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 $\Phi^\ast$?


2 Answers 2


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


Allow me to finish answering 4 and 5 for completion.

A very typical situation that we use a Gelfand triple in theory of elliptic PDEs is when we consider the weak formulation of a $2$nd order elliptic PDE with vanishing Dirichlet Boundary condition. Here we will consider the trial/test space: $ H^1_0 (\Omega) $ where $\Omega$ is an open domain with sufficiently smooth boundary.

In this context we identify $ \big( L^2 (\Omega) \big)^* = L^2 (\Omega) $ via $L^2$ inner product and we have the embedding relation

$$ H^1_0 (\Omega) \hookrightarrow L^2 (\Omega) = \big( L^2 (\Omega) \big)^* \hookrightarrow \big( H^1_0 (\Omega) \big)^* = H^{-1} (\Omega). $$

Also in this context you can see that it makes sense to have $ H^* = \big( L^2 (\Omega) \big)^* \hookrightarrow \Phi^* = \big( H^1_0 (\Omega) \big)^* $ as functionals defined using $L^2$ functions are obviously bounded functionals on $ H^1_0 (\Omega) $ but functionals defined on $ H^1_0 (\Omega) $ can further depend on weak derivatives of the functions which will not make sense for an $L^2$ function in general.

Answer to question 4:

In most of the context, the test functions space $C^\infty_0 (\Omega)$, infinitely differentiable functions with compact support, is dense in the Banach/Hilbert space that we encounter. Therefore, instead of using the norm to describe our topology, equivalently you can use test functions with some suitable conditions:

Let $ ( X, \|\cdot\|_X ) $ be a Banach space. We define $ X^* $ to be space of distributions such that elements in $X^*$ are bounded in the following sense:
$ f \in X^* $ iff there exists a constant such that $ | \int f \phi | \leq C \| \phi \|_X $ for all $ \phi \in C^\infty_0 (\Omega) $.

The integral $ \int f \phi $ should be understood as distribution-test function pairing which is in general not a well-defined integral. Notice that if $ f \in X^* $ by this definition, then $ \int f x $ for $ x \in X $ will also make sense by density of test functions, but without knowing $ f \in X^* $, $ \int f x $ for $ x \in X $ doesn't make sense in general due to how distributions are defined.

You may recall that this is an alternative definition of dual space of $L^p$ spaces and you are right, but this can be used in a much broader context and can give you more tools to deal with the functionals. For example you can replace $ \int f \phi $ by inner product of a Hilbert space and define $H^*$ this way.

In the context of $ H^1_0 (\Omega), L^2 (\Omega) $, we can define $ \big( L^2 \big)^* $ to be distributions such that $ | \int f \phi | / \| \phi \|_{ L^2 (\Omega) } $ is bounded, which is again $L^2$ functions; we can define $ \big( H^1_0 (\Omega) \big)^* $ to be distributions such that $ | \int f \phi | / \| \phi \|_{ H^1 (\Omega) } $ is bounded. And again we can see that $ \big( H^1_0 (\Omega) \big)^* $ contains more distributions since there are less functions to test against $f$ and test functions are better. In fact test functions with bounded $H^1$-norm can absorb first order sigularities via

$$ \int f' \phi = - \int f \phi'. $$

and we can prove that $ \big( H^1_0 (\Omega) \big)^* = H^{-1} (\Omega) $ has the following characterization: If $ f \in H^{-1} (\Omega) $, then there exist $ g, g_1, \ldots, g_d \in L^2 (\Omega) $ such that $ f = g + \sum_{i=1}^d \partial_i g_i $, which at the same time shows that elements of $ H^{-1} (\Omega) $ are in general not functions.

Now, this definition at the same time defines a topology. Since $ f \rightarrow | \int f \phi | $ defines a seminorm on $ X^* $ for any test function $ \phi $, this family of seminorms induces a locally convex topology, which is the so-called weak$^*$ topology. In general context, this definition is inevitable as in a lot of application, like the one I mentioned, elements of dual space cannot be realized as functions. eg. dual space of classical function spaces $C^k$, dual space of $ l^\infty, L^\infty $ are not spaces of functions but spaces of distributions (or measures sometimes).

