linear combination - finite, infinite countable, and continuous

Question

I am a beginner student of functional analysis. We learn that, if $X$ is a vector space over $\mathbb{F}$ of finite dimension, it means it can be generated from a finite base, $V \subset X$, which means, for some $x \in V$ and some $a_k \in \mathbb{F}$, we can represent $x$ as a finite linear combination

$$x = \sum_{k=1}^N a_k \; v_k$$

On the other hand, for instance, in the field of Hilbert spaces, lets suppose we have an infinite countable vector space and an infinite orthogonal basis, so we can write

$$x = \sum_{k=1}^\infty a_k \; v_k$$

My question is if it is possible to generalize the concept of linear combination like this

$$x = \int_{-\infty}^{\infty} a(t) \; v(t) dt$$

meaning, $v(t)$ would be a continuum base that generates an infinite uncountable Hilbert space. I am thinking about, for instance, the Fourier transform as a sort of generalized continuum linear combination of complex exponentials $e^{iwt}$, weighted by a real or a complex function $a(t)$. Besides, I've read that $e^{iw_1 t} \perp e^{iw_2 t}$, $w_1, w_1 \in \mathbb{R}$, if we consider $\langle x(t),\;y(t)\rangle = \int_{-\infty}^{\infty} x(t)y(t)\;dt$.

Does this make sense? If yes, where can I read about it?

Thank you.

Joao

Answer 1

This is not a complete answer, but is to long for a comment.

There are several issues with such a definition of basis. In general, we can expect an uncountable basis to be indexed by some set $\Omega$ and if we want a definition of basis as you wish, we need to integrate over $\Omega$, that is $\Omega$ can be, for example, a measure space, and the integral used to define the expresion of a vector as a ''linear combination'' of vectors in this basis is a Bochner integral.

But let me be more concrete. Imagine we have a basis $\{v(t)\}_{t\in I}$, where $I$ is some interval, of a Hilbert space $H$ and we have the property that for all $v\in H$ there exists a function $a:I\to\mathbb{F}(=\mathbb{R},\mathbb{C})$ such that $$ v = \int_I a(t)v(t) \; dt. $$ In the case of Hilbert spaces we are specially interested in the case of orthonormal bases, so let's assume we have some sort of Gram-Schmidt procedure in this case so we can effectively obtain an orthonormal basis (or perhaps we need to invoke Zorn's lemma). So assume that $\langle v(t),v(s)\rangle = \delta_{ts}$, there $\delta_{ts}$ is the Kronecker symbol. Then we have, for $r\in I$, an expression of the form $$ v(r) = \int_I a(r,t)v(t)\; dt $$ and consequently $$ 1 = \|v(r)\|^2 = \left\langle \int_I a(r,t)v(t)\; dt,\int_I a(r,s)v(s)\; ds \right\rangle = \int_{I}\int_I a(s,t)a(r,s)\langle v(t),v(s)\rangle \; dt ds, $$ where I used the expeted bilinearity property of the inner product. Now let $\Delta=\{(t,s)\in I\times I \; | \; t=s\}$ and let $$ \chi_\Delta(t,s) = \begin{cases} 1 & \text{if } (t,s)\in\Delta,\\ 0 & \text{if } (t,s)\not\in \Delta. \end{cases} $$ Then we obtain $$ 1 = \int_I\int_I a(r,t)a(r,s)\chi_\Delta(t,s)\; dtds, $$ but $\Delta$ is a set of measure zero, so this integral equals zero, that is $1=0$, which is absurd.

Thus at least we cannot hope to have such a theory of bases for Hilbert spaces with the property of having orthonormal basis.

Now, I cannot see the Fourier transform as an analog of a basis. To being with, the function $x\mapsto e^{itx}$ is not an element of $L^2(\mathbb{R})$ so we cannot expect this functions to be a basis of $L^2(\mathbb{R})$. So we need to restrict to functions in some bounded closed interval, $[-\pi,\pi]$ say. In this case we do have $x\mapsto e^{itx}$ as elements of $L^2([-\pi,\pi])$, but they are not orthogonal because under the inner product $$ \langle f,g\rangle = \int_{-\pi}^{\pi} f(x)\overline{g(x)}\; dx $$ they are not orthogonal: actually if $s\neq t$ we have $$ \langle e^{itx},e^{isx}\rangle = \frac{2\sin((t-s)\pi)}{t-s} $$ (I hope I made no mistake) which is different from zero unless $t-s$ is an integer. Thus we again arrived the issue that we loss an orthonormal basis.

I think this is maybe the tip of the iceberg (and maybe I'm wrong). I don't know if there are another patologies with such a notion of basis, and I know no reference about it.

To finish my (long) comment: We do have uncountable ortonormal basis for Hilbert spaces, but the definition is different: Any such basi $(v_i)_{i\in I}$ has the property that for all $v\in H$ there exists a countable subset $I'\subseteq I$ such that $$ v = \sum_{i\in I'} \langle v,v_i\rangle v_i. $$ (This is actually not the definition but a consequence of it.)