# Dual Bases: Finite Versus Infinite Dimensional Linear Spaces

Motivation

I know that in a finite dimensional Euclidean space $\Bbb{E}^n$, for every basis $G=\{g_1,g_2,…,g_n\} \subset \Bbb{E}^n$ we can define a dual basis $G’=\{g^1,g^2,…,g^n\} \subset \Bbb{E}^n$ such that

$$g_i \cdot g^j = \delta_{i}^{j} \tag{1}$$

Also, it can be proved that such a basis exists and it is unique. The main advantage of dual bases is that when we write an arbitrary vector as a linear combination of the original basis $G$ then we can obtain the coefficients of the linear combination by just using the orthogonality property of $G$ and $G’$ like below

\begin{align} x &= \sum_{i=1}^{n}x^i g_i \\ x \cdot g^j &= \sum_{i=1}^{n} x^i g_i \cdot g^j = \sum_{i=1}^{n} x^i \delta_{i}^{j} = x^j \\ x &= \sum_{i=1}^{n}(x \cdot g^i) g_i \end{align} \tag{2}

Now, let us go into the infinite dimensional space of infinitely differentiable functions $f(x)$ over the interval $[-a,a]$. We all know that the eigen-functions of the Sturm-Liouville operator can form an orthonormal basis for such functions with proper boundary conditions at the end points. However, this is really a nice operator, a self-adjoint one which produces orthonormal eigen-functions. In some Partial Differential Equations (PDEs), we encounter non-self-adjoint operators and we should expand our solution and boundary data in terms of their eigen-functions which unfortunately are not orthogonal anymore!

Just to give an example, consider the following biharmonic boundary value problem (BVP)

$$\begin{array}{lll} \Delta^2 \Phi=0 & -a \le x \le a & -b \le y \le b \\ \Phi(a,y)=0 & \Phi_x(a,y)=0 & \\ \Phi(-a,y)=0 & \Phi_x(-a,y)=0 & \\ \Phi(x,b)=f(x) & \Phi_y(x,b)=0 & \\ \Phi(x,-b)=f(x) & \Phi_y(x,-b)=0 & \\ \end{array} \tag{3}$$

where we have the symmetry $f(-x)=f(x)$. Also, for the sake of continuity of boundary conditions at the corners we require that

$$f(a)=f(-a)=f'(a)=f'(-a)=0 \tag{4}$$

Then solving this BVP leads to the following eigen-value problem

$$\left( \frac{d^4}{dx^4}+2\omega^2\frac{d^2}{dx^2}+\omega^4 \right)X(x)=0 \\ X(a)=X(-a)=X'(a)=X'(-a)=0 \tag{5}$$

which has a non-self adjoint operator. The eigen-functions are known as Papkovich-Fadle eigen-functions. They can form a basis for the infinite dimensional space of infinitely differentiable functions $f(x)$ satisfying $(4)$ over the interval $[-a,a]$. As I told before, these eigen-functions are not orthogonal and this makes finding the coefficients $c_i$ of the expansions

$$f(x)= \sum_{i=1}^{\infty} c_i X(\omega_i;x) \tag{6}$$

really difficult leading to solve an infinite system of linear algebraic equations for the unknown coefficients $c_i$!

Questions

$1.$ Is there a dual basis thing for the basis $X(\omega_i;x)$’s that can make the computation of $c_i$’s easier? To be more specific, is there some basis $Y(\omega_j;x)$ such that

$$\int_{-a}^{a} X(\omega_i;x) Y(\omega_j;x) dx =\delta_{ij} \tag{7}$$

which can be considered to have the similar role of $g^j$. If such a thing existed then we could compute the $c_i$’s by using $(6)$ and the orthogonality in $(7)$ easily

$$\int_{-a}^{a} f(x) Y(\omega_j;x) dx = \sum_{i=1}^{\infty} c_i \int_{-a}^{a} X(\omega_i;x) Y(\omega_j;x) dx = \sum_{i=1}^{\infty} c_i \delta_{ij} = c_j \tag{8}$$

$2.$ If the answer to question $1$ is YES, how it can be computed? If NO, Why?

#### Solutions Collecting From Web of "Dual Bases: Finite Versus Infinite Dimensional Linear Spaces"

General Remarks on the dual space of $L^2$

It seems reasonable to consider your operator as an operator in $\mathcal H= L^2([-a,a]\times[-b,b])$, which is the Hilbert space of all measurable functions where $$\langle f,f\rangle=\int_{[-a,a]\times [-b,b]}f(x,y)\overline{f(x,y)}dx dy < \infty.$$

This Hilbert space has a nice relation to its dual, namely you can identify it with itself:

For $f \in \mathcal H$, we can define an element of the dual via

$$f'(g):= \langle g,f \rangle =\int_{[-a,a]\times [-b,b]}g(x,y)\overline{f(x,y)}dx dy,$$

defined on all $g\in \mathcal H.$

Vice versa, for every $f \in \mathcal H’$ there’s an $f \in \mathcal H$ such that the equation above holds (cf., e.g. Conway, A Course in Functional analysis, 1990, Theorem 3.4 (The Riesz Representation Theorem)).

Note that neither $C^0$ nor the space of piecwise continuous functions equipped with this innner product is a Hilbert space. They are subsets of $L^2$, and thus, their duals are even larger than the dual of $L^2$. (Here, I’m assuming a bounded domain $[-a,a]\times [-b,b]$ and only finitely many jump points for the piecewise functions).

Your differential operator (let’s call it $A$) can be considered as an operator in $\mathcal H$. In this case, you need to specify the domain of $A$ (as a subset of $\mathcal H$) such that the functions $f$ are in $\mathcal H$ and $Af$ still lies in $\mathcal H$. The vector space of piece-wise continuous functions is a subspace of $\mathcal H$, but the first derivative doesn’t need to exists. (I’m no PDE expert, but I’d expect some subset of $C^1$ with some additional constraints that the second derivative also exists.
In the case of Sturm-Liouville differential expressions (like $-(pf’)’+qf$), one usually takes functions $f$ which are absolutely continuous (i.e. the first derivative is locally summable), where $pf’$ is absolutely continuous and $-(pf’)’+qf$ lies in $L^2([a,b])$.