New answers tagged

1

Show by induction that all derivatives of $f(t)=\exp(-1/t)$ are of the form $f^{(n)}(t)=p_n(1/t)f(t)$ with polynomials $p_n$.Then conclude that $f$ extended by $0$ for $t\le 0$ is a smooth function and note that your function is a composition of $f$ with a polynomial.


1

After some search on the internet, I found the following explanation, which I think can solve my puzzle: $\int f(x)\delta(x) dx = \int f(0)δ(x) dx = f(0) \int δ(x) dx = f(0)$, since $f(x)δ(x)$ and $f(0)δ(x)$ are the same since they are both zero for every $x \neq 0$, and $\int \delta (x)= 1$ by the property of the Dirac delta function.


0

Actually I have found another solution, more satisfactory to my taste (although maybe not 100% complete). Inspired by the answer of this question on integrating the $\text{sinc}$ function, I thought of the following procedure: Write the exponential as a sum of sines and cosines $$\int_0^\infty\text{d}x\frac{e^{i k ...


0

"In Schwartz's Théorie des distributions, chapter III, Théorème VII : $\mathcal D$ is a Montel space, where bounded sets are relatively compact. Then the weak and strong topologies, restricted to bounded sets, coincide, and convergent sequences are the same in these two topologies (and also in weaker Hausdorff topologies). In more concrete terms: whenever a ...


2

Your functional assigns to every test function the value $0$. Since $$ \mathrm{e}^{-x}x^2\varphi(x)=0, $$ if $x$ is sufficiently large.


1

A quick remark on your solution of (2): You need to say that $\phi = 1$ on $[0;1]$ or something like that, as $\phi$ has to be independent of $n$. I know it's pedantic, but this is how you can loose marks on exams. For (3): you actually want to show that the map $$T(\phi) = \lim_{n \to \infty} \int_{0}^{\frac{1}{n}} n^2f(x)dx - n \delta(\phi)$$ is a ...


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

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

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

We can write the foloowing: take a test function $\psi$ and integrate it against $\nabla \phi_s$: $$\int_{\Bbb R^n}\nabla\phi_s(x)\psi(x)dx = \int_{supp\,\nabla\phi_s}\nabla\phi_s(x)\psi(x)dx=\int_{supp\,\nabla\phi_s}\nabla(\phi_s(x)\psi(x))dx-\int_{supp\,\nabla\phi_s} \phi_s(x)\nabla\psi(x)dx=\int_{\partial \{supp\,\nabla\phi_s\}} \phi_s(x)\psi(x)\cdot ...


1

As some people pointed to you you should consider the delta sequence $$ f_n(x) = \frac{n}{\pi}\frac{1}{1+n^2x^2}$$ Which gives $$\lim_{n\to \infty} <f_n(x), \phi(x) > =<\delta(x), \phi(x) >$$ Note: you can put $\epsilon =\frac{1}{n}$ and consider the limit as $\epsilon$ goes to 0.


1

You need to have small supports of $t\mapsto \psi(t/\varepsilon_m)$. Of course, the derivatives become large and you have to deal with this in the proof of Borel's lemma.


1

Here's another approach, based on an old result of M. Kac (If $X$ and $Y$ are independent random variables such that $X+Y$ is independent of $X-Y$, then $X$ and $Y$ are normally distributed with the same variance.). Let $B'$ be a second Brownian motion independent of $B$ (make a product space construction if necessary), and define $\langle ...


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 $\phi \in \mathcal{D}$. ...


1

I found answer to my question here The solution is using the easily obtained identity $$\lim_{\varepsilon\rightarrow0+}\int_{\mathbb{R}\backslash(-\varepsilon,\varepsilon)}\frac{\varphi(x)}{x}\ \mathrm{d}x=\lim_{\varepsilon\rightarrow0+}\int_{\varepsilon}^{+\infty}\frac{\varphi(x)-\varphi(-x)}{x}\ \mathrm{d}x$$ As $\varphi\in\mathbb{C}^\infty$ the mean ...


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 ...


0

The main difficulty, I think, comes form building such a distribution $T$. One can take $T=\delta_0$. In this case $\langle T,g_a\rangle=g(0)$ is a constant sequence. Another example would be to take $T\in L^1(\Bbb R)$. In this case $\langle T,g_a\rangle\to0 $ as $a\to\infty$. One of the examples of such distributions of first order is $p.v.(1/x)$, ...


