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4

Intuitively, because there a bump functions $\varphi$ with $\varphi(0) = 1$ but which are zero except on an arbitrary small interval $[-\epsilon,\epsilon]$ around $0$. That, however, is incompatible with the requirement that for some fixed $u$, $$ \int_{\mathbb{R}^n} u(x)\varphi(x) \,dx = \varphi(0) = 1 $$ essentially because you can make that integral ...


3

Proof by @fgp is very nice. Here it's alternative proof: Suppose that exist function $u(x) \in L^1_{\text{loc}}(\mathbb{R}^n)$ such that $$\int_{\mathbb{R}^n} u(x)\varphi(x)\, dx = \varphi(0), \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}^n).$$ If $\varphi \in \mathcal{D}(\mathbb{R}^n)$ then also $x_1 \varphi(x)=x_1 ...


3

I think this is a great question. This is the direction I would try. So I presume that you are able to prove this if $f$ is compactly supported (since multiplication by $f$ would then be a continuous operation from $\mathcal S \to \mathcal S$). Secondly, I presume that there is a theorem that says if $\mu, \nu \in \mathcal S'$, and $\mu(\phi) = \nu(\phi)$ ...


3

Define $$T_\varepsilon(\varphi):=\int_{-\infty}^{+\infty}\frac{\varepsilon}{\varepsilon^2+x^2}\varphi(x)\mathrm dx.$$ Then, after the substitution $x=\varepsilon t$, we obtain $$T_\varepsilon(\varphi)=\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\varphi(t\varepsilon)\mathrm dt.$$ Using dominated convergence, we obtain that for each $\varphi\in\mathcal D(\mathbf ...


2

Obviously, the first convergence implies the second. If the supports of all $v_k$ are in the same compact set and uniform convergence on this set is given, the convergence in $\mathcal{S}$ follows immidiately as the polynomials are bounded on this set. The second convergence does not imply the first in general, an example are smoothly cut-off $e^{-x^2}$, ...


2

Without doing any complex integration, we can easily see that $$\lim_{\epsilon \to 0} \frac{1}{x + i\epsilon} = \lim_{\epsilon \to 0} \frac{x - i\epsilon}{x^2 + \epsilon^2} = \lim_{\epsilon \to 0} \frac{x }{x^2 + \epsilon^2} -i \lim_{\epsilon \to 0} \frac{\epsilon}{x^2 + \epsilon^2} $$ Applying the limit to the first term gives $\frac{1}{x}$, while the ...


2

Well, to answer your question: let $$f:\Bbb R^n\to \Bbb R, \quad \text{continuous},\quad \lim_{|x|\to\infty}f(x)=0.$$ Fix $\varepsilon >0$; then by definition of the limit there exists $R>0$ such that $|f(x)|<\varepsilon$ whenever $|x|>R$. On the compact set $\{|x|\le R\}$ the function $|f|$ is continuous, hence attains its supremum and infimum ...


2

What is probably meant is that the derivative of $f$ with respect to $x^j$ in sense of distributions can be represented by an $\mathbb L^\infty$ function. This means that there exists a function $g\in \mathbb L^\infty$ such that for any $\varphi\in\mathcal DV(\Omega)$, $$-\int_\Omega f(x)\partial _j \varphi(x)\mathrm dx=\int_{\Omega}g(x)\varphi(x)\mathrm ...


2

It's the standard "continuous implies bounded" argument: Let \begin{equation} \Vert\phi\Vert_N:= \sum_{|\alpha|,|\beta|\leq N} \sup |x^\alpha\partial^\beta \phi|. \end{equation} If your inequality fails then for every $N$ then there is a $0\neq\phi_N\in\mathscr S$ s.t. $$|\langle u,\phi_N\rangle|\geq N\Vert\phi_N\Vert_N.$$ Let $\psi_N=\phi_N/(N ...


1

We will answer negatively to this question. Note that the elements in $A(\Bbb{R})$ are all uniformly continuous functions. So, to find a counter example, we only need to consider some discontinuous function $f$. Let $f(x)={\bf 1}_{[-1,1]}(x)$ then it is easy to see that $f\in X_p$ for every $p\in(1,+\infty]$ but $f\notin A(R)$. So $X_p\not\subset ...


1

$\nabla f$ is the gradient, $\nabla^2 f$ is the Hessian matrix $f_{ij}$, in general $\nabla ^j f$ is the "tensor" with entries $f_{i_1\cdots i_j}$, where I used lower indices to mean partial derivatives. So in that passage it really mean $$||f||_{C^k} = \sum_{j=1}^k \sum_{i_1, \cdots i_j}\sup_{x\in \mathbb R^d} |f_{i_1\cdots i_j}(x)|$$


1

In the finite-dimensional case, there's only one topological vector space in each dimension (up to continuous isomorphism), so there's no ambiguity about what's meant by an "$n$-dimensional manifold," at least in the topological sense. But in the infinite-dimensional case, things are very different. There isn't one definition of "infinite-dimensional ...



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