Dual space of continuous functions and meaning of $\langle u,\phi \rangle$

Question

Definition. Let $D(\Omega)$ be the vector space of $C^\infty$ functions with compact support in $\Omega$ and $D^\prime$ be its dual space. A linear mapping

$$ u:D(\Omega) \ni \phi \mapsto \langle u,\phi \rangle_{D^\prime, D} \in \mathbb{R},\text{or} \hspace{1mm}\mathbb{C} $$ is a distribution on $\Omega$ if and only if ...

---

My question is specifically about the notation $\langle \cdot,\cdot \rangle$. Does $\langle u,\phi \rangle$ mean inner product of $u$ and $\phi$? It cannot be a inner product since $u$ and $\phi$ are in different spaces and according to inner product they must be in the same space. So what does $\langle u,\phi \rangle$ represent(or mean)?

More importantly, a concrete example for my question above would be really helpful!

Answer 1

Let $U, V, W$ be three vector spaces over $F$.

Then the map $B:U\times V\to W$ which is linear in the both slot is called a bilinear map.

$B_u:V\to W$ and $B_v:U\to W$ is linear for all $u\in U$ and $v\in V$ respectively.

If $W=F$ , then the bilinear map $B:U\times V\to F$ is called a bilinear pairing.

We denote $B(u, v) $ by $\langle u, v\rangle$.

---

Example: $V$ be a vector space and $V'$ is it's dual.

Then $\langle \cdot,\cdot\rangle :V'\times V\to F$ defined by $$\langle f, v\rangle=f(v) $$

is a bilinear pairing.

---

In particular ,

$D(\Omega)$: space of all test functions.

$D'(\Omega) $ : Continuous dual of $D(\Omega) $

$\langle \cdot,\cdot\rangle :D'(\Omega)\times D(\Omega)\to \Bbb{R}$ defined by $$(u,\varphi)=\langle u, \varphi\rangle$$

is a bilinear pairing.

---

Suppose $f\in L^{1}_{\textrm{loc}}(\Omega)$ then $ \langle u_f, \varphi\rangle=\int_{\textrm{supp}(\varphi) }f(t) \varphi(t) dt$

Answer 2

$ \langle u, \phi \rangle$ is simply the value of $u$ at $\phi$. For example if $\mu$ is a finite Borel measure on $\Omega$ then $\langle u, \phi \rangle=\int_{\Omega}\phi d\mu$ defines distribution $u$, a special case of this being $\langle u, \phi \rangle=\int_{\Omega} f(x)\phi (x) dx$ where $f$ is an integrable function on $\Omega$ w.r.t Lebegue measure.

Answer 3

it's just a mark, which represents $u(\phi)$

Answer 4

Let me mention a few comments.

Let $E$ be a real vector space, and $F$ a subspace of the set of linear forms on $E$.

1. It is important to notice that

$\begin{array}{rcl} E \times F &\rightarrow &\mathbb{R}\\ (v,\lambda) &\mapsto &\lambda(v)\\ \end{array}$

is a bilinear map. Moreover, it is quite common, in bilinear algebra, to denote bilinear maps by brackets, or parentheses, etc. so this justifies a bit the notation.

2. Consider now a subspace $G$ of the space of linear forms on $F$; moreover, assume that for every $v \in E$, $eval_v := \lambda \mapsto \lambda(v)$ is in $G$. That is, $G$ can be thought as a kind of bidual of $E$.

We then trivially have, for all $v,\lambda$, $\langle eval_v,\lambda\rangle = \lambda(v) = \langle \lambda ,v\rangle$. That is, if we identify $v$ and $eval_v$, we get, for all $v$ and $\lambda$, $\langle \lambda,v\rangle = \langle v,\lambda\rangle$. That is, in this precise sense, $\langle \cdot,\cdot \rangle$ is kind of symmetric!

3. Assume moreover that $E$ comes with an inner product $\langle \cdot,\cdot\rangle_E$. Now, for every $v \in E$, we can build a linear form $\lambda_v$ by the formula $w \mapsto \langle v,w\rangle_E$. Keeping using this notation, we have, for all $v,w \in E$, $\langle \lambda_v,w\rangle = \langle v,w\rangle_E$. If we identify $v$ with $\lambda_v$, we just get, for all $v,w \in E$, $\langle v,w\rangle_E = \langle v,w\rangle$.

4. Mentally speaking, it can sometimes be fruitful to think of linear forms on a vector space $E$ endowed with an inner product as generalized vectors. The notation $\langle \cdot,\cdot \rangle$ helps.

Now, take $E := D(\Omega)$ be the space of continuous, compactly-supported functions on $\Omega$ and $F$ be the space of continuous linear forms on $E$, that is, $F$ is the space of distributions. Define $\langle \cdot,\cdot\rangle_E$ to be the integral of the product. Then all I've said applies. Moreover, distributions were invented to serve as generalized functions! More precisely, if one is not able to prove that some PDE's have solutions, it is sometimes easier to prove that they have generalized solutions, that is, there are generalized functions that are solutions to a variant of the PDE, and only then (sometimes) prove that these generalized functions are in fact real ones. As far as I understand, this is the philosophy behind distributions.

In summary, at a first glance it might look that $\langle \lambda,v\rangle$ is just a fancy variant for $\lambda(v)$; but the graphical symmetry of the notation $\langle \lambda,v\rangle$ sticks much better to the intuition of generalized things.