However, this topology is not really important in this context. Wiki simply wanted to elaborate that dual of $\Phi$ is understood in distributional sense in general.

Answer to question 5:

The last few sentences that you quoted means the following:

First, as explained above, elements of $\Phi^*$ are defined as distributions and they correspond to linear functionals by

$ \phi \mapsto \int f \phi $ or $ \langle f, \phi \rangle $ in wiki's notation.

Among all the elements of $\Phi^*$, if $ f \in \Phi^* $ and can be realized as above with $ \phi \rightarrow v \in H $, then as the discussion above suggests, $ | \int f v | / \| v \|_H $ is bounded and $ f \in H^* $. For this to be possible, $\Phi$ has to be a dense subspace of $H$ otherwise we cannot extend $ \int f \phi $ to $ \int f v $ using continuity.

To my knowledge, "faithfully represented as distribution" simply means that this representation of elements in $H^*$ via distributions is unique, i.e. every linear functional in $H^*$ corresponds to a unique distribution. The article stated this so that when we identify $H^*$ with $H$, we at the same time identified the space of the distributions defined by $ v \mapsto \langle f, v \rangle $ and this allow us to identify $H^*$ as subspace of $\Phi^*$, i.e. $H^*$ embeds naturally into $\Phi^*$ as spaces of distributions.


I will describe why this triple construction is important.

By the explanation above, if $ Y \subset X $ are two Banach spaces (or normed vector spaces) then we have $ X^* \hookrightarrow Y^* $ naturally. This embedding makes perfect sense in the distributional sense.

What if now both $ X, Y $ are Hilbert spaces? as you've seen in our example on top, both $ H^1_0 (\Omega), L^2 (\Omega) $ are Hilbert spaces. We know that we can always identify dual of a Hilbert space to itself via Riesz Representation Theorem. But this will then mean that $ Y \subset X \cong X^* \subset Y^* \cong Y $ which implies that $ X \cong Y $. Well this seems wrong and not wrong at the same time so what's happening?

On one hand, two separable Hilbert spaces being isomorphic to each other is a known fact since you can just construct an isomorphism using coefficients of orthonormal basis. On the other hand, this is strange in that it seems to conflict the conclusion we have above and the problem is hidden in the identifications $ X \cong X^* $ and $ Y \cong Y^* $. We are using different inner products when we apply Riesz Representation Theorem in these two identification.

Therefore, in order to keep the inclusions we had as in the case of Banach spaces, we need to fix an inner product instead of using both. There are two parts in the definition of this triple worth emphasizing: first, when we embed $\Phi$ into $H$, the inclusion map is natural, i.e. the identity map, not an isomorphism contructed using coordinates etc; second, we only identified $ H^* \cong H $, not $ \Phi^* \cong \Phi $ also.

We have very good reason of doing this. In the example above in elliptic PDEs, if we identify both $ \big( L^2 (\Omega) \big)^* \cong L^2 (\Omega) $ and $ \big( H_0^1 (\Omega) \big)^* \cong H_0^1 (\Omega) $. In view of $ \langle \cdot, \cdot \rangle_{ H_0^1 (\Omega) } $, elements of $ \big( H_0^1 (\Omega) \big)^* $ are again functions in $ H_0^1 (\Omega) $, but in view of $ \langle \cdot, \cdot \rangle_{ L^2 (\Omega) } $, elements of $ \big( H_0^1 (\Omega) \big)^* $ are distributions in $ H^{-1} (\Omega) $. In the other word, we are identifying some $ H^{-1} (\Omega) $-distributions to $ H_0^1 (\Omega) $-functions. And we do not like this. If a function is a $ H_0^1 (\Omega) $-function, we want it to be a $ H_0^1 (\Omega) $-function through out the calulation. We don't want a function to be a $ H_0^1 (\Omega) $-function on LHS and a $ H^{-1} (\Omega) $-distribution on RHS in an equation. That will make the calculation inconsistent and complicated.


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