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Such a function does not exist. Indeed, let $f\colon[a,b]\to\mathbb R$ be an increasing differentiable function. Then $$f(x)-f(a)=(HK)\int_a^x f'(t)\,dt$$ for all $x$, where $(HK)\int$ is the Henstock …

As written in my paper [1], the inequality $$P(f^*>r) \le 2\exp\big(-r^2/2(p-1)\big) $$ in Theorem 3 in [1] for martingales in $\mathcal{X}=L^p$ can be compared with the inequality $$ P(f^*>r) \le …

If $f=1_{[0,1]}$, then for real $x$ we have $$f^*(x)=\frac1{1-x}1(x<0)+1(0\le x<1).$$ So, for $\lambda=1/2$ the left-hand side is $$\tfrac12\,m(\{x\colon f^*(x)>\tfrac12\})=\tfrac12\,m((-1,1))=1,$$ wh …

Without the assumption that $C$ be closed the answer is no. Indeed, suppose such $C_m$'s exist. Then without loss of generality $|C_m|\le1/3^m$ for all $m$. Let now $C:=\bigcap_{m=1}^\infty ([0,1]\set …

$\newcommand\op\oplus$Apparently, here $\mu$ is the Lebesgue measure. Identify the interval $[0,1)$ with the (say) unit circle in a standard manner. Slightly more elementarily, for $x$ and $K$ in $(0, …

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\ep}{\epsilon} \newcommand{\si}{\sigma} \newcommand …

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\ …

For $x\in(0,1)$, let \begin{equation} f(x):=\frac1{x\ln^2\frac ex}. \tag{1} \end{equation} Then $f\ge0$ and $\int_0^1f(x)\,dx=1$. On the other hand, $\ln f(x)\sim\ln\frac ex$ as $x\downarrow0$. So …

$\newcommand{\Om}{\Omega}\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\pOm}{\partial\Om}$Let $n:=N$. Consider the following "cone" condition: the boundar …

$\newcommand\ep\varepsilon$The conjunction of the following conditions is enough: The $f_n$'s are uniformly bounded: $|f_n|\le M$ for some real $M>0$ and all $n$; $X$ is Polish; $f_n\to f$ uniformly …

$\newcommand\Om\Omega\newcommand\ga\gamma$ I think all you need to do is clarify/formalize the terms you are using. Given a measurable space $(X,E)$, let us say that random variables (r.v.'s) $A_1$ an …

This inequality cannot hold because of the homogeneity matter. Indeed, take any $u$ with $\int u^3\in(0,\infty)$ and $\int u^2+\int(u')^2<\infty$. Then replace $u$ by $cu$ for a real number $c$, and l …

$\newcommand\ep\epsilon$The answer is no. Indeed, assume that $$\int_0^1|f''_\ep(x)|\,dx\le\int_0^1|f_\ep(x)|\,dx\tag{0}$$ for $\ep\in(0,1)$. Let $$M:=\max_{0\le x\le\ep}|f_\ep(x)|.$$ Then $M\ge f_\ep …

Apparently, a relevant source here is H. J. Keisler and A. Tarski, From accessible to inaccessible cardinals. Fundamenta mathematicae, vol. 53 (1964), pp. 225--308, a review of which is given at https …

Direct calculations show that $$(f_2*f_0)(y)=\frac{1}{4} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} \left(y^2+1\right), $$ $$(f_1*f_0)(y)=\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} y, $$ $$(f …