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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 ...


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 $(\ell^1,\sigma(\ell^1,\ell^\infty))^*=\ell^\infty$.


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 ...


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Hint: ${s\over{s+t}}x+{t\over{s+t}}y\in A$ if $x,y\in A$ and $A$ is convex, thus $(s+t)({s\over{s+t}}x+{t\over{s+t}}y)=sx+ty\in (s+t)A$


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 ...


1

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

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 ...



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