# Eigenvalues and eigenvectors of the q-Bernstein operator

The Bernstein operator maps $f\in C[0,1]$ to its Bernstein polynomial $B_n f.$ The eigenvalues and eigenfunctions of the Bernstein operator on $C[0,1]$ have been described in [1]. Similar description has been obtained for the $q$-Bernstein polynomials in [2]. The study of $q$-Bernstein polynomials in the case $0<q<1$ leads to the following definition.

Definition. Let $0<q< 1.$ The limit q-Bernstein operator on $C[0,1]$ is given by: $$B_{\infty,q}:f\mapsto B_{\infty,q}f,$$where $$(B_{\infty,q}f)(x) = \left\{ \begin{array}{ll} \displaystyle \prod_{j=0}^{\infty}(1-q^jx)\cdot \sum_{k=0}^{\infty}\frac{f(1-q^k)\,x^k}{(1-q)\dots (1-q^k)}, & x\in [0,1),\\ f(1), & x=1. \end{array} \right.$$

Problem. Find all $f\in C[0,1]$ so that $$B_{\infty,q}f=\lambda f,\;\;\lambda \in {\bf C}\setminus \{0\}.$$

Conjecture: If $B_{\infty,q}f=\lambda f,\;\lambda \neq 0,$ then $f$ is a polynomial and $\lambda\in \{q^{m(m-1)/2}\}_{m=0}^{\infty}.$

Remark. The conjecture has been proved under some additional conditions on the smoothness of $f$ at 1 (for example, for $f\in {\rm Lip}\,\alpha$) in [3], Corollary 5.6.

[1] S. Cooper, S. Waldron, The Eigenstructure of the Bernstein Operator, J. Approx. Theory, 105, 2000, 133-165.

[2] S. Ostrovska, M. Turan, On the eigenvectors of the q-Bernstein operators, Mathematical Methods in the Applied Sciences, Vol 37, Issue 4 (2014), pp. 562-570.

[3] S. Ostrovska, On the improvement of analytic properties under the limit $q$-Bernstein operator, J. Approx. Theory, 138, 2006, 37-53.

Again, I doubt that what I'm saying will be new to you, but let's close this question as well (though not completely).

In general the conjecture is false and there are plenty of "bad" eigenfunctions (at least, for small enough $$q$$).

Lemma: Let $$A=(a_{i,j})_{i,j\ge 1}$$ be an infinite matrix with complex entries that defines a bounded operator $$T$$ in $$c_0$$, i.e. $$\sup_{i}\sum_j|a_{i,j}|<+\infty$$ and $$a_{ij}\to 0$$ as $$i\to\infty$$ for all $$j$$. Suppose that there is an infinite sequence $$J_i$$ of disjoint finite sets of positive integers such that

$$\bullet 1\notin J_i$$ for any $$i$$;

$$\bullet \left|\sum_{j\in J_i}a_{ij}\right|\ge \frac 12$$ for all $$i$$;

$$\bullet \sum_{j\notin J_i}|a_{ij}|<\frac 14$$ for all $$i$$.

Then for every $$\lambda\in\mathbb C$$ with $$|\lambda|<\frac 14$$, there exists $$f\in c_0$$ such that $$Tf=\lambda f$$.

Proof: Consider $$f$$ of the following kind:

$$\bullet f_1=1$$

$$\bullet f_j=z_i$$ for all $$j\in J_i$$

$$\bullet f_j=0$$ for all $$j\notin \{1\}\bigcup\cup_i J_i$$

Then the equation $$Tf=\lambda f$$ can be rewritten as $$\left(\sum_{j\in J_i}a_{ij}\right)z_i + \sum_{i'\ne i}\left(\sum_{j\in J_{i'}}a_{ij}\right)z_{i'} - \lambda z_{I(i)}= -a_{i1}+\lambda\delta_{i1}$$ where $$I(i)$$ is defined as the unique index $$I$$ such that $$i\in J_I$$ if it exists and this term is just absent otherwise, and $$\delta_{i1}=1$$ if $$i=1$$ and $$0$$ otherwise.

Due to the assumptions of the lemma, the diagonal terms dominate when $$|\lambda|<\frac 14$$ and the right hand side is in $$c_0$$, so the system has a solution $$z\in c_0$$.

It the original setup, we have $$f_j=f(1-q^{j})$$. $$f_0$$ is set to $$0$$ (the corresponding row is $$(1,0,0,0,\dots)$$) and the remaining matrix is approximately (up to a factor $$1+O(q)$$) just $$a_{ij}=(1-x_i)x_i^j$$ where $$x_i=1-q^{i}$$. The sets $$J_i$$ are just $$(q^{-i+\frac 12},q^{-i-\frac 12})\cap\mathbb N$$. The verification of all conditions (for small enough $$q$$) is left to the reader.

This all makes me suspect that the point spectrum of $$B_{\infty,q}$$ as an operator in $$C([0,1])$$ may be the entire open unit disk (plus $$\{1\}$$, of course) for all $$q\in(0,1)$$ but I cannot prove that much and I am not betting anything on such a strong "counterconjecture".

• Thank you so much indeed for this solution! I apologize for the delay with my response - it took me a while to find a suitable condition on the entries of the matrix S in $Sz=-a_{j1}$ and to solve the exercise for the reader. The answer per se is also very interesting since some "bad" functions appeared in other problems related to Bq. Now they emerge as eigenfunctions. Thank you again for your interest. – Deepti Jun 4 at 9:30