Let $n<p < \infty$. We want to prove Morrey's inequality, which states that there exists a constant $C=C_{n,p}$ such that \begin{equation} \|u\|_{C^{0,\gamma}(\mathbb{R}^n)} \leq C\|u\|_{W^{1,p}(\mathbb{R}^n)} \quad \forall u\in C^1(\mathbb{R}^n), \text{ with }\gamma := 1-n/p. \end{equation} I'm only stuck on one part of the proof, which seems self-contained. So I'll include here only a sketch of the proof.
The proof of Evans begins by first establishing the inequality $$\overset{\circ}{\int_{B(x,r)}} |u(y)-u(x)| dy \leq C \int_{B(x,r)} \frac{|\nabla u(y)|}{|y-x|^{n-1}}dy$$ Where $\overset{\circ}{\int_A}:= \frac{1}{m(A)}\int_A$ denotes the "average" over the set $A$, with $m$ the Lebesgue measure.
Then, we use this inequality as follows: \begin{align} |u(x)| &\leq \overset{\circ}{\int_{B(x,1)}} |u(x)-u(y)| dy + \overset{\circ} {\int_{B(x,1)}} |u(y)| dy \\ &\leq C \int_{B(x,1)} \frac{|\nabla u(y)|}{|y-x|^{n-1}}dy + C \|u\|_{L^p(B(x,1))}\\ &\leq \vdots \end{align}
QUESTION: I think this is a very simple question, maybe stupid. But I have no idea how we have the estimate $$\overset{\circ}{\int_{B(x,1)}} |u(y)| dy \leq C\|u\|_{L^p(B(x,1))}. $$ Basically, I need to show that $\overset{\circ}{\int_{B(x,1)}} |u(y)|dy = \frac{1}{m(B(0,1))}\|u\|_{L^1(B(x,1))} \leq C \|u\|_{L^p(B(x,1))}$, right? The whole point of Morrey's inequality is to make the constants independent of $u$, but for this estimate in particular, I have no idea how to find such a constant $C$ that is independent of $u$. Any help would be appreciated, thanks in advance!
