22

Integrate by parts:
\begin{align}
\int_x^{x+1}\sin(e^t)dt
& =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-2t}d\sin e^{t}\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-e^{-2(x+1)}\sin e^{x+1}\\
& \hphantom{={}}+e^{-2x}\sin e^x+2\...

16

The answer is no. This is because, if $f: \mathbb R \rightarrow \mathbb R$ is a continuous, nowhere differentiable function, then $f \!\restriction\! Q$ is nowhere differentiable for any dense $Q \subseteq \mathbb R$.
To see this, fix $x \in \mathbb R$ and, aiming for a contradiction, let us suppose $f \!\restriction\! Q$ is differentiable at $x$, say with ...

11

There are elementary necessary conditions on real functions to be the derivative of a real function. For instance, Darboux's theorem states that a derivative must satisfy the intermediate value theorem. The Baire category theorem implies that a derivative is continuous on a dense set. It allows to find functions with no antiderivative easily.
There is also ...

10

This problem is related to the higher dimensional multiplication table problem (specifically, the three dimensional version). From the work of Tenenbaum, Ford, and Koukoulopoulos this is now well understood in many cases. From their work, one can show that the number of $n$ up to $x$ with triangular divisors is
$$
\ll \frac{x}{(\log x)^{\alpha}}(\log \...

8

Permit me to include this nice image from Day & Li
to illustrate @Igor's point that "in general on a surface, it [the cut locus] is a graph not a tree."
The source point $p$ is on the cat's forehead,
the other side in this rear-view.
Dey, Tamal K., and Kuiyu Li. "Cut locus and topology from surface point data." In ...

7

It can't be done. Here's a counterexample with $n = 2$ and $m = 1$.
For each $k \in \mathbb{N}$, define a function $f_k: [0,1] \to [0,1]$ by linearly interpolating between the values $f_k(\frac{i}{k}) = \frac{i}{k}$ ($0 \leq i \leq k$) and $f_k(\frac{i}{k} + \frac{1}{k^2}) = \frac{i+1}{k}$ ($0 \leq i \leq k -1$). They look like staircases.
I claim there is ...

7

Here is a method that will allow one to find the exact upper and lower bounds on $g(z)$ over $z>0$ with any degree of accuracy.
Take any real $z>0$. Since
\begin{equation*}
\frac1y=\int_0^\infty dt\,e^{-y t}
\end{equation*}
for any real $y>0$, we have
\begin{align*}
\frac{g(z)}z
&=\int_z^{e z} dy\, \frac{\sin y}y \\
&=\int_0^\infty ...

5

In view of the answer by Carlo Beenakker and the comment by Alexandre Eremenko, it appears that what you actually had in mind is the following question:
By the mean value theorem, for each $t\in(0,1]$ there is some $\xi_t\in(1,e)$ such that
\begin{equation*}
r(t):=\frac{\sin et-\sin t}{(e-1)t}=\cos(\xi_t t). \tag{2}
\end{equation*}
(Since $\cos u$ ...

4

this is a plot of $\frac{\sin e\,t-\sin t}{e\,t-t}-\cos\xi \,t$ as a function of $\xi$ for $t=1.1$; the curve does not cross zero in the interval $[\tau,e]$, so I would conclude that (1) does not hold.

4

$\newcommand\ep{\epsilon}$$\newcommand\de{\delta}$We shall be assuming that $\ep\in(0,1/e]$. Note that $l(x):=(\ln x)/x$ is decreasing in $x\ge e$.
So, for $x\ge e$ we have
$$(\ln x)/x\le\ep\iff x\ge x_\ep,$$
where $x_\ep\in[e,\infty)$ is the root of the equation
$$l(x_\ep)=\ep.$$
Letting
$$y:=y_\ep:=\frac1\ep\,\ln\frac1\ep\ge e,$$
we have
$$l(y)=\ep\...

4

If you want the smallest, try
$$x = -LambertW(-\epsilon)/\epsilon = 1+\epsilon+{\frac{3}{2}}{\epsilon}^{2}+{\frac{8}{3}}{\epsilon}^{3}+{\frac{125}{24}}{\epsilon}^{4}
+O \left( {\epsilon}^{5} \right)
$$

4

Of course not, because this equation is far from being elliptic. Actually, it is even under-determined, in the sense that you have only one equation, for $n^2$ unknowns (where the matrix is $n\times n$).
Let me however give you a result in this direction, that I discovered two years ago, which has important consequences in various domains.
Let $A$ be ...

