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Suppose that the support of $T$ is not compact. Then this support intersects with infinite number of intervals of the form $[n,n+1]$, $n\in\Bbb Z$. Denote the indices of intervals that intersect with the support of $T$ by $n_k$, $k\in \Bbb Z$. Now, $\forall k\in \Bbb Z$ there exist a test function $\phi_k$ with support in $[n_k,n_k+1]$ such that ...

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$u\in H^{-s}$, therefore by definition of $H^{-s}$, $u\in S'$, the space of tempered distributions (I did not realize this fact when asking the question). Therefore $(1+|\xi|^2)^s\bar{\hat{u}}=0$ in the sense of tempered distributions. If we multiply by $(1+|\xi|^2)^-s$, and since multiplying by this function with this crescence rate let S' stable, we find ...

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You have $$\int_{\Bbb R^d} u(x)\phi(x)dx = \int_{\Bbb R^d} \hat u(\xi)\bar{\hat \phi}(\xi)d\xi =\int_{\Bbb R^d} \hat u(\xi) (1+|\xi|^2)^{-s/2} \bar{\hat \phi}(\xi)(1+|\xi|^2)^{s/2}d\xi.$$ Now you have $\hat u(\xi) (1+|\xi|^2)^{-s/2}$ is a $L^2$ function. Take, for example, arbitrary $\phi\in S$, therefore, functions of the form $\bar{\hat ... 1 First note that the integral you've written (that the author assumes is zero) is precisely the inner product$\langle \phi, u\rangle_{H^s}$. Now pick an element$u$in the orthogonal complement of$D$in$H^s$. Our goal is to prove that$u$is zero, and hence$\overline D = H^s$. (This implies the result by the standard fact that for a Hilbert space, ... 1 For your first question, the dirac delta function$\delta(x)$is undefined at$x=0$, you can think of it as being 'infinite' there, but this isn't strictly true. Thus$f(x)$is undefined at$x=\frac{1}{2}$. Your intuition in terms of 'blips' is correct however. For your second question, note that: $$\int_{-\infty}^{+\infty}\delta(x)dx=1,$$ but ... 0 First Hint The first integral can be computed directly using substitution and the fact that$(\arctan x)'= 1/(1+x^2)$. Second Hint For the weak convergence you need to prove that$\int_{-\infty}^\infty f(x)\phi_n(x)\, dx \to f(0)$for any given smooth function$f$with compact support. A common trick is to use the fact that$\phi$integrates to$1$in ... 0$\phi_n(x) = \frac{1}{\pi} \frac{n}{1+(nx)^2} \\ \text{because} \frac{d}{dx} \arctan x = \frac{1}{1+x^2}\text{,}\\ \text{then} \int_{-\infty}^{\infty} \phi_n(x) = \frac{1}{\pi} \arctan x \mid_{-\infty}^{\infty}\\ \text{and because} \lim_{x \rightarrow \pm \infty} \ \arctan x = \pm \frac{\pi}{2}\\ \text{then} \int_{-\infty}^{\infty} \phi_n(x) = \frac{1}{\pi} ...

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Consider, for fixed $s$, the function $$g_s(\tau) = \exp(\tau + s)\sinh \tau$$ we have $$g_s'(\tau) = \exp(\tau +s)(\sinh\tau + \cosh\tau) = \exp(2\tau + s)$$ Hence, $g_s$ is monotone, therefore we have by substitution for $\phi \in C_c^\infty(\def\R{\mathbf R}\R)$: \begin{align*} \int_\R f_1(s,\tau)\phi(\tau)\, d\tau &= \int_\R ...

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I'll use a measure theoretic approach (as you are clearly not interested in densities): Let $\mu$ be a probability measure on $\mathbb{R}$ and assume that $$\int x^2 d\mu(x)=0$$ This implies that the integrand $x^2=0$ $\mu$-almost everywhere. But because $x^2\neq 0$ on $\mathbb{R}\setminus \{0\}$ we must have $\mu(\mathbb{R}\setminus \{0\})=0$. Thus, all ...

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This is an interesting question as it delves into a more general distributional version of the change of variables formulas for integration of functions. This exact question is answered by Theorem 6.1.3 of Hormander's Analysis of Linear Partial Differential Operators, Vol. 1. For convenience, I'll reprint it here: Theorem 6.1.5 (Hormander Vol. 1): If $g$ ...

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A function in $\mathcal{S}$ decays faster than the reciprocal of any polynomial at infinity and is bounded. (This is usually built into the definition.) So its magnitude is bounded by some $M_1$ on the ball of some radius $R$ centered at the origin, and by $M_2/|x|^N$ outside this ball. Choose $N$ sufficiently large (depending on the dimension and $p$) to ...

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Recall: If $f \in L^1_{\def\loc{{\operatorname{loc}}}\loc}(\Omega)$, we define the corresponding distribution $T_f \colon \mathcal D(\Omega) \to \mathbf C$ by $$\def\<#1>{\left<#1\right>}\<T_f, \phi> = \int_\Omega f\phi\, dx$$ If now $f \in L^1_\loc(\Omega)$ is given, the question wants you to show that $\tilde T_f = T_{\tilde f}$, using ...

