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Let $n,k\colon\mathbb{R}\to\mathbb{R}$ be real functions such that function $N$ given by $N(x)=n(x)-ik(x)$ is a holomorphic function in the upper half-plane. Also I know some additional properties of these functions:
1. $n(x)=n(-x)$ if $x\in\mathbb{R}$;
2. $k(x)=-k(-x)$ if $x\in\mathbb{R}$;
3. $k(x)>0$ if $x>0$;
4. limit $\lim_{x\to\infty}(n(x)-ik(x))$ exist.

Unfortunately I don't know functions $n$ and $k$. I know only some approximate values of $n_i\simeq n(x_i)$, $k_j\simeq k(y_j)$ for finite number of points $x_i,y_j\in [a,b]\subset\mathbb{R}$. Which methods can I use in order to get approximate values of the function $N$ on the whole region $[a,b]$? Where can I read about them?

Update (example data and an experiment):

 x_j      n(x_j)
 0.000  1.36364  
10.204  1.32231  
17.346  1.23967  
18.367  1.19835  
27.551  0.53719  
31.632  0.53719  
37.755  0.82644  
42.857  0.90909  
47.959  0.95041  
50.000  1.32231  
51.020  1.36364  
53.061  2.14876  
56.122  2.60331  
61.224  2.47934  
64.285  2.06612  
67.346  1.40496  
70.408  1.36364  
83.673  1.94215  
90.816  1.98347  
102.04  1.94215  

y_j     k(y_j)
0.000   0.00095  
3.061   0.00288  
8.163   0.00425  
11.22   0.00870  
14.28   0.01562  
24.48   0.09638  
27.55   0.18459  
30.61   0.40261  
32.65   0.52213  
47.95   1.47684  
50.00   1.68181  
53.06   1.47684  
71.42   0.12499  
78.57   0.07931  
102.0   0.04715  

I've tried to fit a $n(x)$ by $\Re(f(x))$ with $f(x)=1+\sum_i\frac{c_i^2}{x-a_i-i b_i^2}$ with $c_i,a_i,b_i\in \mathbb{R}$. Here is the result of the fitting:
alt text
Not a good fitting :(. This is actually why I ask this question. But note, I haven't used the data I have for k(x). Let's check, how is it approximated by $\Im(f(x))$ (with coeffitients from the previous fitting).
alt text
It's better, than nothing.

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Maybe the question is not clear: your unknown function is defined on (a part of) the upper complex plane $\Im z \ge 0$, and you know only a discrete set of values of its trace at $\Im z=0$? If so, I think holomorphy plays no role –  Piero D'Ancona Oct 14 '10 at 8:12
    
It is defined on the whole half-plane $\Im z\geq 0$. And yes, I know only something about the trace of the function. But from holomorphy I know some relations, for example this: en.wikipedia.org/wiki/Kramers%E2%80%93Kronig_relation –  Fiktor Oct 14 '10 at 9:44
    
Well yes, if the function is in a Hardy space then the two components are Hilbert transforms of each other. What I mean is that this might not be helpful for your problem. Think of this related example: any periodic function $\sum c_ne^{inx}$ with $c_n=0$ for $n\le 0$ and $c_n$ growing at most polynomially (so it can be a distribution) is the trace of the holomorphic function $\sum c_ne^{inx-ny}$. But the two sequences of coefficients $\Re c_n$ and $\Im c_n$ are completely independent of each other –  Piero D'Ancona Oct 14 '10 at 10:16
    
Yes, they are independent, but both can be calculated given only the real part of $f(x)=\sum c_n e^{inx}$ (if $c_n\to 0$ fast enough): $c_n=\lim_{N\to\infty} \int_{-N}^{N} \Re(f(x))e^{inx}/N dx$. I understand, that function cannot be determined by its values in the finite number of points. Nevertheless we are measuring, for example, temperature and then we are interpolating it between the points. Ok, we use some assumptions on the dependence of temperature on time. But it works. –  Fiktor Oct 14 '10 at 12:13
1  
Is the upper bound $M_1$ for $n(x)$ known, or just that it exists? The latter follows from continuity and the fact it's approaching $1$ at $\pm \infty$. –  Dylan Thurston Oct 17 '10 at 13:21

1 Answer 1

up vote 3 down vote accepted

If I understand correctly, we can make a change of variables from $z$ in the upper half plane to $\zeta$ in the unit disk. Then the function $N(\zeta)$ is continuous on the unit circle $\zeta = \exp(i \theta)$ and has a unique harmonic extension to the unit disk. If $N$ is a favorable function, then this harmonic extension is complex analytic. A function in this class is one with a Taylor series in $\zeta$ with a radius of convergence at least 1. Or, it is a Fourier series in $\theta$ which uses only terms $\exp(i k \theta)$ with $k \ge 0$, and whose coefficients satisfy a convergence condition.

In your question, in addition, the real and imaginary parts of $N(\zeta)$ satisfy certain inequalities, and they are supposed to be as close as possible to certain data points. Let's assume a least-squares error model for these data points. If you put all of this information together, the result is that the Fourier coefficients $a_k$ lie in a certain convex region, and you want to minimize a certain quadratic objective function. Thus, your question is a convex programming problem. You can make this into a practical computational convex programming problem by assuming that $k \le 100$ (say) and also discretizing the unit circle on which $\zeta$ lies. Or, there are also numerical methods to impose your constraints (1) and (2) without discretizing the unit circle, but they make the convex programming problem more complicated.

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Concretely, Greg's suggestion amounts to fitting your original function $N(x)$ by $\sum_{k\ge 0} a_k \left(\frac{i-z}{i+z}\right)^k$. I might try doing this naively first, and hoping that the additional convex constraints are satisfied automatically. –  Dylan Thurston Oct 17 '10 at 5:32
    
But, without the linear constraints, it's just Lagrange interpolation. Maybe allowing a huge number of terms in the fit will lead to trouble even with the inequalities, I'm not sure. –  Greg Kuperberg Oct 17 '10 at 7:07
    
I hoped for better results, but nevertheless your method works (number of (real) coefficients = 25--50% of the (total) number of data points). Some tests are in this Mathematica 7.0 file: goo.gl/65Bt. I've done a computational experiment: I've taken a function, data points from it (I've added small noise), and interpolated it by these points, using your method. On the following images dotted line is initial function. Re: goo.gl/2IQK, Im: goo.gl/vFPW. Often coefficients do not obey inequalities, so I am constraining them. Thank you for your answer. –  Fiktor Oct 19 '10 at 12:56

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