ln \left(1-e^{-i (\...

1
vote

Accepted

### On the additive property of the subdifferential of lower semicontinuous functions

In part (P3) of Definition 2.1 in the paper you linked, it is also required that $g$ be $\partial$-differentiable at $x$ (meaning that both $\partial g(x)$ and $\partial(-g)(x)$ are nonempty), which ...

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