# Tag Info

1

In various branches of mathematics it turns out to be profitable to shift attention from what some objects are to what they do. In case of functions, we shift attention from (locally integrable) functions $f$ being some set of ordered pairs, etc to what it does: it defines a linear functional on the set of test functions, by $$\varphi \mapsto \int f\varphi ... 2 We indeed obtain that the limiting distribution is 0 because x\mapsto x^3\varphi(x) is a smooth integrable function with integrable derivative. So Riemann-Lebesgue lemma applies. 2 The integral with e^{-ixy} works for functions that are integrable on the real line (are in L^1(\mathbb R), but the domain of the Fourier transform extends way beyond that. How do we compute it when the integral diverges? Either by some approximating process, or by using general properties of the transform such as its interplay with derivatives. I'll use ... 1 Hint: You can use the sifting property of the delta function$$ \int_{-\infty}^{\infty} \delta(x-x_0) f(x) dx = f(x_0) . $$Added: Following my last comment we advance as$$ \hat{f}(y)= \int_{-\infty}^{\infty} e^{-ixy} = \delta(y) \implies \frac{d\hat{f}(y)}{dy}= \int_{-\infty}^{\infty}(-ix) e^{-ixy} = \delta'(y) \longrightarrow (*). $$So by ... 4 For U \subset V, we have a natural (continuous) injection$$\iota^U_V \colon \mathscr{D}(U) \hookrightarrow \mathscr{D}(V).$$Its transpose,$$\rho^V_U \colon \mathscr{D}'(V) \to \mathscr{D}'(U)$$is called the restriction of the distributions on V to distributions on U. For regular distributions, that corresponds to the restriction of the locally ... 2 I can show (1), but I don't know if \Delta(\Phi\circ u) is a measure. The equality (1) has to be understood in the sense of distributions, so we must figure out how (\Phi'\circ u)\Delta u acts on test functions. We know how \Delta u acts: T(\phi)=\int u\Delta \phi. Since u is a W^{1,2} function, T is a particularly nice distribution: its value ... 0 1) You can use your norms to define a neighborhood base of zero, i.e. the sets U_{N,\epsilon}:=\{\varphi \in D(\mathbb{R}) : \|\varphi\|_N<\epsilon\} as a local base. This topology is obviously metrizeable via a Frechét metric of the form \rho(\varphi, \psi) = \sum_{N\in \mathbb{N}} \frac{\|\varphi - \psi\|_N}{2^N(1+\|\varphi - \psi\|_N)}  2)I dont ... 2 We have a lemma stating the following: Given a succession of L^1(\mathbb{R}^n) functions f_j, such that$$ f_j(x)=j^nf(jx), $$then the corresponding function-type distribution converge to C\delta(x) as j\to \infty, where:$$ C=\int_{\mathbb{R}^n}f(y) \, d^ny. $$Here we have:$$ \frac{1}{\pi} \frac{\sin^2\alpha x}{\alpha x^2} = ...

1

Your attempt went astray at this step: $$\left|\sum_{n \epsilon N} \psi(x) -\psi(0)\right| =\sum_{n \epsilon N} |\psi(x) -\psi(0)| \tag{wrong}$$ The triangle inequality gives $\le$, but this does not help you demonstrate your claim (since you wanted to show the sum on the left is large). Instead, consider a partial sum: $$\sum_{n=1}^M (\psi(x) -\psi(0)) ... 0 Thanks to @mathematican For any ψ with ψ′=1 on [0,1] the sum isn't even finite. 2 No, local integrability is not enough for pointwise convergence. Take f=\chi_{[0,1]}, then f_\epsilon\to 1/2 at 0 and at 1. Continuity of f is enough. Indeed, when \epsilon is small, all the values of f involved in the definition of f_\epsilon(x) come from a small neighborhood of x. By virtue of continuity, they all lie close to f(x). ... 0 The Fourier transform of the delta distribution is a constant function. Thus, if you Fourier transform, multiply with a Schwartz function and Fourier transform back, you get a Schwartz function. 1 Both proofs are done by definition Definiton of distribution: for each compact K there exists N_K\ge 0 and c_K\ge0 such that for any test function \phi  with support in K we have |\langle T,\phi\rangle|\le C_K \sum_{j=0}^{N_K}\|\phi^{(j)}\|. A: take a compact set K and a test function \phi with support in K. Clearly, there exist two ... 1 Hint for Part B: Let n \geq 1. Then it is easily seen that T = (-1)^n \delta^{(n)}_n on a neighbourhood of n and so the order of T is at least the same as that of \delta^{(n)}_n, which is equal to the order of \delta^{(n)} since the order is invariant by translation. Now show that this order is exactly n. Since n can be chosen arbitrarily it ... 1 Note that,$$ \frac{\partial}{\partial x} \delta(x) = - \delta(x) \frac{\partial}{\partial x}, $$which implies that,$$ \frac{\partial^2}{\partial x^2} \delta(x) = \frac{\partial}{\partial x} \frac{\partial}{\partial x} \delta(x) = - \frac{\partial}{\partial x} \delta(x) \frac{\partial}{\partial x} = + \delta(x) \frac{\partial}{\partial x} ...

