# Tag Info

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.

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 $\displaystyle \langle \varphi, \widehat{1} ... 0 The distributional derivative of the Dirac delta distribution is the distribution$\delta′$defined on compactly supported smooth test functions$\varphi$by $$\delta'[\varphi] = -\delta[\varphi']=-\varphi'(0).$$ The first equality here is a kind of integration by parts, for if$\delta$were a true function then $$\int_{-\infty}^\infty ... 2 Using integration by parts we get:$$\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 ... 0 You must know continuous immersion in locally convex spaces, in this case we can also speak of Fréchet spaces for$\mathcal{S}(\mathbb{R}^n)$or$L^1(\mathbb{R}^n)$, instead$C_{c}^\infty(\mathbb{R}^n)$is a LCS. Precisely theorem (1) of this question: Continuous inclusions in locally convex spaces The problem is that you have to consider the appropriate ... 0 While$C_c^\infty(\mathbb R^n)$fails to be first countable, it is still true that for any linear map$T: C^\infty_c(\mathbb R^n) \to X$where$X$is a topological vector space whose topology is induced by a family seminorms (i.e., a locally convex TVS), we have that$T$is continuous iff$T$is sequentially continuous. It's immediate that ... 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 ... 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 ... 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 ... 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 ...

3

$\delta$ 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 ...

0

The standard reference is de Rham's Differentiable Manifolds, which develops the whole theory in terms of currents. But more quickly, see the beginning of Section 7.3 of Nicolaescu's notes http://www3.nd.edu/~lnicolae/Lectures.pdf A current is a functional, so it must eat differential forms and spew out numbers. This integral $\int_X a\wedge j_e$ is nothing ...

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 ...

0

Actually it was (as you can check in the lecture notes you've referenced) $$\frac 1 {2\pi} \int_{\mathbb R} \int_{\mathbb R} e^{-ixt} g(x)\ dt\ dx.$$ If you integrate by $t$ first then it can be rewritten as (I'll assume below that integration interval is always $\mathbb R$) $$\int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}} \int \frac{1}{\sqrt{2\pi}} ... 0 Take g outside the inner integral on the left and you have the fourier transform of 1. This is the dirac delta, a distribution that integrates to 1 and is infinity at x. In this sense only the value of g where the dirac delta is non-zero becomes relevant. The point is then moving with x in your case. 1 Suppose that x \notin \newcommand{\supp}{\mathrm{supp}}\supp(\mu). Then there is a ball B(x,r) with \mu(B(x,r)) = 0. If \phi is any function compactly supported in B(x,r) then T_\mu(\phi) = 0. Thus B(x,r) \cap \supp(T_\mu) = \emptyset, so that x \notin \supp (T_\mu). On the other hand, suppose that x \notin \supp (T_\mu). Then there ... 2 The fundamental solution, as mentioned, satisfies -u''+k^2 u = \delta_{y}(x). To the left or to the right of y, the fundamental solution satisfies -u''+k^2u=0. The fundamental solution needs to be continuous across y, and, in order to have the \delta function behavior, there is a discontinuity in the first derivative of the solution with a jump of ... 0 You are basically finished - note that you can write$$ \sum_{j = 1}^{m}[f(x_j^+) - f(x_j^{-})]\phi(x_j) = \langle \sum_{j = 1}^{m}[f(x_j^+) - f(x_j^{-})]\tau_{x_j}\delta, \phi\rangle $$where \langle\tau_{x_j}\delta, \phi\rangle = \phi(x_j) denotes the Dirac measure at x_j. Now combine with$$ \int_{-\infty}^{\infty}\frac{df}{dx}\phi(x)d x = \langle ...

3

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 $\mu((a,b]) = F(b) - F(a)$ ...

1

Typically a "fundamental solution" or "Green's function" is a function that solves the equation $-u'' + k^2u = \delta(x).$

0

In the end it turns out that the key was a physical consideration. These correlation functions $\langle0|\phi(x)\phi(y)|0\rangle$ are Lorentz-invariant, which implies they only depend on the distance between $x$ and $y$. With that in mind the calculation is simple. From now on write $\langle0|\phi(x)\phi(0)|0\rangle=f(x)$. In the case of $F$, let $g(k)$ be ...

0

By taking the complementary, showing that the supports are equal is equivalent to show that the biggest annihilating sets are equal. So it is enough to show that an annihilating set for $\mu$ is an annihilating set for $T_{\mu}$ and conversly. That was the suggestion of your professor.

2

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 $\phi \in ... 1 The author indeed means$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, ... 3 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, ... 0 You have only to check that$\langle T, \varphi_1 + \varphi_2\rangle=\langle T, \varphi_1 \rangle +\langle T, \varphi_2\rangle$is false in general with the definition of$\langle T,\varphi\rangle=|\varphi(0)|$. Briefly said: the absolute value blocks the linearity. Your last question deals implicitely with the meaning of$|\delta|$; but we have ... 0 A fruitful way to consider (and sometimes to work on)$\delta$distribution is as the limit when$s \rightarrow \infty$of (gaussian density) functions$f_s(x)=\frac{1}{s \sqrt{2 \pi}}e^{-x^2/(2s^2)}$. Thus$\delta'$and$\delta''$can be considered as the limit of functions$f'_s$and$f''_s$resp. In order to answer your question [see figure below] ... 0 Oh, it turns out to be pretty easy. The equation simply says that in$x \in [0, c)$it has the form$y''(x) - a y(x) + b = 0$and in$x > c$:$y''(x) - a y(x) = 0$. Boundary conditions give trivial solution on$x > c$, so I can treat$\theta$-function on$[0, c)$as a constant to obtain: $$y(x) = b\theta(c - x)(e^{-\sqrt a x} + 1)$$ 0 Observe that$\mathrm{Supp}(f\phi)\subset\mathrm{Supp}(\phi)$. Since$\mathrm{Supp}(\phi)$is compact and$\mathrm{Supp}(f\phi)$is closed,$\mathrm{Supp}(f\phi)$is compact. Furthermore, the smoothness of$f\phi$comes from the product rule of differentiation. 0 Convergence of the$\phi_n$in$\mathscr{D}$implies that the$\phi_n$are convergent in$\mathscr{D}_K$where$K$is a compact subset of$\Bbb{R}^N$. Rudin Functional Analysis 6.5. In$\mathscr{D}_K$, the topology is given by a collection of sup norms, see Rudin 6.2, and so convergence of$\phi_n$is the same as convergence in all those norms, i.e., ... 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 ... 2 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 ... 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$\alpha<\frac12, we may compute \begin{align} \int_0^1\int_0^1|k(x,y)|^2\ \mathsf dy\ \mathsf dx &= \int_0^1\int_0^1 L(x,y)^2|x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\\\ &\leqslant C \int_0^1\int_0^1 |x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\ &= 2C\int_0^1\int_0^x (x-y)^{-2\alpha}\ \mathsf ... 1 Take a positive, nonzerou \in C^{\infty}_c(\mathbb{R}^d)$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) $$\|u_{\epsilon}\|_{L^{2^*}} = \Big(\int |u_{\epsilon}|^{2^*}\Big)^{\frac{1}{2^*}} = \epsilon^{\frac{d}{2^*}}\|u\|_{L^{2^*}}$$ $$\|\nabla u\|_{L^2} = ... 2 Defining \phi_\lambda(x)=\phi(\lambda x) for smooth \phi, the requirement is$$ 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.

Top 50 recent answers are included