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1

The behaviour of $f(\phi)$ is easiest to control when only one of the terms in the defining sum can be nonvanishing. To this end, for every $k\in \mathbb{N}$ we define $K_k := [k-1,k+1]$. For $\phi \in D(K_k)$ we then have $f(\phi) = \phi^{(k)}(k)$. It is straightforward to see that the order of the restriction of $f$ to $D(K_k)$ is at most $k$, and when we ...


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You can try induction over m, starting with a j-th power of a bump-function. This yields a sequence of bounded functions that has unbounded first derivatives. From $m\to m+1$ this is a bit more difficult.


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Let $C=\max\{\|L\|_{\infty}^2,1\}$. Assuming that $\alpha<\frac12$, we may compute \begin{align} \int_0^1\int_0^1|k(x,y)|^2\ \mathsf dy\ \mathsf dx &= \int_0^1\int_0^1 L(x,y)^2|x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\\\ &\leqslant C \int_0^1\int_0^1 |x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\ &= 2C\int_0^1\int_0^x (x-y)^{-2\alpha}\ \mathsf ...


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Take a positive, nonzero $u \in C^{\infty}_c(\mathbb{R}^d)$ and consider $\{u_{\epsilon}\}$, where $u_{\epsilon}(x) = u(x\epsilon^{-1})$. For simplicity, let $s = 1$ and notice that (by a change of variables) $$\|u_{\epsilon}\|_{L^{2^*}} = \Big(\int |u_{\epsilon}|^{2^*}\Big)^{\frac{1}{2^*}} = \epsilon^{\frac{d}{2^*}}\|u\|_{L^{2^*}}$$ $$\|\nabla u\|_{L^2} = ...


2

Defining $\phi_\lambda(x)=\phi(\lambda x)$ for smooth $\phi$, the requirement is $$ u(\phi_{\lambda})=\lambda^{-m-d}u(\phi)\quad\forall\phi\in C_0^{\infty}. $$ If $u$ happens to be a continuous function (and hence $u(\phi)=\int u(x)\phi(x)$), this is equivalent to what you wrote.


0

Your computations are exact but you can bypass them. In fact, there is a general result that says that, till the $n$th derivative (including it), a $C^n$ function has identical distributional derivatives and ordinary derivatives. In this case, where in fact, we deal with a $C^{\infty}$ function ($\cos(|x-2)|)$ is identical with $\cos(x-2)$), there will be ...


2

The first result you need is the following: (i) If $u \in \mathcal{D}'(\mathbb{R}^n)$ and $\varphi \in \mathcal{D}(\mathbb{R}^n)$, then $u \ast \varphi \in \mathcal{E}(\mathbb{R}^n)$ and $(u \ast \varphi) \ast \psi = u \ast (\varphi \ast \psi)$ $\forall \psi \in C^{\infty}_c(\mathbb{R}^n)$ We show that $\mathcal{D}(\mathbb{R}^n)$ is dense in ...


2

$\int \rho_{n,\epsilon}(x)dx = \int \rho_{\epsilon}(x)dx=\epsilon^{d}\int\rho(x)dx$ and the latter integral is considered of limit $\frac{1}{2^{j}}\searrow 0$ of Riemann summation of which the partition is equilateral cube of length $\frac{1}{2^{j}}$. That is, $$\int \rho(x)dx=lim_{j\searrow ...


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First of all, this depends on the topologies, I assume you mean the weak$^*$ topologies $\sigma(\mathscr S',\mathscr S)$ and $\sigma(\mathscr D',\mathscr D)$. Note that $r$ has dense range (it is the transposed of the inclusion $\mathscr D \hookrightarrow \mathscr S$) and $r$ is injective (because the inclusion $\mathscr D \hookrightarrow \mathscr S$ has ...


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In this case the Haar measure is simply the Lebesgue measure. So you have $$ \hat \mu(\varepsilon)=\int_{S^2} e^{-i\varepsilon x}dx. $$


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Hint: Remember that $\varphi \in S(\Bbb{R})$ or $\varphi \in C_c^{\infty}(\Bbb{R})$. Either way $\lim\limits_{x\to\pm\infty} x\varphi(x)=0$


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The integral equals $\epsilon^n\widehat{f}(\epsilon z) e^{-ia\epsilon z}$, and thus the sum can be viewed as a Riemann sum for $\int_{\mathbb R^n}\widehat{f}(t)e^{-iat}\, dt=f(a)$.


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You have answered almost everything by yourself. Let $\tau$ be the usual topology on $\mathcal{D}$ and $\sigma$ the subspace topology of $\mathcal{S}$ on $C_c^\infty$. You are asking for $\tau=\sigma$. You constructed a sequence in $\mathcal{D}$ converging with respect to $\sigma$ but not with respect to $\tau$. This shows $\tau\nsubseteq\sigma$. To see ...


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I think the following books are useful and very interesting. A Guide to Distribution Theory by Strichartz, Introduction to the Theory of Distributions, by Friedlander, Distributions Theory and Applications, by Duistermaat and Kolk. Is this for your own interest or for your thesis? Or both? Just curious.


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I wonder if the point is that the distribution can be chosen to have compact support. This is true if you consider the functional as acting on functions whose support is contained in a fixed bounded set.


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Does Rudin really use the notation $C^\infty_{loc}$? This makes little sense to me; I can't imagine how $C^\infty_{loc}$ could be anything other than $C^\infty$. Anyway, hint: The topology is uniform convergence of all derivatives on compact sets. That is to say, the topology induced by the seminorms $$\rho_{K,N}(f)=\sup_{x\in K}\sum_{|\alpha|\le ...



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