1

Given a test function $\phi$, decompose it into even part $\phi_e(\omega)=\frac12(\phi(\omega)+\phi(-\omega))$ and odd part $\phi_o(\omega)=\frac12(\phi(\omega)-\phi(-\omega))$ and odd part. So, $\phi=\phi_e+\phi_o$. We have $$\int \frac{\cos \omega T}{j \omega} \phi_e(\omega)\,d\omega = 0$$ by symmetry. Since $\phi_o(0)=0$, we can write ...

2

For two distributions $S\in \mathcal{D}'(\mathbb{R}^k)$ and $T\in \mathcal{D}'(\mathbb{R}^n)$, we define the distribution $S\otimes T \in \mathcal{D}'(\mathbb{R}^{k+n})$ by $$(S\otimes T)[\varphi] := S\left[x \mapsto T[\varphi(x,\,\cdot\,)]\right].$$ Verification that that is a distribution, and that also $(S\otimes T)[\varphi] = T\left[y \mapsto ... 0 Does the line integral$\int_C u(x,y)\ dC$exists? If$C$is a rectifiable curve, then yes. Note that for a general smooth function$\varphi$the zero set$\varphi=0$can be quite horrible: it can be an arbitrary closed subset of$\mathbb R^2$. But if you assume that$\nabla \varphi\ne 0$on$C$, then the implicit function theorem implies that$C$is a ... 2 I just responded to my own questions by looking at Robert Israel's answer. The function is $$H_\delta=\frac{1}{1+{e^{{\frac {4 x\delta}{ x^2-\delta^2 }}}}},\;\;{\mbox{for}}\;\;|x|<\delta.$$ 1 As in my comment, the goal is to show that there is a constant$c$such that$\langle T, \phi \rangle = c \int_{-\infty}^\infty \phi(x)\,dx$for all test functions$\phi$. First of all, let's identify the constant$c$. Let$\psi$be your favorite test function having$\int_{-\infty}^\infty \phi(x)\,dx = 1$and set$c := \langle T, \psi \rangle$. Next: for ... 0 The answer is yes. It is known that$\frac{1}{4\pi|\boldsymbol{x-x_0}|}$is the Green's function of Laplace's equation in 3-D. Thus$\Delta(\frac{1}{4\pi|\boldsymbol{x-x_0}|})=\delta(\boldsymbol{x-x_0})$. Due to the property that the vector Laplacian of a vector is equal to the vector whose components in Cartesian coordinates are the scalar Laplacians of the ... 2 Remember that$\langle \delta,f\rangle = f(0)$. Because$x_i(0) = 0, we're done. 1 How about using the Standardmollifier and massage it a bit: \begin{align} f(x) = \begin{cases} 0& \text{ if } x\leq 0\\ e^{1+\frac{-1}{1-(x-1)^2}}& \text{ if }0<x<1 \\1 & \text{ if }x\geq 1\end{cases} \end{align} 2 Recall that $$f(t) = \left\{ \begin{array}{lr} e^{-1/t} & t >0\\ 0 & t \leq 0\\ \end{array}\right.$$ is smooth. Theng(t) = \frac{f(t)}{f(t)+f(1-t)}$is smooth, is equal to$1$for$t \geq 1$, and equal to$0$for$t \leq 0\$. This is the standard construction; you can probably find it in most any intro book to smooth manifolds, for ...

Top 50 recent answers are included