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Suppose $f:\mathbb{R}^d\to \mathbb{C}$ is $m$ times continuously differentiable and compactly supported. Then for $\mu=(\mu_1,\dots,\mu_d)\in\mathbb{N}_0^d$ with $|\mu|=\mu_1+\dots+\mu_d\le m$, we know the Fourier transform of the partial derivative $\frac{\partial^\mu f}{\partial x^\mu}$ is $$\mathcal{F}\left[\frac{\partial^\mu f}{\partial x^\mu}\right](\xi)=\int_{\mathbb{R}^d}\frac{\partial^\mu f}{\partial x^\mu}(x)e^{-ix\cdot\xi}\,dx=(-i\xi)^\mu \mathcal{F}f(\xi),\quad\xi\in\mathbb{R}^d.$$ Moreover, the Paley-Wiener theorem says that $\mathcal{F}\left[\frac{\partial^\mu f}{\partial x^\mu}\right]$ and $\mathcal{F}f$ are both analytic functions. I am interested in whether the Fourier transform has an anti-derivative analog. To be specific, suppose $\eta:\mathbb{R}^d\to\mathbb{C}$ is a compactly supported and continuous function such that its Fourier transform $\mathcal{F}\eta$ satisfies $$\lim_{\xi\to 0}\frac{\mathcal{F}\eta(\xi)}{(-i\xi)^\alpha}=L$$ for some $\alpha=(\alpha_1,\dots,\alpha_d)\in\mathbb{N}_0^d$ and $L\in\mathbb{C}$. Define $$\phi(\xi):=\begin{cases}\frac{\mathcal{F}\eta(\xi)}{(-i\xi)^\alpha}, &\xi\neq 0,\\ L, &\xi=0.\end{cases}.$$ My questions: Is it necessary that $\phi$ is the Fourier transform of some compactly supported function (distribution) $f$, and $\frac{\partial^\alpha f}{\partial x^\alpha}=\eta$? If not, what additional assumptions on $\eta$ are needed to make this claim true? Intuitively, I think the answer should be positive. However, when I try to write down the proof formally, I have some trouble. By the definition of $\phi$, it is clear that $\phi$ is analytic. However, to apply the Paley-Wiener theorem, there are restrictions on the growth of $|\phi(\xi)|$ as $|\xi|$ gets large, and this is where the problem arises. Any suggestions are welcomed. Thanks!

A subset A of a normed linear vector space X is called balanced if $\alpha A \subset A$ forall scalar $\alpha$ s.t $|\alpha| \leq 1$. Noted that $\epsilon B_X=\{x \in X: \|x\|<\epsilon\}$ and $A+B=\{a+b: a\in A, b\in B\}.$ This is my question: "Given that $M$ is a convex, balanced subset of a normed linear vector space X. Can we have $$\forall \epsilon >0, \exists a>1: aM \subset cl(M+\epsilon B_X)$$ " It holds when $M$ is bounded (not necessary convex or balanced) by $K>0$: For every $\epsilon>0$, take $a=1+\dfrac{\epsilon}{K}>1$ then $$x=\left(1+\dfrac{\epsilon}{K}\right)m=m+\dfrac{\epsilon}{K}m \in M + \epsilon B_X \in cl(M+\epsilon B_X)$$ (because $\left\|\dfrac{\epsilon}{K}m\right\|=\dfrac{\epsilon}{K}\|m\| < \epsilon.$) But for $M$ is unbounded, I don't know if this statement is true?

Let $A_1,A_2 \subseteq \mathcal M$ be two substructures of a structure $\mathcal M$ with signature $S$. What is a sufficient condition on the signature $S$ that ensures the union $ A_1 \cup A_2 $ is also a substructure of $\mathcal M$? My guess is, that if $A_1 \cap A_2$ is a substructure and the closures satisfy $\bar A \cap \bar A_2$, then $A_1 \cup A_2 $ should be a substructure. Does anyone have a reference or proof for this? Additionally, if such a condition does not hold, how can one prove that there exists a structure and subsets that contradict this conclusion? Thanks in advance for any help!

