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3

First the answer for distributions, then tempered distributions: If $f$ is a distribution with $\Delta f=0$ then $f$ is actually a smooth function, which is to say the nullspace of the Laplacian is the space of harmonic functions. Sketch, assuming the domain is $\Bbb R^d$: Fix $\phi\in C^\infty_c$ with $\int\phi=1$, and such that $\phi$ is radial ($\phi(x)$...

2

Hint $\phi$ has compact support and $e^{\frac{x}{n}} \to 1$ uniformly on compact sets. P.S. Note that your integration by parts approach cannot help: Indeed, since $e^{x/n} \to 1$ on $supp(\phi')$, we have $$\int_{\mathbb R} e^{x/n} \phi'(x) dx \to \int_{\mathbb R} \phi'(x) dx =0$$ with the last equality following from the fact that the antiderivative of $\... 2 Let me handle the case$d = 1$for simplicity of notation. Since$H$is a Hilbert space,$F$is represented by an element$g \in H$. Thus, for all$\phi \in \mathcal{D} \subset H$with$\operatorname{supp} \phi = K$we have $$|F(\phi)| = \left| \left<\phi, g \right>_H \right| \leq ||g||_H ||\phi||_H = ||g||_H \left( \int_{\Omega} \phi \cdot \phi + \... 2 The trick is explained here at the end of page 1 : If f is continuous and bounded by C, with$$A = \sup_x |(1+x^2) \phi(x)|$$you get$$\sup_x |(1+x^2) \phi(x)f(x)| \le A C$$i.e.$$|\phi(x)f(x)| \le \frac{AC}{1+x^2}$$and \displaystyle F(\phi) = \int_{-\infty}^\infty f(x) \phi(x) dx is a tempered distribution since$$\left|\int_{-\infty}^\infty f(x) ... 2 What prevents you from using Kullback-Leibler divergence (KL divergence) as a measure of distance from the uniform distribution? I do agree with you on the fact that KL divergence is not a true measure of "distance" because it does not satisfy (a) symmetry, and (b) triangle inequality. Nonetheless, it can serve as a criterion for measuring how far/close a ... 2 Total variation distance, also known as statistical distance, is a good metric (very stringent). (Note that up to a factor$2$, it's equivalent to$\ell_1$distance between the vectors of probabilities.) It also has a nice interpretation in terms of closeness of events.$\ell_2$will be much more forgiving towards small differences, and put the emphasis on ... 2 Let$f\in C^\infty_c(\Omega)$. Then$f$is supported in a compact set$K$and$|f|$attains a maximum$C$in this$K$. Thus $$\int_{\Omega} |f|^p dx = \int_K |f|^p dx \le \int_K C^p dx = \text{Vol}(K) C^p.$$ Thus$f\in L^p$for all$p$. Indeed$C^\infty_c(\Omega)$is dense in$L^p$for all$1\le p <\infty$. 1 A simple counterexample is$f(x)=e^{-\sqrt{|x|}}$(OK, that's not smooth at$0$, but just smooth it out near$0$since all we care about is the behavior for$x$large). Much more generally, suppose$f_0,f_1,f_2,\dots$are any countable collection of Schwartz class functions. Then we can find a Schwartz class function$f$such that for all$n$there exists$...

1

let's take $\Omega = [0,1]$, $\varphi,f \in C^\infty_c((0,1))$. Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$ $\varphi,f$ have their support strictly inside $(0,1)$ so $f'(0)\varphi (0) = f'(1)\varphi (1)= 0$ and $$\langle f'',\varphi \rangle =-\langle f',\... 1 In your formula, \arccos u is not necessarily the principal branch; there's summation over all x_0 such that \cos(x_0)=0, so when considering each such term, we use \arccos(0)=x_0. Your (correct) approach is what should be done, and the remaining computation of second derivative is not hard: you only need it at u=0. Claim:$$\frac{d^2}{du^2}\bigg|...

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No, this map is not injective. Indeed, the inclusion $C^\infty(\Omega)\to \mathcal{D}'(\Omega)$ mapping a function to the regular distribution it generates is injective and the $i$-th partial derivative of a regular distribution generated by some $f\in C^\infty(\Omega)$ is the regular distribution generated by $\partial_i f$. So if $f,g\in C^\infty(\Omega)$ ...

1

You can approximate any function in the Lebesgue space arbitrarily well by a differentiable function. Taking a converging sequence of such approximating functions you can show that $Vf_n$ is Cauchy in the Sobolev space, hence has a limit in the Sobolev space. Since $V$ is continuous as operator into the Lebesgue space the Sobolev limit equals $Vf$

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Since the mapping $$\iota\colon\mathcal{D}\to H$$ is a continuous linear operator (which is the "generalization" of bounded operators to the setting of locally convex spaces), there is a (continuous) transpose $$\iota^t \colon H' \to \mathcal{D}'$$ satisfying $$\langle \varphi, \iota^t(T)\rangle = \langle \iota(\varphi), T\rangle$$ for all \$\varphi\in\...

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