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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 ...

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

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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 $(s+t)({s\over{s+t}}x+{t\over{s+t}}y)=sx+ty\in (s+t)A$

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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 ...

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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 $e^{-2i \pi ... 1 You have answered almost everything by yourself. Let$\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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