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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 ...


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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 $f,g\in C^\infty(\Omega)$ ...


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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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