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This is wrong. Take, for example, $p_\epsilon(x)=\frac1\epsilon \varphi(x/\epsilon)$ and $\varphi(x)=\varphi_0(x-1)-\varphi_0(x+1)$, where $\varphi_0$ is a smooth cap supported on $(-1,1)$. Then $$ \i …

A simple example is given by $$ f(t,x)=\cases{\exp(-(x-t^{-2})^2),&$t\neq 0$,\\0,&$t=0.$} $$ For each fixed $t\neq 0$, $\int f(t,x)\,dx$ is a Gaussian integral equal to $\sqrt{\pi}$, while for $t=0$, …

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) b …

For any $\alpha>0$, put $$f(x)=\begin{cases}-e^{-\frac{1}{x^2}},&x<0\\x^\alpha,&x>0\end{cases}\quad \text{and}\quad F(x)=\begin{cases}x,&x<0\\e^{-\frac{1}{x^2}},&x>0\end{cases},$$ the composition $F\c …

You can always find a map $\Phi$ (unique up to shifts) such that $\Im\mathfrak{m}(\Phi(z)-z)$ is bounded, but, in general, $\Re\mathfrak{e}(\Phi(z)-z)$ may be unbounded. Let $h$ be the bounded harmon …

This is clearly false as stated, since a necessary condition is that $g(x)\to g(\xi)$ as $x\to\xi$ a. e. in the sphere, but if $g$ is merely $L^2$, this may well fail for every point. The convergence …

Consider first the $d=2$ case. Then, $u$ is a real part of an analytic function. We can write $$u(z)=\frac12\sum_{n=0}^{\infty}(a_nz^n+\overline{a}_n\overline{z}^n)$$ and $$\partial_\nu u(z)=\frac12\s …

Here's a complex analysis proof. For $|\theta|\leq\pi,$ we have that $$ F(t)=\frac{\cos(\theta t)\pi}{\sin (\pi t)} $$ is an odd meromorphic function for $t\in\mathbb{C}$ with simple poles at $k\in\ma …

The composition may have points of differentiability. Let $f_0(x)=x$ for $x\geq 0$ and $f_0(x)=2x$ for $x<0$. Let $g_0(x)=2x$ for $x\geq 0$ and $g_0(x)=x$ for $x<0$. Then none of them is differentia …

This is more of a long comment than answer. First, the analogous statement for the first derivative is already non-trivial, although not very difficult, see Aull, Charles E. "The first symmetric deriv …

Both your functions are analytic, and in this situation there is a general approach called Newton-Puiseux series (the results in Wiki are formulated for polynomials, but the theory applies more genera …

It is not clear what you mean by "enclosed by". If this is just the set of points where $f_i(x)>\lambda$, then the answer is "yes", since $$\int_\Omega fdx=\int_0^\infty A(\lambda)d\lambda.$$ This is …

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 $\mu_{E_n}$ i …

On the question of continuity, much more is true: the Green's function is continuous under convergence of domains in the sense of Carathéodory, which is the weakest reasonable topology of planar domai …

One can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$ and any $p>2$. Given a function $f\in L^p(K)$, we can define its Cauchy transform $$ \left(\mat …