New answers tagged

0

I could solve the problems: We show that $\pi$ is open onto its image $\pi(S)$. Let $\{s\} \subseteq S$, then we have $\{\delta_s\} = \pi(s) \cap (\operatorname{ev}_{1_{\{s\}}})^{-1}(\mathbb C \setminus \{0\})$, where $\operatorname{ev}_{1_{\{s\}}}(\varphi) = \varphi(1_{\{s\}})$ for all $\varphi \in G_S$. Hence $\{\delta_s\}$ is open in $\pi(S)$ and $\pi(S)...


0

The functorial property of the inductive limit makes life much easier: The inclusion $\mathscr D(\Omega) \hookrightarrow C(\Omega)$ where the latter space is endowed with the topology of convergence on all compact sets is continuous because so are all $\mathscr D(K)\hookrightarrow C(\Omega)$. Since $C(\Omega)$ is Hausdorff (trivially: if $f\neq g$ there is $...


0

As for the first question, the singleton $\{\varphi_1\}$ is closed because you have just shown that the complement $\mathcal{D}(\Omega)\setminus\{\varphi_1\}$ is open. I suspect that the Hausdorffness argument is not correct, as the equivalent of this in $\mathbb{R}$ would be that $\varphi_1$ and $\varphi_2$ are points, and the set $\varphi_2+U$ would be the ...


2

As has been mentioned in some comments, this is the definition of distributional derivative. In general, if $\alpha \in \mathbb{N}^n$ is a multi-index, the distributional derivative of order $\alpha$, of a distribution $T \in \mathcal{D}'(\Omega)$ is defined by placing $(\star)$ $\displaystyle \langle D^\alpha T, \varphi \rangle =(-1)^{|\alpha|} \langle T, ...


0

In fact there is an easier way to look at the problem: Let $U$ be such an overlapping area. Consider the pairing $\, C^\infty(U) \times C_c^\infty(U) \to \mathbb{C}\\ \, (u,\phi) \mapsto \langle u,\phi \rangle$ All we need to show is that this is non degenerate: (by contraposition) If $u \neq 0$, then $u$ does not vanish on an open set $V$. Take any bump ...


1

For every point $x\in V_p\cap V_q$, you can use bump functions $\varphi_n$ (with integral $1$) supported in $(1/n)$-neighborhood of $x$ to conclude that $$u_p(x) = \lim_{n\to\infty} \langle u_p, \varphi_n\rangle = \lim_{n\to\infty} \langle v_p, \varphi_n\rangle = v_p(x)$$ Put another way, $\varphi_n$ converge to the Dirac delta at $x$ in the sense of ...


2

Your computation is correct. Someone dropped the minus sign when stating the exercise. Remembering the calculus formula $(1/x)=-1/x^2$ would help prevent that. The calculus formula doesn't prove the result for distributions, but it must be consistent with it in the sense that differentiation away from singularities is the same as classical derivative.


1

One can consider the distributional derivative of any locally integrable function, which includes all continuous functions. This doesn't mean there's much to say about them other than "they exist". am I correct in understanding that the distributional derivative of the brownian motion is a measure No, it is not a measure. The distributional ...


1

Ok let me see if I can't answer your question from a mathematical perspective in a way that might be helpful for an engineer. The first thing you need to understand is that $\varphi_{n}(x)=\frac{1}{\sqrt{2\pi}}e^{ikx} , \, k \in \mathbb{Z}$ forms an orthonormal sequence in $L^{2}([-\pi,\pi]))$. Now, anytime you take the inner product of two orthonormal ...


0

One way to see such a thing is by approximating your pure $\sin x$ function/signal by truncated or otherwise modified functions whose Fourier transforms exist as integrals, and look at the limit of such things. For example, truncating $\sin x$ to be $0$ outside expanding intervals $[-2\pi N,2\pi N]$. We can integrate explicitly $$ 2i\int_{-N}^N e^{-i xt}\,\...


-2

Devraj - $$ \int_{-\infty}^{\infty} \frac{e^{2 \pi f_{0} i t}-e^{-2 \pi f_{0} i t}}{2i} e^{2 \pi i k t} dt =\frac{1}{2i} \int_{-\infty}^{\infty} -e^{-2\pi i(f-f_{0})t}+e^{2 \pi i(f+f_{0})} dt=\frac{1}{2i}\left[ \delta(f+f_{0})-\delta(f-f_{0}) \right]$$


0

I am also learning this stuff at the moment so I'll let you know what I think. I don't see how you would use closed graph theorem, in fact, you would need either $\mathcal{S}=\mathcal{S}(\mathbb{R}^d)$ to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied ($\mathcal{S}$ is normable or compact, at least I don't think so). ...


0

The first two examples you mention appear respectively in Fourier and Cauchy who specifically mention that the epsilon is infinitesimal. Of course they didn't have distribution theory. Nor is distribution theory necessary to make this work; one can use a modern theory of infinitesimals where one can define Dirac delta functions possessing local values (of ...


