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4

You are right, the partial derivatives of distributions on higher-dimensional spaces are defined as you surmised, $$\frac{\partial T}{\partial x_i} \colon \varphi \mapsto -T\biggl[\frac{\partial \varphi}{\partial x_i}\biggr],$$ more generally $$D^{\alpha} T \colon \varphi \mapsto (-1)^{\lvert\alpha\rvert} T[D^{\alpha}\varphi]$$ for higher derivatives. ...


4

You have $$ e^{-|x|} = \begin{cases} e^{-x} & \text{if } x>0, \\ e^x & \text{if } x<0, \\ 1 = e^0 = e^{-0} & \text{if } x = 0. \end{cases} $$ The function above is continuous. So $$ \frac d {dx} e^{-|x|} = \begin{cases} -e^{-x} & \text{if } x>0, \\ e^x & \text{if } x<0. \end{cases} $$ This is undefined at $x=0$. Notice the ...


4

Subtracting $1$ (whose Fourier transform is Dirac delta, up to a normalization constant) gives $-1/(1+x^2)$, whose Fourier transform integral does converge, and can be evaluated by residues: a constant multiple of $e^{-|x|}$.


4

A mollifier is a function $f$ that you convolve with another function $g$ to get a function which is "close" to $g$ but "nicer". For instance $f$ might be a general $L^1$ function and $g*f$ might be a smooth, compactly supported approximation to $f$. Really a mollifier is not one function but a sequence, or even sometimes a one-parameter continuous family. ...


3

The ''function'' as it is has no right to be called a distribution, since it has non-summable singularities in the form $1/y$ at $x=\pm1$: therefore I'm going to assume that your teacher understood it as Cauchy principal value: $$ u(x) = PV \frac{x}{x^2-1}. $$ According to your book, a function $f\in L^1_{\text{loc}}(\mathbb R)$ is a tempered distribution in ...


3

Coming from a measure-theoretic point of view, one can write the following: $\int_{-\infty}^{+\infty}\delta(x-a)dx = \int_{\mathbb R} 1 d \mu_a(x)$, where $\mu_a$ denotes the point measure with mass $1$ at $a$ and $0$ everywhere else. Now, if we want to integrate over an interval $I$, we integrate the characteristic function $\chi_I(x)$ over $\mathbb R$: ...


3

Convergence in the sense of distributions is defined as follows: a sequence of distributions $\tau_n\in \mathscr D'(\mathbb R)$ is said to converge to a distribution $\tau\in\mathscr D'(\mathbb R)$ if, for every test function $\varphi\in C^{\infty}_0(\mathbb R)$, there holds $$ \lim_{n\to\infty}\langle \tau_n, \varphi\rangle =\langle \tau, \varphi\rangle, $$ ...


2

If we set $$\phi := \sum_{j=1}^n \alpha_j \phi_j \in \mathcal{D},$$ then by the linearity of the integral $$\alpha_1 \langle W,\phi_1 \rangle + \ldots + \alpha_n \langle W, \phi_n \rangle = \langle W,\phi \rangle.$$ Consequently, it suffices to show that $\langle W,\phi \rangle$ is Gaussian for all $\phi \in \mathcal{D}$. Fix $\phi \in \mathcal{D}$. ...


2

Step 1 : Let us define $$\Gamma(x,y):=\begin{cases}\frac{1}{(2-n)\omega_{n-1}}|x-y|^{2-n} & \text{if }n\geq3 \\ \frac{1}{2\pi}\ln|x-y| & \text{if }n=2 \end{cases}$$ for all $x,y\in\mathbb{R}^n$, $x\neq y$, where $\omega_{n-1}$ is the Lebesgue measure of the unit sphere. We show that $$\Delta_x\Gamma(x,.)=\Delta_y\Gamma(x,.)=\delta_x.$$ Indeed, for ...


2

Remember that $f \in W^{1,1} (\gamma_1)$ means that $f, f' \in L^1 (\gamma_1)$ where $f'$ is understood in the weak sense. Remember also that $f'$ in the Sobolev sense in the same as $f'$ in the distributional sense, therefore let us compute the latter one. Finally, we shall use that $\gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}$. Note, ...


