$\int_{\{ |x|\leq 5C \}} (1+|x|^2)^{ps} dx \leq C_1(C) \int_{\{ |x|\leq 5 \}} (1+|x|^2)^{ps} dx$?

Question

Fix $C>0, p>0$ and $s<0.$

Can we find constant $C_1= C_(C)$ depending on $C$ so that

$$\int_{\{ |x|\leq 5C \}} (1+|x|^2)^{ps} dx \leq C_1(C) \int_{\{ |x|\leq 5 \}} (1+|x|^2)^{ps} dx$$?

If so, is $C_1$ also depend on $p$?

Answer 1

Apply the change of variable $x = Cy$. Assuming this is an $n$-dimensional integral you have $dx = C^n dy$ so that $$ \int_{|x| \le 5C} (1 + |x|^2)^{ps} \, dx = {C^n} \int_{|Cy| \le 5C} ( 1 + |Cy|^2)^{ps} \, dy = {C^n} \int_{|y| \le 5} ( 1 + |Cy|^2)^{ps} \, dy.$$

Now appeal to cases: if $C \ge 1$ then $1+|Cy|^2 \ge 1 + |y|^2$, and since $ps < 0$ you get $$(1 + |Cy|^2)^{ps} \le (1 + |y|^2)^{ps}$$ so the inequality holds with $C_1 = C^n$.

If $0 < C < 1$ then $1 + |Cy|^2 \ge C^2 + |Cy|^2 = C^2(1 + |y|^2)$ so that $$(1+|Cy|^2)^{ps} \le C^{2ps}(1 + |y|^2)^{ps}$$ and the inequality holds with $C_1 = C^{n+2ps}$.