Let consider $1<b<2$, the sets $A=\{ x: |x|\geq2 \} \subset \mathbb{R}^3$, $B=\{x: |x|<2 \} \subset \mathbb{R}^3$ and the expression \begin{equation} |x|^{1-b} \chi_{ A} + |x|^{2-b} \chi_{ B }, \end{equation} where $\chi$ denotes the characteristic function.
Question: Is there an exponent $f(b)$ such that \begin{equation} |x|^{1-b} \chi_{ A} + |x|^{2-b} \chi_{ B } \leq |x|^{f(b)}, \end{equation} for all $x\in \mathbb{R}^3$ ?
I already tried with $f(b)=b-2$ but it is wrong. Do you have some idea how to attack the problem?
The answer below was corrected thanks to @user23571113's comment.
Yes. Any $f(b) \in [1-b,2-b]$ satisfies the requirements. For instance, $f(b) = \frac 32-b$ is OK.
We can work in dimension one, renaming $|x|$ to $r \geq 0$. On the one hand, as $r \to 0$, you want that $$ r^{2-b} \leq r^{f(b)} $$ which is satisfied if and only if $f(b) \leq 2-b$. On the other hand, as $r \to +\infty$, you want $$ r^{1-b} \leq r^{f(b)} $$ which is satisfied if and only if $f(b) \geq 1-b$.