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The definition of Chebyshev system is as follows.

A linearly independent system of $n$ basis functions $\varphi_1,\ldots,\varphi_n$ on $[a,b]$ is a Chebyshev system on $[a,b]$ if every non-trivial linear combination $\displaystyle\sum_{k=1}^n a_k\;\varphi_k$ has at most $n-1$ zeros on $[a,b]$.

Problem

The problem is to show that for any real $0=\lambda_0 < \lambda_1 < \ldots < \lambda_n$, the basis functions $\left\{x^{\lambda_i}\right\}_{i=0}^n$ form a Chebyshev system on $(0,\infty)$.

Approach

The first thing that I should prove is that $\left\{x^{\lambda_i}\right\}_{i=0}^n$ form a basis of the space $C(0,\infty)$, right? I don't know how to see that.

Assuming that they form a basis. I think that the proof of being a Chebyshev system is equivalent to prove that

$$\det\left(\begin{matrix}1 & x_1^{\lambda_1} & \ldots & x_1^{\lambda_n}\\ \vdots & \vdots & & \vdots\\ 1 & x_{n+1}^{\lambda_1} & \ldots & x_{n+1}^{\lambda_n}\end{matrix}\right)\neq 0$$

for all pairwise distinct $x_1, \ldots, x_{n+1}\in(0,\infty)$ because it means that there is no linear combination that evaluated in one of those points is zero.

I need a bit of guidance in this exercise. Thank you

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    $\begingroup$ The $x^{\lambda_i}$ won't, in general, form a basis of $C(0,\infty)$ I think. For example, $\lambda_i = 1 - 2^{-i}$ fullfills your requirements, yet $x^2$ can't be approximated by this system because all the $x^{\lambda_i}$ are in $O(x)$. $\endgroup$ – fgp May 23 '13 at 10:34
  • $\begingroup$ Sorry, what is $O(x)$? $\endgroup$ – synack May 23 '13 at 10:36
  • $\begingroup$ Informally speaking, the set of all functions which grow at most as fast as $x$. Formally, $f(x) \in O(g(x))$ if there's an $M$ and an $x_o$ such that $|f(x)| \leq M|g(x)|$ for all $x \geq x_0$. Instead of $f(x) \in O(g(x))$ people often write $f(x) = O(g(x))$. $\endgroup$ – fgp May 23 '13 at 10:38
  • $\begingroup$ Chebyshev system need not to be a basis of the space. $\endgroup$ – leshik May 23 '13 at 10:45
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As I mentioned in the comment, you do not need to show that $x^{\lambda}$ forms a basis of $C(0,\infty).$ Obviously, it is not going to be true since you cannot span $C(0,\infty)$ using finite number of functions. There is a beautiful theorem due to Muntz which sates that $\sum\frac{1}{\lambda_i}=\infty$ is a equivalent to the system $span\{x^{\lambda_n}\}^{\infty}_{n=0}$ being dense in $C[a,b],$ $a>0.$

As to the original problem, just use Rolle's theorem and induction on $n.$ Indeed, base case is trivial. Now to prove the step, we assume that $f(x)=\sum_{i=0}^ka_ix^{\lambda_i}=0$ has $k+1$ zeros in $(0,\infty).$ By Rolle's theorem, $f'(x)=\sum_{i=0}^k(\lambda_i)a_ix^{\lambda_i-1}=x^{\lambda_1-1}(\sum_{i=1}^k(\lambda_i)a_ix^{\lambda_i-\lambda_1})$ has at least $k$ zeros in $(0,\infty).$ This contradicts to the fact that $\{x^{\lambda_i-\lambda_1}\}_{i=1}^{k}$ forms a Chebyshev system.

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  • $\begingroup$ How does $\{x^{\lambda_i-\lambda_1}\}_{i=1}^{k}$ not being a Chebyshev system imply that $\{x^{\lambda_i}\}_{i=1}^{k}$ is a Chebyshev system? Also, in your derivative, the indeterminate coefficient should vanish, right? Instead, you get that the indeterminate coefficient is $\left(\lambda_0\;a_0\;\frac{1}{x}\right)$. Thank you $\endgroup$ – synack May 23 '13 at 12:37
  • $\begingroup$ Another doubt: you're applying Rolle's theorem without verifying the theorem's assumptions: i) the interval must be a closed interval and ii) f(a)=f(b) $\endgroup$ – synack May 23 '13 at 14:16
  • $\begingroup$ Assuming that the statement holds for $n=k-1$ we want to prove for $n=k.$ Now the proof goes by contradiction, namely, if for $n=k$ the corresponding polynomial has too many zeros ($k$ in our case), then its derivative has $k-1$ zeros (between each pair of consecutive roots there is a root of the derivative). Now we use our assumption that for $n=k-1$ to conclude that we have at most $k-2$ zeros and reach a contradiction. $\endgroup$ – leshik May 23 '13 at 15:14
  • $\begingroup$ I understand, thanks. But we should assume that for $n=k$ we have $k+1$ zeros in order to negate the statement. Because we have $k+1$ basis functions and thus the definition says that any non-trivial linear combination has at most $k$ zeros. $\endgroup$ – synack May 23 '13 at 16:02
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If $0 < x_0 < x_1 < \ldots < x_n$ and $\lambda_0 < \lambda_1 < \ldots < \lambda_n$, then the determinant of the generalized Vandermonde matrix $$ \left( \begin{matrix} x_0^{\lambda_0} & x_0^{\lambda_1} & \ldots & x_0^{\lambda_n}\\ x_1^{\lambda_0} & x_1^{\lambda_1} & \ldots & x_1^{\lambda_n}\\ \vdots & \vdots & \ddots & \vdots\\ x_{n}^{\lambda_0} & x_{n}^{\lambda_1} & \ldots & x_{n}^{\lambda_n} \end{matrix} \right) $$ is strictly positive due to a special case of Theorem 1 published in the article [Shang-jun Yang, Hua-zhang Wu, Quan-bing Zhang, 2001].

Also, you may find interesting articles that are devoted to the construction of unique normalized Bernstein-like bases in extended Chebyshev Müntz spaces. For more details consider, e.g., the article [Marie-Laurence Mazure, 1999].

(Hopefully, you can download the cited papers.)

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