# linear combination - finite, infinite countable, and continuous

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

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.)

• Thank you Albert, for clarifying my confusion. When we take the Fourier Series and extend the function period to infinity, the integration interval is not finite anymore. Besides, if we calculate the Fourier Transform of a complex exponential, we get a scaled and shifted dirac delta, which, despite being a generalized function, it could never be zero. May 29 at 19:01