A paper I am going through claims the following Sobolev inequality (working always in $\mathbb{R}^3$): for a fixed $R > 0$, $$ |f(x)|^2 \leq C\sum_{|\alpha| = 0}^2 R^{-3}\int_{B_R(x)} R^{2|\alpha|}|\partial_x^\alpha f(y)|^2\, dy. $$ My question is how to prove this? I have tried as follows; however I get the wrong powers of $R$.
I have tried proceeding from the usual Sobolev inequality $H^2 \hookrightarrow L^\infty$ on $\mathbb{R}^3$ and multiplying by a cutoff function $\phi_R$, where $\phi_R(x) = \phi(x/R)$ and $\phi$ is a bump function which is $1$ near $0$ and $0$ outside of $B(0, 1)$. Then $\partial^\alpha \phi_R(x) = R^{-|\alpha|}\partial^\alpha \phi(x/R)$, and since $\phi$ is smooth we have $L^\infty$ bounds on these derivatives and can pull them out of the integral after using the Leibniz rule. Since $$ \partial^\alpha (f\phi_R) = \sum_{\beta \leq \alpha} C_{\alpha\beta} \partial^{\alpha - \beta} f \partial^\beta \phi_R, $$ we have $|f\phi_R| \lesssim |f|$, as well as $$ |\partial_i (f\phi_R)| \lesssim |\partial_i f| + R^{-1}|f|, $$ and $$ |\partial_{ij} (f\phi_R)| \lesssim |\partial_{ij}f| + R^{-1}(|\partial_i f| + |\partial_j f|) + R^{-2}|f|. $$ To check the lemma, then, we can check just at $x = 0$ (otherwise translate); then using that $f(0) = f\phi_R(0)$, we have by the usual Sobolev inequality $$ |f(0)|^2 \leq C\sum_{|\alpha| \leq 2} \int_{\mathbb{R}^3} |\partial_x^\alpha (f\phi_R)|^2\, dy. $$ Then one plugs in the bounds above in terms of inverse powers of $R$. However one can see that the powers are "off" from the desired estimate on each term; this is seen easiest on the second derivative. The desired result has a power of $R$ on the second derivative term on the right-hand side; however my approach here has no power of $R$ on the second derivative term.
Any ideas are appreciated!