2

There is a "standard" locally convex topology on $\mathcal{D}$ generated by semi-norms (see the answer to your other question), and pretty much every user of generalized stochastic processes uses it (e.g. Gelfand & Vilenkin). It's always better to choose your topology before going on to work with continuity, dual space, etc. Local convexity is about ...


2

The proof is essentially the same, the only difference is that with $\varphi \in \mathscr{S}(\mathbb{R}^3)$ we can't use one fixed outer radius $R$ of the spherical shell over which we integrate, since Schwartz functions generally don't have compact support. But Schwartz functions decay rapidly as $\lVert x\rVert \to \infty$, and thus if we let the radius of ...


2

Since $\phi$ has compact support, there is an $N$ such that $n\ge N\implies T_n=T$. This makes your last argument correct. Another aay too do it is to observe that if the support of $ºphi$ is contained in $[_N,N]$, then $$\begin{align} ...


2

For a linear functional $u$ on normed spaces, continuity is just $||u(x)||\le C||x||$. Sequences don't have to enter into this particular formulation, but continuity defined in terms of sequences can be shown to be equivalent to the condition just stated (e.g., $||u(x_n)-u(x)||=||u(x_n-x)||\le C||x_n-x||$ so $u(x_n)\to u(x)$ if $x_n\to x$ in the norm ...


1

Well, it isn't strange that $\delta$ appears. Generally, the derivative (in the sense of distribution) of a piecewise $C^1$ function $f$, possibly having jumps (discontinuities) at $x_0<x_1<\ldots<x_n$ is $$f'=f'_{\rm class}+\sum_{i=0}^n J_f(x_i)\delta_{x_i}$$ where $J_f(x_i)=f(x_i^+)-f(x_i^-)$ is the jump of $f$ at $x_i$, $\delta_{x_i}$ is a Dirac ...


1

Its the distributional divergence, that is for $v \in L^2(\Omega)$ we have $v = \operatorname{div}u$ iff $$\int_\Omega v \, \varphi \, \mathrm{d}x = -\int_\Omega u \cdot \nabla \varphi \, \mathrm{d}x \quad\forall \varphi \in C_0^\infty(\Omega).$$ To be compared with https://en.wikipedia.org/wiki/Integration_by_parts#Higher_dimensions.


1

For fixed compact $K \subset \def\R{\mathbf R}\R^d$, we topologize $C^\infty(K)$ by the semi-norms $$ \|u\|_{\alpha,K} := \sup_{x \in K} \def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}\abs{D^\alpha u(x)}, \qquad \alpha \in \def\N{\mathbf N}\N^d $$ The usual topology on $\def\D{\mathcal D}\D := C^\infty_c(\R^d)$ is the direct limit topology of the ...


1

Of course, as it was stated above, the equality holds in the sense of distributions. In order to approach this same issue, i.e. showing that the equality follows from the theory of distributions, one can define the derivative of a function-type distribution (a distribution which in fact is a function in $L^1_{\text{loc}}(\mathbb R)$), as the limit, in the ...


1

As it was already explained above, the sequence of function-type distributions, i.e. distributions in $\mathscr D'(\mathbb R)$ which are in fact functions in $L^1_{\text{loc}}(\mathbb R)$, $$ \tau_a=\theta(x)\frac{x^{1-a}}{\Gamma(a)}, \text{ for }a>0, $$ does not converge to $\delta(x)$ as $a\to 0.$ A very quick, informal way to see this is to note that, ...


1

Before starting my answer, let me remark that all your questions concern a general distribution $T$, and isn't really related to the fact that $T=X\chi_E$. In point 2, the usual definition is indeed with any test function (it is a trivial definition: $T=0\Leftrightarrow(\forall\phi\in C^\infty_c, T(\phi)=0)$; then the specificity of $T=X\chi_E$ rely on the ...


1

The only way this can be true for bounded $f$ is when $f \equiv 0$. Letting $a = 0, h = 1$ in (3) gives that $\int_{-\infty}^\infty K(t)dt = \int_{-\infty}^\infty K(-t)dt = M$ for some $M < \infty$. So, making the substitution $u =\frac{a-t}h$, we get: $$\int_{-\infty}^\infty K\left(\frac{a-t}h\right) dt = h\int_{-\infty}^\infty K(u)du = Mh$$ If $f \le ...