I have a decomposition of a Hilbert space $H$ into $H = \ker T \oplus (\ker T)^\bot$. On both these subspaces I have a linear operator which is bounded. Let's call them $A$ and $B$ respectively. So $A$ is the operator on $\ker T$ and $B$ is the operator on $(\ker T)^\bot$. I want to show that the sum is also a bounded linear operator from $H\to H$. Define $R\colon H\to H$ by $R(x+y) = Ax +By$. It is clear that $R$ is linear. For boundedness I showed the following: $$\|R\| = \sup_{\|z\| = 1} \|Rz\| = \sup_{\|x+y\| = 1} \|R(x+y)\| = \sup_{\|x+y\| = 1} \|Ax+By\|\leq \sup_{\|x+y\| = 1} \|Ax\| + \|By\| \leq \sup_{\|x\|=1 }\|Ax\| +\sup_{\|y\| = 1}\|By\|.$$ I am not sure if this chain of (in)equalities is correct. Moreover, I have to show that $R$ is well-defined. And I don't know what well-defined means in this context... What do I have to show to prove that $R$ is well-defined?

Here is the Rudin's book: The Calculus of Residues 10.41 Definition A function $f$ is said to be meromorphic in an open set $\Omega$ if there is a set $A \subset \Omega$ such that (a) $A$ has no limit point in $\Omega$, (b) $f \in H(\Omega-A)$, (c) $f$ has a pole at each point of $A$. Note that the possibility $A=\varnothing$ is not excluded. Thus every $f \in H(\Omega)$ is meromorphic in $\Omega$. Note also that (a) implies that no compact subset of $\Omega$ contains infinitely many points of $A$, and that $A$ is therefore at most countable. If $f$ and $A$ are as above, if $a \in A$, and if $$ Q(z)=\sum_{k=1}^m c_k(z-a)^{-k} $$ is the principal part of $f$ at $a$, as defined in Theorem 10.21 (i.e., if $f-Q$ has a removable singularity at $a$ ), then the number $c_1$ is called the residue of $f$ at $a$ : $$ c_1=\operatorname{Res}(f ; a) . $$ If $\Gamma$ is a cycle and $a \notin \Gamma^*,(1)$ implies $$ \frac{1}{2 \pi i} \int_{\Gamma} Q(z) d z=c_1 \operatorname{Ind}_{\Gamma}(a)=\operatorname{Res}(Q ; a) \operatorname{Ind}_{\Gamma}(a) . $$ This very special case of the following theorem will be used in its proof. 10.42 The Residue Theorem Suppose $f$ is a meromorphic function in $\Omega$. Let $A$ be the set of points in $\Omega$ at which $f$ has poles. If $\Gamma$ is a cycle in $\Omega-A$ such that $$ \operatorname{Ind}_{\Gamma}(\alpha)=0 \quad \text { for all } \quad \alpha \notin \Omega, $$ then $$ \frac{1}{2 \pi i} \int_{\Gamma} f(z) d z=\sum_{a \in A} \operatorname{Res}(f ; a) \operatorname{Ind}_{\Gamma}(a) . $$ Proof. Let $B=\left\{a \in A: \operatorname{Ind}_{\Gamma}(a) \neq 0\right\}$. Let $W$ be the complement of $\Gamma^*$. Then $\operatorname{Ind}_{\Gamma}(z)$ is constant in each component $V$ of $W$. If $V$ is unbounded, or if $V$ intersects $\Omega^c$, (1) implies that $\operatorname{Ind}_{\Gamma}(z)=0$ for every $z \in V$. Since $A$ has no limit point in $\Omega$, we conclude that $B$ is a finite set. The sum in (2), though formally infinite, is therefore actually finite. My questions: Why do we have $A$ is at most countable? How can I deduce that $B$ is a finite set from $A$ has no limit point?