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$\delta$ is NOT a function. $\delta$ is what's called a distribution. A distribution is a "linear functional" $\delta\colon C_c^{\infty}(\Bbb{R})\to\Bbb{R}$. Where $C_c^{\infty}(\Bbb{R})$ are infinitely differentiable functions that vanish outside of some interval. We typically write (although this is absolutely an abuse of notation, this is not really an ...

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The Cantor distribution shows that the singular part of a Radon-Nikodym decomposition need not be a sum of point masses. Let $F: [0,1] \to [0,1]$ be the Cantor function. Since $F$ is monotone increasing, we can construct a measure $\mu$ on the Borel sets of $[0,1]$ by defining $\mu(\varnothing) = 0$, $\mu(\{0\}) = F(0) = 0$, and $\mu((a,b]) = F(b) - F(a)$ ...

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The starting point is that the limit $$\lim_{\delta \to 0}\,\int_{\delta<|x|<1} \frac{1}{|x|}(\phi(x)- \phi(0))\,dx\tag1$$ is finite, because the integrand is bounded by the Mean Value Theorem : $|\phi(x)- \phi(0)|\le |x|\sup |\phi'|$. This tells you what $c_\delta$ should be: it's $-2\log\delta$. To complete the proof that this is a distribution, ...

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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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For $(1)$, to prove that $\mathcal D(U)$ with $\{\| \cdot \|\}_{m \in \mathbb N}$ is a locally convex topological vector space, you need to show that $\{\|\cdot\|_m\}_{m \in \mathbb N}$ is a separating family of semi-norms on $\mathcal D(U)$. The question already says "norms" in it, so maybe you just have to show that it's separating, i.e., given $\phi \in ... 2 The requirement of$\phi$having compact support allows us to interpret every locally integrable function$f$as a distribution, because the integral$\int f\phi$is guaranteed to converge (and have the right kind of continuity properties) as long as$f$is locally integrable. Allowing$\phi$to have unbounded support comes with a tradeoff: some ... 2 Upon request in the comments: There is a large class of distributions which are given by integration against locally integrable functions. Specifically, given a locally integrable$f$and a smooth compactly supported$g$, one can define$T_f(g)=\int_{-\infty}^\infty f(x) g(x) dx$. This leads to a common abuse of notation, where we write the same thing for ... 2 The fundamental solution, as mentioned, satisfies$-u''+k^2 u = \delta_{y}(x)$. To the left or to the right of$y$, the fundamental solution satisfies$-u''+k^2u=0$. The fundamental solution needs to be continuous across$y$, and, in order to have the$\delta$function behavior, there is a discontinuity in the first derivative of the solution with a jump of ... 2 Defining$\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. 2 By definition we have$\displaystyle \langle \varphi, \widehat{\delta} \rangle= \langle \widehat{\varphi}, \delta \rangle = \widehat{\varphi}(0)=(2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{-i x \cdot 0} \varphi(x)dx = (2\pi)^{-n/2} \langle \varphi , 1 \rangle$by arbitrariness of$\varphi$get the identity. Similary$\displaystyle \langle \varphi, \widehat{1} ...

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Using integration by parts we get: $$\lim \limits _{n \to \infty} \int \limits _{-\infty} ^\infty h_n ' (x) \varphi (x) \ \textrm d x = \lim \limits _{n \to \infty} \left( h_n \varphi \Bigg|_{-\infty} ^\infty - \int \limits _{-\infty} ^\infty h_n (x) \varphi' (x) \ \textrm d x \right) = \\ \lim \limits _{n \to \infty} \left( (0 - 0) - \int \limits ... 1 Let us assume that f_k converges to F \in D'(\Omega) in the sense that$$\int_\Omega f_k \, v \, \mathrm{d}x \to F(v) \qquad\forall v \in D(\Omega).$$Now, since W^{1,q}(\Omega) is reflexive, you get f \in W^{1,q}(\Omega) and a subsequence such that f_{n_k} \rightharpoonup f in W^{1,q}(\Omega). In particular, f_{n_k} \rightharpoonup f in ... 1 Using that \mathcal{F}\circ\mathcal{F}(f)(x)=f(-x) and in the comments you mentioned that you know that \hat\delta =1, we get$$ \hat 1 = \hat{\hat \delta} =\delta(-\cdot) =\delta $$with some normalizing coefficient depending on your Fourier transform. 1 Note that in order to show that the function f \ast g is in L^1 we just have to integrate its absolute value over the variable, in this case over x. For the detailed calculation consider \int_{\mathbb{R}^n}| (f \ast g)(x)| dx=\int_{\mathbb{R}^n}| \int_{\mathbb{R}^n} f(y) g(x-y) dy| dx \leq \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(y) g(x-y)| dy ... 1 Typically a "fundamental solution" or "Green's function" is a function that solves the equation -u'' + k^2u = \delta(x). 1 There are so many books on the theory of distributions. Probably it is also useful to use more than one. I see to put some order. To study the topology of the space of test functions and other, you can see in "A First Course in Sobolev Spaces" by G.Leoni, but also in "Functional Analysis" by W.Rudin (here you can find many results on topological vector ... 1$$ \hat f(\nu)=\int_{-\infty}^{\infty}\frac{H(x-1)}{x}\mathrm e^{-i\nu x}\mathrm d x=\int_{1}^{\infty}\frac{\mathrm e^{-i\nu x}}{x}\mathrm d x=\Gamma(0,i\nu)=\mathrm E_1(i\nu)=-\mathrm{ci}(\nu)+i\,\mathrm{si}(\nu) where \Gamma(a,z) is the Incomplete Gamma function, \mathrm E_1(x) is the \mathrm E_n(x)-Function for n=1 and \mathrm{ci}(x) and ... 1 There is a familiar function whose distributional derivative is H: namely, g(x)=(x+|x|)/2. (By the way, the value H(0)=1/2 is irrelevant, since changing the value at one point does not change the distribution at all.) So the problem becomes (xT_f-g)'=0, which implies xT_f-g = c. From here, T_f should be ((|x|+x)/2+c)/x, which is more precisely ... 1 Suppose that x \notin \newcommand{\supp}{\mathrm{supp}}\supp(\mu). Then there is a ball B(x,r) with \mu(B(x,r)) = 0. If \phi is any function compactly supported in B(x,r) then T_\mu(\phi) = 0. Thus B(x,r) \cap \supp(T_\mu) = \emptyset, so that x \notin \supp (T_\mu). On the other hand, suppose that x \notin \supp (T_\mu). Then there ... 1 The author indeed means f=f*h_{1/R}. To achieve this, on the Fourier transform side we want \hat f = \hat f \widehat{h_{1/R}}. And this is possible if \widehat{h_{1/R}} is equal to 1 on the support of \hat f. Choosing h in the way described in the hint achieves that. When differentiating a convolution, we decide where the derivative goes. So, ... 1 A distribution is a continuous linear functional F:\mathscr{D}\to\mathbb{R}. F is well-defined in our case because f is locally integrable and each \phi is compactly supported, and F is linear since integration is linear. Thus it is enough to check continuity. To do this, it is enough to show that for each compact set K there is a real number ... 1 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. 1 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 ... 1 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} = ...

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