$5$ questions on the definition of the Gelfand triple

Solution 1:

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 $$\Phi\hookrightarrow H\cong H^\ast\hookrightarrow\Phi^\ast\;.$$