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The operation $$h \mapsto \int_{a}^b f(x, z) h(z) dz$$ is called an operator with kernel $f$ https://en.wikipedia.org/wiki/Integral_transform (no relationship with the other meaning of the word in linear algebra). You have noticed that, instead of taking $\mathbb{R}$ in its whole, I have taken bounds $a,b$. One of the important kernels is $e^{-2i \pi ... 1 First of all, this depends on the topologies, I assume you mean the weak$^*$topologies$\sigma(\mathscr S',\mathscr S)$and$\sigma(\mathscr D',\mathscr D)$. Note that$r$has dense range (it is the transposed of the inclusion$\mathscr D \hookrightarrow \mathscr S$) and$r$is injective (because the inclusion$\mathscr D \hookrightarrow \mathscr S$has ... 2 Here is a way of getting lots counterexamples. Let$\widetilde{X}$be your favorite infinite-dimensional complete space, let$X\subset \widetilde{X}$be a dense subspace, let$v\in\widetilde{X}\setminus X$, and let$L$be the span of$v$. Then we can take$Y=\widetilde{X}/L$, and the quotient map$\widetilde{\varphi}:\widetilde{X}\to Y$is not injective ... 1 You have answered almost everything by yourself. Let$\tau$be the usual topology on$\mathcal{D}$and$\sigma$the subspace topology of$\mathcal{S}$on$C_c^\infty$. You are asking for$\tau=\sigma$. You constructed a sequence in$\mathcal{D}$converging with respect to$\sigma$but not with respect to$\tau$. This shows$\tau\nsubseteq\sigma$. To see ... 4 A dense set$D$of$X$is such that the closure of$D$equals$X$. Or equivalently, every non-empty open set contains a pont of$D$. So the points of$D$are in a sense "close" to all points of$X$, we can "approximate" points of$X$by points in$D$. The name separable is somewhat unlucky (what can be separated, exactly?). It probably has an historic ... 0 This is not true. Let$X=Y=[0,1]$and let $$f(x)=\begin{cases} [0,1] \mbox{ for } x=0\\ \left\{ \sin x^{-1} \right\} \mbox{ for } 0<x\leq 1\end{cases}$$ then consider a sequence$x_{n}=\left(\frac{\pi}{2} + n\pi\right)^{-1} .$2 No they do not coincide e.g. for$X=c_0$. Then$X^*=\ell^1$and$X^{**}=\ell^\infty$. Then$\sigma(\ell^1,c_0)$is strictly coarser than$\sigma(\ell^1,\ell^\infty)$because$(\ell^1,\sigma(\ell^1,c_0))^*=c_0$and$(\ell^1,\sigma(\ell^1,\ell^\infty))^*=\ell^\infty$. 1 Tychonoff's theorem in general point-set topology states that an arbitrary product of compact topological spaces, endowed with the product topology, is again compact. Since the unit ball in a Hilbert space is weakly compact,$X$becomes compact. Your conclusion is correct I believe. 1 It suffices to allow countable intersections. Given$a\in \mathbb R$notice$\{a\}=\bigcap\limits_{n=1}^{\infty}(a-\frac{1}{n},a+\frac{1}{n})$. Being a countable intersection of open sets,$\{a\}$is open. Now take$A\subseteq \mathbb R$and notice$A=\bigcup\limits_{a\in A}\{a\}$. Being a union of open sets,$A$is open. So every subset of$\mathbb R\$ is ...