$5$ questions on the definition of the Gelfand triple Let $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a Hilbert space over $\mathbb F\in\left\{\mathbb R,\mathbb C\right\}$, $\left\|\;\cdot\;\right\|$ be the norm induced by $\langle\;\cdot\;,\;\cdot\;\rangle$ and $\Phi$ be a subspace of $H$.

Question 1: Why can we find a finer topology $\tau$ on $\Phi$ such that $$\iota:(\Phi,\tau)\to(H,\left\|\;\cdot\;\right\|)\;,\;\;\;x\mapsto x\tag 1$$ is continuous?
Question 2: Why is it no loss to assume that $\Phi$ is dense in $(H,\left\|\;\cdot\;\right\|)$?

Now, let $$\Phi^\ast\stackrel{\text{def}}=\left\{f:\Phi\to\mathbb F\mid f\text{ is continuous and linear}\right\}\tag 2$$ denote the dual space of $\Phi$. Then, for all $f\in\Phi^*$ there is exactly one $\phi\in\Phi$ such that $$f\equiv\langle\;\cdot\;,\phi\rangle\tag 3$$ by the Fréchet-Riesz representation theorem.
Let me quote from the Wikipedia article about the Gelfand triple:

We consider the inclusion of dual spaces $H^\ast$ in $\Phi^\ast$. The latter, dual to $\Phi$ in its 'test function' topology, is realised as a space of distributions or generalised functions of some sort, and the linear functionals on the subspace $\Phi$ of type $$\phi\mapsto\langle v,\phi\rangle$$ for $v$ in $H$ are faithfully represented as distributions (because we assume $\Phi$ dense).

I can't make much sense of this paragraph.

Question 3: In $(1)$ we had considered the inclusion of $\Phi$ in $H$. Why do we now consider the inclusion of $H^\ast$ in $\Phi^\ast$? Moreover, given the definition of the dual space in $(2)$, we won't have $H^\ast\subseteq\Phi^\ast$ unless $\Phi=H$. So, what is meant by inclusion here?
Question 4: What do they mean by 'test function' topology? Is that just a fancy name for $\tau$?
Question 5: I have no idea what they mean in the last sentence. I'm not familiar with distributions. Is this somehow related to $(3)$? And why do we need the density of $\Phi^\ast$?

 A: Answer to question 1: Let $\left|\;\cdot\;\right|$ be a norm on $\Phi$. Since $\iota$ is linear, it is continuous if and only if $$\left\|\phi\right\|\le c|\phi|\;\;\;\text{for all }\phi\in\Phi\tag 1$$ for some $c>0$. Since $\Phi$ is a subset of $H$ we can choose $\left|\;\cdot\;\right|$ to be the restriction of $\left\|\;\cdot\;\right\|$ to $\Phi$ and $c=1$. Now we can choose $\tau$ to be the topology induced by $\left|\;\cdot\;\right|$. It's clear, that the topology of open sets in $(\Phi,\left\|\;\cdot\;\right\|)$ is contained in $\tau$, i.e. $\tau$ is finer and hence $\iota$ is a continuous embedding.

** Answer to question 2**: I'm unsure why they state, that it's no loss to assume that $\Phi$ is dense in $(H,\left\|\;\cdot\;\right\|)$. However, they might mean the following:

Let $(\;\cdot\;,\cdot\;)$ be the inner product on $\Phi$ derived from $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$. Then, there is a Hilbert space $\tilde H$ containing a dense subspace $\tilde\Phi$ such that $\tilde H$ is unique up to isometric isomorphy and $\tilde\Phi$ is isometrically isomorph to $\Phi$. $$\tilde H=:\overline\Phi^{(\;\cdot\;,\cdot\;)}$$ is called completion of $\left(\Phi,(\;\cdot\;,\cdot\;)\right)$.

Now, we might be willing to replace $(H,\Phi)$ by $(\tilde H,\tilde\Phi)$. In the mentioned sense, it's "no loss" to assume the density. At least if the main object of interest is $\Phi$ (and not $H$).

Answer to question 3: Clearly, $$\left.f\right|_{\Phi}\in\Phi^\ast\;\;\;\text{for all }f\in H^\ast\;.$$

Let $(X,\left\|\;\cdot\;\right\|_X)$ be a normed space $\Rightarrow$ $$\left\|f\right\|_{X^\ast}:=\sup_{\left\|x\right\|_X=1}|f(x)|$$ is a norm on $X^\ast$. 

I will assume, that $\tau$ is generated by a norm on $\Phi$. According to the Wikipedia article, we should be able to prove the following result without this assumption. Maybe someone else is able to provide an answer targeting this issue.
Since $$\iota^\ast:H^\ast\to\Phi^\ast\;,\;\;\;f\mapsto\left.f\right|_{\Phi}\tag 2$$ is linear and $$\left\|\iota^\ast(f)\right\|_{\Phi^\ast}\le\left\|f\right\|_{H^\ast}$$ by definition of the supremum, $\iota^\ast$ is continuous. Now, we need to prove, that $\iota^\ast$ is injective:


*

*Let $f\in H^\ast$ and $g:=\left.f\right|_{\Phi}$

*Since $\Phi$ is dense in $(H,\left\|\;\cdot\;\right\|)$, for all $x\in H$, there is a sequence $(\phi_n)_{n\in\mathbb N}$ such that $$\left\|\phi_n-x\right\|\stackrel{n\to\infty}\to 0$$ and hence (since $f$ is continuous) $$|g(\phi_n)-f(x)|\stackrel{n\to\infty}\to 0$$

*Thus, $f$ is uniquely determined by its values on $\Phi$, i.e. $\iota^\ast$ is injective


So, we can conclude, that $H^\ast$ is continuously embedded into $\Phi^\ast$, $$H^\ast\hookrightarrow\Phi^\ast\;,\tag 3$$ which is most probably what they mean by "$H^\ast\subseteq\Phi^\ast$".

Question 4 and Question 5 are still open and might be answered by someone else. However, let me repeat the fact that for each $f\in H^\ast$ there is exactly one $x=x(f)\in H^\ast$ such that $$f\equiv\langle\;\cdot\;,x\rangle$$ and hence $$H^\ast\to H\;,\;\;\;f\mapsto x(f)\tag 4$$ is injective. Since $\langle\;\cdot\;,x\rangle\in H^\ast$ for all $x\in H$, $(4)$ is even bijective. Thus, we can identify $H^\ast$ and $H$ and summarize $$\Phi\hookrightarrow H\cong H^\ast\hookrightarrow\Phi^\ast\;.$$
