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This question is motivated by Fourier and Laplace transforms. Recall that both transforms are intuitively representation of a function as a combination of $g_k(t) = e^{kt}$, i.e., a function $f(t)$ can be written as $\int F(k) g_k(t) dk$, where the domain of $k$ is anything you want it to be.

Now I want to pick different functions for expansion, say $h_k(t) = \cos(ke^{-t})$ for example, and I try to express $f$ on some domain by $f(t) = \int \hat F(k)h_k(t) dk$. What can I say about $\hat F(k)$? What might be conditions to impose on $f$ and/or $h_k$ that make calculation of $\hat F(k)$ possible? For example, when is it true that $\hat F(k) = \int \frac{f(t)}{h_k(t)}w(t) dt$ for some $w(t)$? (This is "heuristically" true for both Fourier and Laplace transforms, so I definitely accept heuristic suggestions.)

The reason I want to do this is because I expect, for this particular example, that representing $f$ by $\hat F$ would simplify the calculation involving this linear operator: $M(f)(t) = e^{2t}(f''(t) + f'(t))$. This is because $h_k$ is an eigenvector of $M$. I believe the motivation for Fourier and Laplace transforms is similar as $e^{kt}$ is an eigenvector of the first derivative operator.

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@LePressentiment Please stop editting posts which are older than 6 months with such trivial (and often incorrect) edits. –  Daniel Rust Aug 11 '13 at 11:10
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1 Answer

That one can do Fourier transforms has largely to do with the fact that the "basis" in which one expands a Fourier series is an orthogonal one. More precisely, one has that for a vector space $V$ with inner product $\langle,\rangle$, if $\{e_k\}$ is a basis of $V$ that is pairwise orthogonal, we can write any vector $v\in V$ as $$ v = \sum_{k} \hat{v}_k e_k $$ where $$ \hat{v}_k = \frac{\langle v,e_k\rangle}{\langle e_k,e_k\rangle} $$

(Note that in the case of the Fourier transform, the function $e^{ikx}$ is not really in the inner product space $L^2$; but morally this is how things work.)

So the first questions you need to ask yourself is: are the functions $\cos(k e^{-t})$ mutually orthogonal and do they form a basis?

  • Orthogonality: most likely, your functions are not $L^2$ orthogonal. This means that you should not blindly copy the formula above for a Fourier decomposition. A quick illustration of the fact is the following: consider the basis for $\mathbf{R}^2$ given by $\{(0,1), (1,1)\}$. This basis is not orthogonal under the standard inner product. Consider the vector $(2,3)$. Its inner products are $$ (0,1)\cdot (2,3) = 3 \qquad (1,1)\cdot (2,3) = 5 $$ So naively your procedure would give the "expansion" $$ (2,3) "=" 3(0,1) + \frac{5}{2}(1,1) = (5/2,11/2)$$ To understand why this gives a problem you are encouraged to think about how projection operators are defined in Linear Algebra, and how one generally expresses a vector inside an arbitrary basis.

  • Basis: this opens another can of worms. Do you know that all the functions you are interested in can be approximated by finite linear combinations of your proposed basis vectors? If not, it would not make sense to try to use those functions as a basis of a Fourier-like expansion.


As an aside, consider the change of variables $u = e^{-t}$. Then $$\frac{d}{dt} = \frac{du}{dt} \frac{d}{du} = - u \frac{d}{du} $$ and so $$ \left(\frac{d}{dt}\right)^2 = u\frac{d}{du}\left( u \frac{d}{du}\right) = -\frac{d}{dt} + u^2 \left(\frac{d}{du}\right)^2 $$

This means that for the question you are considering, you are much much better off just first do the above change of variables. For in the variable $u$, your operator $M$ is nothing more than $M(f) = f_{uu}$ and you can use the standard machinery of Fourier/Laplace transforms without having to worry about constructing from scratch a integral transform to suit your needs.

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Orthogonality: Of course the collection $\cos(ke^{-t})$ does not seem orthogonal. I cannot, however, say that there exist no "nice" restriction of $k$ and $t$, and some weight function that will make the collection orthogonal. For example, the functions $e^{st}$ do not seem orthogonal in the context of Laplace transform if $s$ takes on real values. The good thing comes when $s$ is restricted to be on a vertical line in a complex plane. –  Tunococ Sep 4 '12 at 21:17
    
@Tunococ: in terms of the Laplace transform, note that the Laplace and inverse Laplace transforms are not symmetric, which tells you that the Laplace transform is not simply a "projection on an orthogonal basis" the same way Fourier transform is. –  Willie Wong Sep 5 '12 at 10:41
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Also, if you read my last paragraph below the cut carefully, you'd see that in fact $\cos k e^{-t}$ are orthogonal in a weighted $L^2$ space. (Hint: the weight is just the Jacobian function...) –  Willie Wong Sep 5 '12 at 10:43
    
Oh you're right. What I'm interested in, though, is this asymmetry of Laplace transform which actually turns out to be pretty nice. The form of the inverse does not have to be like what I claimed, many integral transforms have inverses like that. That means I'm not expecting orthogonality in the usual sense. –  Tunococ Sep 5 '12 at 21:32
    
I do have a lot to thank you :) It seems like now I'm more interested in identifying the transform with the domain. As you suggested, using $\cos(ke^{-t})$ would be "essentially the same" as Fourier transform if I'm interested in $t \in \mathbb R$. So, is it possible to give a classification result of transform based on the domain (and codomain) alone? –  Tunococ Sep 5 '12 at 22:00
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