any idea why this is true? I am not able to figure out.
Given $f \in W^{1,p}(B(x, R))$, we want to prove that $$ \left| \frac{1}{|B_{2^{-l}}(x)|} \int_{B_{2^{-l}}(x)} f - \frac{1}{|B_{2^{-l-1}}(x)|} \int_{B_{2^{-l-1}}(x)} f \right| \leq C (2^{-l})^{-n} \int_{B_{2^{-l}}(x)} \left| f - \frac{1}{|B_{2^{-l}}(x)|} \int_{B_{2^{-l}}(x)} f \right|. $$
Let $$(f)_{x,r} = \frac{1}{\vert B_r(x)\vert}\int_{B_r(x)}f(y) \, dy . $$ Re-writing your inequality in this notation, we wish to prove that $$ \big \vert (f)_{x,2^{-\ell}} - (f)_{x,2^{-\ell-1}} \big \vert \leq C(n) 2^{n\ell}\int_{B_{2^{-\ell}}(x)}\big \vert f(y) - (f)_{x,2^{-\ell}}\big \vert \, dy $$
Note that for any constant $a\in \mathbb R$, $$(f+a)_{x,r} = (f)_{x,r}+a .$$ Since $(f)_{x,2^{-\ell}}$ is a constant, we can bring it inside the integral to obtain $$(f)_{x,2^{-\ell-1}} - (f)_{x,2^{-\ell}} = \big (f -(f)_{x,2^{-\ell}} \big )_{x,2^{-\ell-1}} = \frac 1 {\vert B_{2^{-\ell -1}}(x)\vert}\int_{B_{2^{-\ell -1}}(x)} \big ( f(y) - (f)_{x,2^{-\ell}} \big ) \,d y . $$ It follows that \begin{align*} \big \vert (f)_{x,2^{-\ell}} - (f)_{x,2^{-\ell-1}} \big \vert &\leq \frac 1 {\vert B_{2^{-\ell -1}}(x)\vert}\int_{B_{2^{-\ell -1}}(x)} \big \vert f(y) - (f)_{x,2^{-\ell}} \big \vert \,d y. \end{align*} Then $ \vert B_{2^{-\ell -1}}(x)\vert = \vert B_1 \vert 2^{-(\ell+1)n} = C(n) 2^{-\ell n }$, so $$\big \vert (f)_{x,2^{-\ell}} - (f)_{x,2^{-\ell-1}} \big \vert \leq C(n) 2^{n\ell}\int_{B_{2^{-\ell -1}}(x)} \big \vert f(y) - (f)_{x,2^{-\ell}} \big \vert \,d y. $$ (This is actually sharper than the required inequality since $B_{2^{-\ell -1}}(x) \subset B_{2^{-\ell }}(x)$).
