# Laplace transform/hilbert space

1. Jul 25, 2006

### jbusc

I was just thinking back to my advanced linear algebra class and I remember the prof mumbling something about the fourier series being an orthonormal basis for the hilbert space of square-integrable real-valued functions, lebesgue something, etc., and demonstrating the properties of the integral inner product and projections into that space. It all made sense at the time, I think. :) Sorry if I'm kinda vague, I can provide more info if needed.

I am curious, though, if something similar exists for laplace transform. It certainly looks like it fits the form, but I have been unable to find any detailed analysis of such. Anyone have insight?

2. Jul 25, 2006

### quasar987

3. Jul 25, 2006

### quasar987

Some of the following may be nonsense but...

The way I interpret the article is that $\{exp(inx)\}_{n\in \mathbb{Z}}$ is a complete orthonormal basis for the space of real valued square integrable smooth function on $(-\pi,\pi)$ of domain $(-\pi,\pi)$ with inner product defined by

$$\langle f,g \rangle =\int_{-\pi}^{\pi}f(x)g(x)dx$$.

This allows us to write, for any function f of that space,

$$f(x) = \sum_{n\in \mathbb{Z}}\langle f,\exp(inx) \rangle \exp(inx)$$

Using the language of vector spaces, <f,exp(inx)> is the projection of the vector f in the direction of the unit vector $\exp(inx)$, i.e. the component of f in the direction of $\exp(inx)$

More generally,

$$\left\{ \exp \left( in\frac{2\pi}{P}x \right) \right\} _{n\in \mathbb{N}}$$

is a complete orthonomal basis for the space of real valued periodic functions of period P.

For functions that are NOT periodic, but that have the property that the integral

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)\exp (-in\omega)dx = F(\omega)$$

converges, we can write them in a kind of "continuous" form of a Fourier series, i.e. as its Fourier transform:

$$f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}F(\omega)\exp (inx)d\omega$$

Now let's make the analogy with the Laplace transform. Suppose a function $f:(0,+\infty)\subset D \rightarrow \mathbb{R}$ can we written as the Laplace transform of some function a(t):

$$f(s) = \int_0^{+\infty} a(t) \exp(-st)dt$$

Note that $\exp(-st)=(e^{-s})^t$. Make the substitution $e^{-s}=x$ and the equation becomes a restriction of f to the positive real line:

$$f|_{\mathbb{R}^+} = f(x) = \int_0^{+\infty} a(t) x^t dt$$

This is a kind of "continuous" form of a power series, right?

...the analogue is the set $\{x^n\}_{n\in \mathbb{N}}$, which is a complete orthonormal basis for, say, the space of real valued function developable in a Taylor series of convergence radius R and of domain (-R,R), with the inner product defined as...

...as what exactly? Also, maybe $\{x^n\}_{n\in \mathbb{N}}$ is not orthoNORMAL, but just orthogonal. What inner product would yield

$$\langle f,x^n \rangle = \frac{f^{(n)}(0)}{n!}$$

and

$$\langle x^m,x^n \rangle = 0$$

except for m=n

??

Last edited: Jul 25, 2006
4. Jul 25, 2006

### jbusc

Ohh I see! I still have to process all that but just skimming it makes sense. Thanks!

5. Jul 25, 2006

### jbusc

Hmm, I vaguely remember this. But isn't $\{x^n\}_{n\in \mathbb{N}}$ neither orthonormal nor orthogonal? You have to apply Gram-Schmidt for that to be true, right?

As for which inner product...I have to think about it for a minute...

hmm.the laplace transform? is that what I'm missing? I think I'm overthinking it...

Last edited: Jul 25, 2006
6. Jul 27, 2006

### quasar987

At least it's a complete set, we know that much! :p

(Also, the Fourier integrals in post #3 are with 'x' in the exponential and not 'n')

Last edited: Jul 27, 2006