I'm studying something about singular value and I came across a problem.
Matrix $A = \begin{bmatrix} a & 1 \\ 0 & a \end{bmatrix}$ is a Jordan matrix with $a>1$.
For an integer $k$, let $r(z) = p(z)/q(z)$ is an irreducible rational function on the complex plane $\mathbb{C}$, where $p$ and $q$ are polynomials of degree $k$, and $p$ and $q$ are not zero on $\mathbb{R}$.
My problem is finding a positive number $K$ (possibly depend on $k$) such that \begin{equation} \|r(A)\|_2 \leq K \sup_{x \in \mathbb{R}} |r(x)|. \end{equation}
Now I have get that \begin{equation} 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 $K_1$ (possibly depend on $k$) such that \begin{equation} |r'(a)| \leq K_1 \sup_{x \in \mathbb{R}} |r(x)|. \end{equation} However, I don't know how to prove or disprove this conjecture. If anyone knows how to do this, I'll thank you in advance.
Now, I think there is no positive number $K_1$(possibly depend on $k$) such that \begin{equation} |r'(a)| \leq K_1 \sup_{x \in \mathbb{R}} |r(x)|. \end{equation}
For example, let $r(z) := \frac{z-a}{z-a+t \mathbf{i}}$, where $t > 0$ is a real number, and $\mathbf{i} = \sqrt{-1} $. Then it is easy to get that \begin{equation} \sup_{x \in \mathbb{R}} |r(x)| = \lim_{x \rightarrow \infty} |r(x)| = 1. \end{equation} However, $r'(z) = \frac{t\mathbf{i}}{(z-a+t \mathbf{i})^2}$ on $\mathbb{R}$, this means that \begin{equation} \sup_{x \in \mathbb{R}} |r'(x)| = |r'(a)| = \frac{1}{t} = \frac{1}{t}\sup_{x \in \mathbb{R}} |r(x)|. \end{equation} If let $t \rightarrow 0_{+}$, then it implies that \begin{equation} \frac{|r'(a)|}{\sup_{x \in \mathbb{R}} |r(x)|} \rightarrow +\infty. \end{equation}