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.



Eigenvalues and eigenvectors of the q-Bernstein operator
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". 


