Generalization of the Cartesian basis in Hilbert space Given the Hilbert space of continuous functions $H = \{ u : [0,1] \to \mathbb{R}\}$ with
inner product : $<u,w> = \int_0^1 u(x)w(x)dx$,
is there a generalization of a Cartesian basis ?
Intuitively, what I want to have is a basis that "picks up" only the value of $u(x)$ at $x$ each time, like (1,0) and (0,1) in $\mathbb{R}^2$
I was thinking something like $B = \{ \delta(x-l), l \in [0,1] \}$, but this is not continuous.
Thanks a lot,
Kostas
 A: I suppose a precise statement of what the OP has in mind would be:
Is there a basis $\{e_x\}_{x\in[0,1]}$ of $H$ such that, once any $f\in H$ is expanded in that basis, the coefficient of each $e_x$ is precisely $f(x)$?
In other words
$$
  f= \sum_{x\in[0,1]} f(x)e_x.
  $$
The difficulty here is the fact that this will often have uncountably many nonzero summands.  While infinite sums make perfectly good sense in normed spaces, no generalization of the theory of series allows for summing uncountably many nonzero terms.
The short answer is thus that no such basis exist.  Nevertheless the intuition behind the desire for the existence of a basis like that should not be totally ignored since it underlies many important developments in Analysis.  A similar situation is expressed by the highly intuitive although totally meaningless expression
$$
\int_a^bf(x)\, dx = \sum_{x=a}^b f(x)\Delta_x.
$$
This idea pops up again in the subtle difference between spectral values, on the one hand, and eigenvalues of a self-adjoint operator on an infinite dimensional space.
In the theory of unitary group representations this also shows up in the delicate problem of decomposing a representation as the direct sumof irreducible ones.
A: The problem is that integrals do not change when you change the integrand at a finite number of points (or, even on a set of measure $0$,) which makes it impossible for the integral inner product to give you a pointwise value for a function. You can retrieve an average value over a small interval with an integral such as $\frac{1}{2\delta}\int_{x-\delta}^{x+\delta}u(x)dx=\int K_{x,\delta}u$, but a point value is not possible; otherwise integral convergence would end up forcing pointwise convergence, which is not the case.