4

The Mehta integral is
$$M_n(\gamma):=E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}
=\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$
So, your fraction under the limit sign is
$$\frac{M_n(1/(2n))}{M_{n-1}(1/(2n)}=\frac{\Gamma(3/2)}{\Gamma(1+1/(2n))}\to\Gamma(3/2)\approx0.886227.$$

4

Suppose you have a power series with coefficients $a_n$
$$ f(z):= \sum_{k=1}^\infty a_k z^k .$$
Then the coefficients of $f^2$ are exactly $c_n$. Also if we denote by $\odot$ the Hadamard multiplication of powerseries (coefficient-wise or equivalently convolution of the boundary values), $c_k^2$ are the coefficients of $f^2\odot f^2$. We apply first Hardy's ...

4

First assume that $E$ is compact. Then, your inequality says that you can approximate it from above by a sequence $E_n$ of convex polytopes with decreasing perimeters. Then, the sequence $\mu_{E_n}$ is weak*-precompact by Banach$-$Alaoglu theorem, so we can assume by passing to a subsequence that $\mu_{E_n}\to \mu$. Hence
$$
\int_E \mathrm{div}\,T=\lim_{n\...

4

Your operator $K$ is a Hilbert-Schmidt operator since its kernel belongs to $L^2$. As a result this is a compact operator whose spectrum contains a sequence of eigenvalues $\\{\lambda_k\not=0\\}$ with finite multiplicities such that $\lim_k\vert \lambda_k\vert=0$. To deal with the self-adjoint case, you can find an orthonormal set $\\{\mathbf e_k\\}$ such ...

3

Most antiderivative of $g$ are not in $L^1(\mathbb{R})$, in your case only one antiderivative $G_0$ will be in $L^1(\mathbb{R})$, the one actually equal to $f$. All the other antiderivatives $G$ are equal to $G_0 + c$, with $c \neq 0$, which is not in $L^1(\mathbb{R})$.
To proof that there exist one antiderivative $G_0$ in $L^1(\mathbb{R})$.
You start by ...

3

For instance, no equivalent distance on $\mathbb{R}^1$ can make Lipschitz the Cantor function $f:[0,1]\to\mathbb{R}^1$.
Suppose by contradiction $d$ is a distance that makes $f$ $L$-Lip.
Let $F_n$ be the closed sets in the usual construction of the cantor set $C=\bigcap_{n\ge\mathbb{N}}F_n$, that is $F_0=[0,1]$ and $F_{n+1}=\frac{1}{3}F_n \cup (\frac{1}{...

3

It is indeed elementary with some slight maneuvering:
Since $s' < r' \leq r$, there exists $\alpha \in (0,1]$ such that \begin{equation}\|f\|_{r'} \lesssim \|f\|_{s'}^{1-\alpha} \, \|f\|_r^\alpha.\label{1}\tag{1}\end{equation}
Hence (by the $S^\theta_{r,s}$ assumption)
\begin{equation}\|f\|_{r'} \lesssim \|f\|_{s'}^{1-\alpha} \, \|f\|_s^{\alpha(1-\theta)}...

3

The ABP estimate indeed holds in your setting. The key is that the concave envelope of $u$ is in $C^{1,\,1}$, so the area formula is valid for its gradient. Assuming for simplicity that $L = \Delta$, that $\Omega = B_1$ and that $\sup_{\partial B_1} u = 0$, the way I would argue is:
Let $\Gamma$ be the concave envelope (the infimum of linear functions ...

2

$\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb{Z}}$
We should, and will, assume that
\begin{equation}
\int_p^{p+1/n}f_n=\int_p^{p+1/n}f \tag{0}
\end{equation}
for all $n\in\N$ and all $p\in\Z/n$. (If (0) is assumed only for $p\in\N/n$, then the conclusion will obviously be false in general.)
The key here is
Lemma 1: If ...

