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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\...

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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 ...

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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 ...

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

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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 Hardy–Littlewood—...

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 \...

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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 $\epsilon>0$.

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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 ...

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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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