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I have a question I am phrasing in terms of expansion of a function in terms of Hermite polynomials but it applies to other expansions as well.

First I establish my convention for Hermite polynomials which is

$$ (-1)^n \partial_z^n \varphi(z)= H_n(z) \varphi(z) $$

where $\varphi(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{z^2}{2}}$. Under this the first few Hermite polynomials are

$$ \begin{eqnarray} H_0 &=& 1 \\ H_1 &=& z \\ H_2 &=& z^2-1 \\ \dots \end{eqnarray} $$

Now we can expand a function in Hermite polynomials

$$ f(z)=\sum_{n=0}^\infty c_n H_n(z) $$

and we can extract the coefficients by

$$ c_n = \frac{1}{n!} \int_{-\infty}^\infty dz~ \varphi(z) H_n(z) f(z) $$

All this is fine. What I want to know what is the problem with the following. Suppose I have the function value only in the interval $z \in [a,b]$ and we compute

$$ \begin{eqnarray} \tilde c_m &=& \frac{1}{m!} \int_{a}^b dz~ \varphi(z) H_m(z) f(z) \\ &=& \frac{1}{m!} \int_{a}^b dz~ \sum_{n=0}^\infty c_n ~\varphi(z) H_m(z) H_n(z) \\ &{\bf \color{blue}{\stackrel{?}{=}}}& \frac{1}{m!} \sum_{n=0}^\infty c_n \int_{a}^b dz~ ~\varphi(z) H_m(z) H_n(z) \\ &=& \frac{1}{m!} \sum_{n=0}^\infty \gamma_{m,n} c_n \end{eqnarray} $$

where I have defined

$$ \gamma_{m,n} := \int_{a}^b dz~ ~\varphi(z) H_m(z) H_n(z) $$

which can be evaluated explicitly. My question is I can have two functions that are the same in the interval $z \in [a,b]$. $\gamma_{m,n}$ are independent of the function and therefore I cannot see how the above can be true. Said differently it seems I can invert the relation and from $\tilde c_m$ get the true $c_n$ which is ridiculous because I should not be able to read of the full function on the real line from the value inside an interval. So what am I doing wrong? I suspect the problem is with the step that I put a question mark on but what is specifically the problem?

Borun Chowdhury
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    How do you know that the mapping from the $c_n$ to the $\tilde c_m$ is invertible? – Hans Lundmark Sep 06 '18 at 13:42
  • Yes I don't know if $\frac{1}{m!} \gamma_{m,n}$ is invertible given that its infinite dimensional matrix. Any ideas on how to check this? – Borun Chowdhury Sep 06 '18 at 13:54
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    Well, I havent thought deeply about your question, but I suspect that in fact the mapping is not invertible in general, for exacly the reason that you give: there are many different functions with the same restriction to $[a,b]$. – Hans Lundmark Sep 06 '18 at 14:01
  • I guess I could simplify the problem by taking a finite box and using $\sin,\cos$ as the basis. Then $\gamma_{m,n}$ would be more tractable but would still be an infinite dimensional matrix and I wouldn't know how to check if its invertible or not. Having said that the possibility of several functions that have same restriction in $[a,b]$ is rigorous enough for me. Thanks. – Borun Chowdhury Sep 06 '18 at 15:10

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You are correct in that a general function cannot be determined by its values on any finite interval, but if you know that the function is given by a power series, for example, then it is determined by its values on any interval, however small. This isn't really all that surprising. If we know that a function is a quadratic polynomial, then it is determined by its values at three points.

saulspatz
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  • Yes but in my case I do not know anything about $f(z)$ to begin with. All I am saying is that any function on the real line can be expanded in Hermites. Now it seems from the way I wrote above that knowing it in an interval allows me to extract all the coefficients (which is clearly wrong). – Borun Chowdhury Sep 06 '18 at 13:44