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From formula (1.15) in Lemma 1.1, you have $$v(t) = v_0 + \int_0^t v'(s) \, \mathrm{d}s.$$ Hence, $$\|v(t)\|_B \le \|v_0\|_B + \| v' \|_{L^1(0,T;B)}.$$ Similarly, you can estimate $\|v_0\|_B$ by using $\|v\|_{L^1(0,T;B)}$. In fact (by using a slightly different proof), one even has $$\|v\|_{L^\infty(0,T;B)} \le T^{-1} \, \|v\|_{L^1(0,T;B)} + ... 1 I am going to gloss over the subtleties of the dirac-delta function which isn't quite a function and get straight to the essential problem: Theorem: if \int_a^b f(x)\,dx = \int_a^b g(x)\,dx, then f and g are equal (almost everywhere). This is what I want to convince you is generally true. In order to do so, I'm going to begin with the following ... 2 The expression " f(x)\delta(x-x_0) = f(x_0)\delta(x-x_0) " is a shorthand. It doesn't mean (in any ordinary sense) that If g and h are functions g(x) = f(x)\delta(x-x_0), h(x) = f(x_0)\delta(x-x_0) then g = h in that g(x) = h(x) for all x. To start, neither g and h are functions \mathbb R \to \mathbb R, because the delta ... 1 The intuitive way to understand this is with the "informal" definition of the dirac delta: It's a "function" (Really a generalized function/distribution) with the property that it is 0 everywhere except at 0, and a sort of "infinity" at 0. The specifics of the definition of the infinity is such that$$\int _a ^b \delta (x)dx=1$$if and only if$$o\in ...

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You can show that $$\int_a^b f(x)\delta(x-x_0) dx = \int_a^b f(x_0)\delta(x-x_0) dx$$ For all real $a$ and $b$. If $a< x_0 <b$ you get $f(x_0)$ at each side. Otherwise $0$

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By definition we can write for any test function $\phi$ $$\langle x\delta_0,\phi(x)\rangle = \langle \delta_0, x\phi(x)\rangle = 0\cdot \phi(0)=0=\langle 0,\phi(x)\rangle,$$ which means that $x\delta_0=0$ in the sense of distributions. In the same spirit, we write $$\langle x\delta'_0,\phi(x)\rangle = \langle \delta_0', x\phi(x)\rangle =- \langle ... 1 For a distribution u and polynomials P,Q, we define the distribution P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u as$$\tag{1} \left(P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u\right) [\varphi]~:=~ u\left[ Q(-\partial)P(M_{{\rm id}_{\mathbb{R}}})\varphi\right], $$where \varphi is a test function; where M_f(g):=fg denotes the multiplication operator ... 0 Given a test function \phi, the goal is to rewrite -\int |x|\phi'(x)\,dx so that it has \phi in it instead of \phi'. Split into two integrals over positive and negative half-axes; then integrate by parts. The result:$$ \int_{-\infty}^0 x\phi'(x)\,dx - \int_0^{\infty} x\phi'(x)\,dx = -\int_{-\infty}^0 \phi(x)\,dx + \int_0^{\infty} \phi(x)\,dx = ...

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You are using that $f$ is uniform continuous when restricted to each open $W$ so that $W\subset\subset U$. Then for all $\delta >0$, there is $\epsilon_0 >0$ so that $$|f(x) - f(y)| <\delta$$ whenever $x, y\in W$ and $|x-y| < \epsilon_0$. Now if $x\in V$. Let $\epsilon_0$ be small so that $|y-x|<\epsilon_0 \Rightarrow y\in W$ (it is in ...

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You could use the equivalent definition of the Schwartz space: $$f\in\mathcal{S}(\mathbb{R})\iff \sup_{x\in\mathbb{R}}|x^iD^jf(x)|<\infty\iff\sup_{x\in\mathbb{R}}|(1+|x|)^nD^kf(x)|<\infty$$For $i,j,k\in\mathbb{N}_0$, $n\geq k$. See here or Folland's Real Analysis. From this it follows readily. Alternatively, ...

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This is not true in the stated generality: if the functions $\psi_k$ do something crazy in the annulus $2^{-k-1}<|x|<2^{-k}$ (e.g., if $\int \psi_k\to\infty$), the sequence $\psi_ku$ need not converge at all. You need some assumptions to ensure the existence of distributional limit. Instead of inventing assumptions, I'll just assume that the limit ...

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Doesn't this follow from definition? $f\in\mathcal{S}(\mathbb{R}) \iff \sup_{x\in \mathbb{R}}|x^jD^kf(x)|<\infty$ for all $j,k\in \mathbb{N}$. So in particular $j=k=0$ implies $f$ is bounded.

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The constant $\pi$ is to normalize the area under the curve to unity. $$\int_{-\infty}^{\infty}{\frac{y}{y^2+x^2}}dx = \pi$$

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Let $f \in L^\infty_{loc}(U)$, and we take $f$ such that its norm is given by its supremum instead of its essential supremum. Now, we have for $x\in U_\epsilon$ $$|f^\varepsilon(x)| \le \int_{B(0,\varepsilon)} | \eta_\varepsilon(y) f(x-y) | \textrm{d}y \le \sup_{z \in B(x, \varepsilon) }|f(z)|\int_{B(0,\varepsilon)} | \eta_\varepsilon(y)| \textrm{d}y = ... 1 By definition, \eta is supported in the unit ball \overline{B(0,1)}, i.e. \eta(x) = 0 for x \in \mathbb{R}^N \setminus \overline{B(0,1)}. Again by definition,$$\eta_{\epsilon}(x) = \epsilon^{-N}\eta(x\epsilon^{-1}).\tag 1 Now I claim that $\eta_{\epsilon}(x) = 0$ for $x \in \mathbb{R}^N \setminus \overline{B(0,\epsilon)}$. Indeed, fix such a $x$, ...

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Note that $C(z) = \frac{1}{z-\zeta}$ is not holomorphic at $z=\zeta$, so you can't quite say that $\partial_{\bar z} C = 0$. Also, you want the integral to give $f(z)$, not just for holomorphic functions (otherwise you can't say that $\partial_{\bar z} C = \delta$. The easiest route is probably through Cauchy's integral formula for $C^1$ functions (whose ...

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