# Tag Info

Just take $f(x)=\dfrac1{\sqrt2-x}$. That will work.
No, there can't be a bound that only depends on $\|A\|$. The problem isn't when $\|A\|$ is large, it's when $\|A^{-1}\|$ is large (which can happen with $\|A\|$ staying bounded). Try $$B = \pmatrix{1 & \epsilon\cr 0 & 1\cr}, \ A = \pmatrix{\epsilon^2 & 0\cr 1 & 1\cr}$$ Then $$A B A^{-1} = \pmatrix{1-\epsilon & \epsilon^3\cr -1/\epsilon ... 1 Try$$f(x) = \begin{cases} 1 & \text{if $x < \sqrt{2}$} \\ 0 & \text{if $x > \sqrt{2}$} \end{cases} $$This not only satisfies your criterion, but f'(x)=0 for all x \in \mathbb Q, and yet f is not constant!. 1 The easiest example might be$$ f(x)=\begin{cases} 1, & x^2<2,\\ 0, & x^2>2.\end{cases}