1

Yes, as you noted, when $\phi$ is not invertable we have to modify the transformation function. Such as, for instance, when $\phi(x)=x^2$, which is a fold mapping two intervals into one (the negative reals and non-negative reals to the non-negative reals), we modify the transformation formula to account for the fact that we have two "inverse" functions: ...


1

Yes, there is such a tempered distribution. In fact, a unique such tempered distribution. Recall that the Fourier transform is an automorphism of $S(\mathbb{R})$, and the inclusion $\iota \colon D(\mathbb{R}) \hookrightarrow S(\mathbb{R})$ has dense image. Therefore $\hat{D}(\mathbb{R}) = \mathscr{F}(D(\mathbb{R}))$ is a dense subspace of $S(\mathbb{R})$. ...


1

Answer to question 1: Let $\left|\;\cdot\;\right|$ be a norm on $\Phi$. Since $\iota$ is linear, it is continuous if and only if $$\left\|\phi\right\|\le c|\phi|\;\;\;\text{for all }\phi\in\Phi\tag 1$$ for some $c>0$. Since $\Phi$ is a subset of $H$ we can choose $\left|\;\cdot\;\right|$ to be the restriction of $\left\|\;\cdot\;\right\|$ to $\Phi$ and ...


1

Considering the integrand as the Fourier transform of a tempered distribution, it makes sense then to write $$ \int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE = -\frac{\partial^2}{\partial t^2}\int_{1}^{+\infty}\frac{\sqrt{\rho^2-1}}{\rho^2}e^{-imt\rho}d\rho. $$ Now, in the complex plane, consider a of circle with centre at $1$, radius $R$ and arc going from $R$ ...


1

The main issue I see here is that since $X,\varphi$ are $H-$valued, how do we define the product $X \varphi$? It seems you need additional structure to even get past this point (i.e., $H$ needs to be a Banach algebra). Assuming an algebra structure, how are you defining the space $\mathcal{L}^p(\lambda;H)$? In particular, what is the norm of this space? ...


1

Let us consider a distribution $T$ on $\mathbb{R}^n$. One can prove that for every compact $K \subset \mathbb{R}^n$, there exists $N$, $c(\Omega,N)$ such that $$ |\langle T, \phi \rangle| \leq c \lVert \phi \rVert_{C^N(\Omega)} $$ for all $\phi \in \mathcal{D}$. The order of the distribution $T$ is the least such $N$ that is good for all compact sets $K$. ...


1

In general, $\mathcal{D}(\Omega):=\bigcup_{K \in \mathcal{K}(\Omega)} \mathcal{D}_K(\Omega)$ (where $\mathcal{K}(\Omega)$ denotes the union of all compacts set content in a open subset $\Omega \subset \mathbb{R}^n$), Note that $\Omega \subseteq \bigcup_{j \in \mathbb{N}} K_j$, where $\lbrace K_j \rbrace$ is an increasing sequence of compact in $\Omega$. Now ...


1

Yes, the Haar measure on any simply connected nilpotent real Lie group is just the push-forward by the exponential of the Lebesgue measure of the Lie algebra. This amounts to showing that ($\sharp$) on a nilpotent Lie algebra, denoting by $\ast$ the BCH law, for each $y$, the map $R_y:x\mapsto x\ast y$ preserves the measure. (Indeed, this proves that the ...


1

For a fixed $x_0$, define $G_{x_0}=|x-x_0|^{-1}$. Then $\nabla^2 G_{x_0}=0$ for $x \ne x_0$. If $\varphi$ is a compactly supported $C^{\infty}$ function on $\mathbb{R}^3$, then $$ \nabla\cdot(G_{x_0}\nabla\varphi-\varphi \nabla G_{x_0})=G_{x_0}\nabla^2\varphi-\varphi\nabla^2G_{x_0}=G_{x_0}\nabla^2\varphi,\;\;\; x \ne x_0. $$ Integrate and apply the ...



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