Let matrix $ A = \begin{bmatrix} a & 1 \\ 0 & a \end{bmatrix}$ is a Jordan matrix with $ -1 < a < 1 $.
Let $ r(z) = \frac{p(z)}{q(z)} $ is any an irreducible rational function, where $p(z)$ and $q(z)$ are polynomials of degree $k$ with real coefficients, and $p(z)$ and $q(z)$ are not zero on the interval $E = \{v^\top Av | v \in \mathbb{R}^2, \|v\| = 1\} = [a-1/2, a+1/2]$.
My problem is finding a positive number $c$ (possibly depends on $k$) such that \begin{equation}\notag \|r(A)\|_2 \leq c \sup_{x \in E} |r(x)|. \end{equation}
Now I have get that \begin{equation}\notag r(A) = \begin{bmatrix} r(a) & r'(a) \\ 0 & r(a) \end{bmatrix} \end{equation} and $\|r(A)\|_2 \leq \sqrt{2}\|r(A)\|_1 = \sqrt{2}(|r(a)| + |r'(a)|)$. Thus I guess that there may exist a positive number $c_1$ (possibly depends on $k$) such that \begin{equation}\notag |r'(a)| \leq c_1 \sup_{x \in E} |r(x)|. \end{equation} However, I don't know how to prove or disprove this conjecture.
Let $k>0.$ Let $1>d>0.$ Let $q(x)=(x+1+d)^k.$ Let $p(x)=1+q(x).$ Then $r(x)=1+(x+1+d)^{-k}$ and $r'(x)=-k(x+1+d)^{-k-1}.$ So on $[-1,1]$ we have $$\frac {\sup |r'|}{\sup |r|}=\frac {kd^{-k-1}}{1+d^{-k}}=\frac {kd^{-1}}{d^k+1}>\frac {kd^{-1}}{2}$$ which can be arbitrarily large if $d$ is close enough to $0.$