1

Here is an example:
Example. Let $g$ be a continuous strictly increasing function such that $\lim_{x \to -\infty} g(x) = -1$ and $\lim_{x \to \infty} g(x) = 1$; for example
$$g(x) = \tfrac{2}{\pi}\arctan(x).$$
For $n \in \mathbf Z$, let $f_n(x) = n$ and $g_n(x) = g(x) + n + \tfrac{1}{2}$. Then $\{f_n\} \cup \{g_n\}$ form a covering: if $h$ is continuous and ...

1

I realise I was too optimistic in my comment: no such ineqaulity holds in general. Indeed, consider $f(x) = \prod_j f_j(x_j)$. Then
$$ K_{(\alpha_1,\ldots,\alpha_d)} f(x) = \prod_j K_{\alpha_j} f_j(x_j) $$
and so
$$ \|f\|_p = \prod_j \|f_j\|_p , \qquad \|K_{(\alpha_1,\ldots,\alpha_d)} f\|_q = \prod_j \|K_{\alpha_j} f_j\|_q $$
So by the usual Hardyâ€“Littlewoodâ€”...

1

$\newcommand{\tb}{\tilde B}$
Let $d:=n$.
The dispersion condition
\begin{equation*}
\lim_{r\downarrow0}\frac{|E\cap B_r(x)|}{|B_r(x)|}=0
\end{equation*}
is of no help, where $|\cdot|$ denotes the Lebesgue measure on $\mathbb R^d$.
More specifically, the following is true:
Theorem Suppose that $f$ is a nonnegative function in $L^1(B_1)$ such that
\...

1

Consider all pairs $(k,k')$ with $\frac{n}{4}\leq k\leq\frac{n}{2}\leq k'\leq\frac{3n}{4}$. There are $\left(\frac{n}{4}\right)^2$ such pairs and for them $\frac{k}{k'}\geq\frac{1}{3}$. So the sum is bounded from below by
$\left(\frac{n}{4}\right)^2\left(\frac{1}{3}\right)^\alpha$,
and thus as $n\to\infty$ is not $O(n^{2-\epsilon})$ for any $\epsilon>0$.

1

You get almost everywhere convergence for both first and second derivatives. In general, no uniform convergence can be expected. Take for example $\Omega=(-1,1)\subset\mathbb{R}$ and $u(x)=|x|$. Then $Du=\frac{x}{|x|}$ is discontinuous and $D^2u=2\delta_0$ in the sense of measures. On the other hand, both $Du_k$ and $D^2u_k$ are continuous functions, hence ...

1

Counterexample: $$\eqalign{g(x) & = 0\cr f(x,\epsilon) &= \cases{1 & for $x \ge \epsilon$\cr
0 & for $x \le \epsilon/2$\cr}}$$
with a smooth interpolation for $\epsilon/2 < x < \epsilon$. Then $f'(x,\epsilon) \to 0 = g'(x)$ pointwise as $\epsilon \to 0+$, $f(0,\epsilon) = g(0)$, but of course $\...

1

The numerator and denominator can also be written as
$$E\left[\exp\left(\frac1n\sum\ln|X_i-X_j|\right)\right]$$
where the numerator has $n(n-1)/2$ summands and the denominator has $(n-1)(n-2)/2$ summands.
Let $\mu$ and $\sigma$ be the mean and standard deviation of $\ln|X_i-X_j|$. Since $X_i-X_j$ is a normal distribution with mean $0$ and standard ...

1

It appears that what you need is the tensor Bezoutian for operator polynomials. Its definition and relation to the counting of the common eigenvalues is briefly reviewed in the following article (Theorem 9), where also references are given for further details:
Lancaster, Peter, Common eigenvalues, divisors, and multiples of matrix polynomials: A review, ...

1

Any distribution $T$ on the real line has an anti-derivative, i.e. there exists a distribution $S$ such that $$S'=T\tag{$\ast$}.$$ Here is a constructive proof: with a given $T$,
define the distribution $S$ by
$$
\langle S, \phi\rangle_{\mathscr D',\mathscr D }=-\langle T, \psi_\phi\rangle_{\mathscr D',\mathscr D },
\quad \text{with}\quad (\psi_\phi)(x)=\...

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