0

Let $f\in L^p(\Omega)$, $1<p<\infty$. $f$ is completely determined by the values of $f(x)$ for almost all $x\in\Omega$; The values of $f(x)$ for almost all $x\in\Omega$ are completely determined by the values of $\int_{\Omega} f(x)h(x)\;dx$ for all $h\in L^q(\Omega)$; The values of $\int_{\Omega} f(x)h(x)\;dx$ for all $h\in L^q(\Omega)$ are ...


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 ...


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 ...


0

In THIS ANSWER, I provided a rigorous development to show that the Dirac Delta can be regularized as $$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$ where $$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} $$ Here, we will use a formal approach that can easily be made rigorous as in the ...


0

It means that if for $f,g \in L^p$ we have $$ \int f h = \int gh$$ for all $h \in L^q$, then $f = g$ (in the usual $L^p$ sense).


0

In the sense of (1), it can be defined for $u_i,v\in L^1_{\text{loc}}(\Omega)$. More generally, it can be defined for arbitrary distributions. There are $u \in L^2(\Omega; \mathbb{R}^d)$, such that there is no $v \in L^1_{\text{loc}}(\Omega)$ with $v = \operatorname{div} u$. But $u$ has always a distributional divergence. If each component of $u$ is weakly ...


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 ...


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

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 ...


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. ...


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 ...


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 ...


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 ...


0

The canonical definition of convolution of two arbitrary distributions with well associated supports is the following. Let $S$ and $T$ be two distributions, respectively, with supports $A$ and $B$, and let $\varphi$ is a test function with support $K$. You define $$ \langle S*T,\varphi\rangle=\langle S_x\otimes T_y,\alpha(x)\varphi(x+y)\rangle, $$ where ...


-1

To your question: Your definition of convolution is a map $$ *:(\mbox{test functions})\times (\mbox{distributions})\to \mbox{distributions}. $$ So you're right, "$\delta * g$" does not make sense, as $\delta$ is not a test function.


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 $e^{-|x|}$.


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

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.


0

Much of the answer depends on the allowed values for the parameters involved. If $x=0$, the expression can be rewritten as $z+iy$, which tends to $z$ as $y\to0$. Suppose $x\neq0$ and let $z=u+iv$, $$ \frac{1}{x+\frac{1}{z+iy}}=\frac{u+i\left(y+v\right)}{\left(1+xu\right)+ix\left(y+v\right)}. $$ If $v\neq0$, as $y\to0$, we get $$ ...


0

It is the (vector) of distributions defined via $$\nabla p : D(\Omega)^n \to \mathbb{R}, \quad \varphi \mapsto -\sum_{i=1}^np(\partial{x_i} \varphi).$$ If $p$ is actually a funcion, the last sum is just $$-\int_\Omega p(x) \, \operatorname{div}(\varphi)(x) \, \mathrm{d}x,$$ and if $p$ is (weakly) differentiable, you have $$-\int_\Omega p(x) \, ...


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} ...


1

($c\equiv 1$) The commutator can be written as the sum of the positive- and negative-frequency part of the Pauli-Jordan function: $$ i[\phi(x),\phi(y)]=D^+(x-y)+D^-(x-y), $$ where \begin{align} D^\pm(x) &= \pm\frac{1}{i(2\pi)^3}\int\frac{d^3k}{2\sqrt{\mathbf k^2+m^2}} e^{\pm i\sqrt{\mathbf k^2+m^2}\,x^0-i\mathbf k\cdot\mathbf x}. \end{align} Performing ...


0

\begin{equation*} \frac{1}{x+\frac{1}{x+iy}}=\frac{x+iy}{1+x(x+iy)}=\frac{x+iy}{1+x^{2}+ixy} \rightarrow \frac{x}{1+x^{2}} \end{equation*} Edit For the correct expression for general $z$ \begin{equation*} \frac{1}{x+\frac{1}{z+iy}}=\frac{z+iy}{1+x(z+iy)}\overset{y\rightarrow 0}{% \longrightarrow }\frac{z}{1+xz} \end{equation*} Distributions do not seem to ...


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 $R$ ...


0

There are a few related concepts here: 1) A bump function is a term for a smooth function with compact support. The set of all bump functions forms a vector space. If these functions are on $\mathbb{R}^n$, then it is often denoted $C_c^\infty(\mathbb{R}^n)$. In distribution theory, this is what is most commonly referred to when one refers to test functions, ...


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. ...


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 ...


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, ...


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

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 ...


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 $$ \lim_{n\to\infty}\langle \tau_n, \varphi\rangle =\langle \tau, \varphi\rangle, $$ ...



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