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 $[a,b]$.


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 $(0,\infty)$.


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

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.