1

I worked this out referring to what you have already done and what Ian suggested. Also refer to https://proofwiki.org/wiki/Scaling_Property_of_Dirac_Delta_Function . To prove $\delta(kx)=\frac{1}{|k|} \delta(x)$ , instead prove ${|k|}\delta(kx)= \delta(x)$, where k is a nonzero real constant. Use the definition of $\delta(x)$ consiting of the first ...


2

My guess is that the author makes a hypothesis that the solution is a distribution which can be represented by a $L^2_{loc}$ function in the sense that there exists a function $g\in L^2_{loc}$ such that for any test function $\phi$ $$\langle u,\phi\rangle=\int_{(0,T)\times \Bbb R^d}g(t,x)\phi(t,x)\,\mathrm dt\,\mathrm dx$$


0

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


1

I assume that the OP actually wants to solve the following ODE $$y'' + y = \delta + \delta'$$ rather than $y'' + y' = \delta + \delta'$. Assuming zero initial conditions and taking the Laplace transform of both sides, $$s^2 Y (s) + Y (s) = 1 + s$$ and, thus, $$Y (s) = \dfrac{s+1}{s^2+1} = \dfrac{s}{s^2+1} + \dfrac{1}{s^2+1}$$ Taking the inverse Laplace ...


1

$$y''(t)+y'(t)=\delta(t)+\delta'(t)$$ $\delta(t)$ is the Dirac function. First integration : $$y'+y=u(t)+\delta(t)+c_1$$ $u(t)$ is the Heaviside function. Seconf integration : The solution of the homogeneous ODE $y'+y=0$ is $y=c\:e^{-t}$ Let $y(t)=f(t)e^{-t}$ $(f'-f)e^{-t}+fe^{-t}=u(t)+\delta(t)+c_1$ $f'=(u(t)+\delta(t)+c_1)e^t$ $f=(e^t-1)u(t)+u(t)+...


0

I'm going to post my answer, even though I think it constitutes only about 80% of a proof. Let $$ Z(t) = Z_c(t) + i Z_s(t)\, , $$ where $$ Z_c(t) = \sum_{k = 1}^{\infty}\cos(k t) $$ and $$ Z_s(t) = \sum_{k = 1}^{\infty}\sin(k t) $$ Now, using the complex exponent version of $\cos(kt)$, $Z_c(t)$ can be written as \begin{align} Z_c(t) &= \frac{1}{2}\left[...


0

This may be justified in the sense that in the space of distributions there are measures. For example, if $\Omega \subset \mathbb{R}^n$ is an open set, the Dirac measure is defined by $\delta_{x_0}(\Omega)=1$ if $x_0 \in \Omega$ and $\delta_{x_0}(\Omega)=0$ if $x_0 \notin \Omega$. Then $\displaystyle \delta_{x_0} (\varphi)=\int_{\Omega} \varphi(x) d \delta_{...


1

Here are some ideas to get you started: If you were using the space $W^{1,p}(\newcommand{\R}{\mathbb R} \R^+)$ you would define the reflection $\bar u(x) = u(-x)$ for $x < 0$ and $\bar u(x) = u(x)$ for $x > 0$. This gives you continuity of $\bar u$ at $x = 0$ from which the absolute continuity of $\bar u$ follows. Suppose instead you are using the ...


1

The first question follows from the fact that $u_m \rightarrow \overline{u}$ in $W^{1,p}(\mathbb{R}^n)$ implies $||u_m-u_l||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du_m-Du_l||_{L^p(\mathbb{R}^n)} \leq C[||Du_m-D\overline{u}||_{L^p(\mathbb{R}^n)} + ||D\overline{u}-Du_l||_{L^p(\mathbb{R}^n)}] \rightarrow 0$ as $m,j \rightarrow \infty$, and $\lbrace u_m \...


4

The identity $$ \int_{-\infty}^{+\infty}dt\ f(t) \delta(t) = f(0)\tag{*} $$ is meaningless without context. Also this notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral. Let's say you are considering $\delta:\mathcal{S}(\mathbb{R})\to\mathbb{R}$ as a tempered distribution on the Schwartz class $\mathcal{S}(\...


1

It is correct provided that one understand the notation $\int_{-\infty}^\infty\delta(x)\phi(x)\ dx$ correctly. This is by no means an integral and $\delta$ is not a function with real variables. Also, one should specify in what function space is the function $\phi$ so that your identities would make sense. Any distribution $T$ is infinitely ...


1

The first question follows by definition, $H_m(\mathbb{R}^n)$ is the completion of $C_{c}^\infty(\mathbb{R}^n)$ iff (by definition) $H_m(\mathbb{R}^n) = \overline{C_{c}^\infty(\mathbb{R}^n)}^{|| \cdot ||_m}$, i.e. $\forall u \in H_m(\mathbb{R}^n)$ exists $\lbrace u_k \rbrace \subset C_{c}^\infty(\mathbb{R}^n)$ such that $u_k \rightarrow u$ in the $H_m$-norm. ...



Top 50 recent answers are included