Consider the sequence of measures $$d\mu_N(x)=e^{-NV(x)}dx$$ on the real axis, where $V$ is continuous and satisfies the growth assumption $$\lim_{|x|\rightarrow\infty}(V(x)-2\log|x|)=+\infty.$$

Then, denote $P_{k,N}$ the $k$-th orthonormal polynomial associated to $\mu_N$, which is known to satisfy a three-terms recurrence relation $$xP_{k,N}=a_{k+1,N}P_{k+1,N}+b_{k,N}P_{k,N}+a_{k,N}P_{k-1,N}.$$

If we take for example the ($N$-independent) Gaussian case, $\mu_N=e^{-x^2/2}$, where is known that $$a_{k,N}=\sqrt{k},\qquad b_{k,N}=0,$$

it then follows by change of variable that for its renormalized version, $\mu_N=e^{-Nx^2/2}$, we have $$a_{k,N}=\sqrt{k/N}, \qquad b_{k,N}=0.$$ Thus, there exists $\epsilon>0$ such that as $N\rightarrow\infty$

$$ \max_{k\geq 0 \; :\;\left|\frac{k}{N}-1\right|\leq \epsilon} |a_{k,N}|=O(1),$$ and of course a similar statement holds for $b_{k,N}$.

The same observation can be done when considering renormalized Laguerre weights, where this time the $b_{k,N}$'s are not identically zero.

I'm looking for a (relatively simple) proof of such statement for general $\mu_N$ as introduced above.

In fact, it can be proved as a consequence of a Riemann-Hilbert asymptotic analysis that there exists functions $a$ and $b$ such that $$ \lim_{k/N\rightarrow s}a_{k,N}=a(s), \qquad \lim_{k/N\rightarrow s}b_{k,N}=b(s),$$ but it would be nice to be able to establish the more modest statement of the "boudedness" with a simpler proof.