7

Just take $f(x)=\dfrac1{\sqrt2-x}$. That will work.


3

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 $$ f(x)=\begin{cases} 1, & x^2<2,\\ 0, & x^2>2.\end{cases}$$


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