Can I perform finite-dimensional inner products with functions as vector components?

Question

I'm an engineer. I'm struggling with infinite-dimensional linear spaces and would appreciate some help.

I will describe my question as an example because I'm unfamiliar with the subject.

Let's say $u(x)$, $v(x)$, $w(x)$, and $r(x)$ are functions of class $C^2$, defined in the interval $[0,1]$. I want to see them as vectors of an infinite-dimensional linear space, call it $\mathfrak{m}_\infty$.

Also, $\mathbf{u}$ and $\mathbf{v}$ are vectors, but from a Euclidean 2-dimension space, $\mathbb{V}_2$.

In $\mathfrak{m}_\infty$, I can define the following scalar product

$\langle u(x),v(x)\rangle=\int_0^1 v(x)u(x) dx$.

In $\mathbb{V}_2$, I can also define a scalar product by

$\langle \mathbf{u},\mathbf{v}\rangle=\mathbf{v}^T \mathbf{u}$.

Now, I want to mix things and define the function $\langle,\rangle$,

$\langle \begin{bmatrix}u(x)\\v(x)\end{bmatrix},\begin{bmatrix}w(x)\\r(x)\end{bmatrix}\rangle=\int_0^1 \begin{bmatrix}w(x)\\r(x)\end{bmatrix}^T\begin{bmatrix}u(x)\\v(x)\end{bmatrix} dx$.

Now, the questions.

1. Is $\langle,\rangle$ a scalar product? If it is, over which space?

2. In case only $r(x)$ is $C^1$($u$, $v$, and $w$ remain $C^2$), how would the answer to the first question change?

I would be pleased if you could also indicate a bibliography so I could study the subject.

Answer 1

I'll try to give a not so technical answer, hoping it will be of help.

In general let $V$ be an inner product space, with inner product $(\cdot,\cdot)$, let $\Omega$ be an appropriate subset of $\mathbb{R}^n$ (appropriate in this context means that it serves for your interests. It can be open, closed, measurable, etc.). You can construct, for example, the space $L^2(\Omega,V)$ that consists of functions $f:\Omega\to V$ that are square integrable (with an appropriante notion of integrability, in this case, I'm thinking of Lebesgue-integrability, but there is no harm, for your purposes to think of Riemann or Darboux integrability). This space is endowed with an inner product as follows: $$ \langle f,g\rangle = \int_\Omega (f(x),g(x))\; dx. $$

In your particular case, you can consider $\Omega=[0,1]$ and $V=\mathbb{R}^2$ with the standard inner product, that is, $(x,y)=x^Ty$. Then a function $v:\Omega\to V$ is a function $f:[0,1]\to \mathbb{R}^2$, that is, it is given by two components: $$ f(x) = \begin{bmatrix} u(x)\\ v(x) \end{bmatrix}. $$

Given another function, say $g(x)=\begin{bmatrix} w(x)\\ r(x) \end{bmatrix}$, we have $$ \langle f,g\rangle = \int_0^1 \begin{bmatrix} u(x)\\ v(x) \end{bmatrix}^T \begin{bmatrix} w(x)\\ r(x) \end{bmatrix} \; dx, $$ as you desire.

You don't need the functions $u,v,w,r$ to be of class $C^1$ or $C^2$, you only need an appropriate notion of integrability. For example, continuity suffices in this context.

All references I know does not provide major information for this, since it is assumed to be of common knowledge. May be, if you are interested in some deep comprehension of the subject, you can search for Bochner spaces, but you will need a certain amount of functional analysis to understand it.

Now, lets get back to the general case. Let's consider the space $C(\Omega,V)$ of continuous functions $f:\Omega\to V$ with the inner product given above. Assume, for simplicity, that $\Omega$ is a closed bounded subset of $\mathbb{R}^n $. It is clear that $$ \langle f,f\rangle \geq 0. $$ If $\langle f,f\rangle = 0$, then $\int_\Omega (f(x),f(x))\; dx = 0$ and since the function $x\mapsto (f(x),f(x))$ is continuous and non-negative, this implies that $(f(x),f(x))=0$ for all $x\in\Omega$, and since $(\cdot,\cdot)$ is an inner product on $V$, this implies that $f(x)=0$ for all $x\in\Omega$.

Symmetry and bilinearity are straightforward, so you indeed have an inner product.