Morrey's inequality. Case $p=\infty$

Question

I work in EDP, and I struggling with the Morrey's Inequality

If $n<p$. Then there exists a constant $C=C(n,p)$ such that
$$\|u\|_{C^{0,1-\frac{n}{p}}}\leq C \|u\|_{W^{1,p}(\mathbb{R}^{n})},$$ for all $u \in C^{1}(\mathbb{R})$.

For the case $n<p<\infty$ is resolved Evans, but the case $p=\infty$ is not resolved. I'm trying to make this case. My attempt is:

1. Prove the following inequality for all $u \in C^{1}(\mathbb{R})$, $r>0$ and $x \in \mathbb{R}^{n}$ $$ \frac{1}{\omega_{n}r^{n}}\int_{B_{r}(x)}|u(y)-u(x)|dy\leq C\int_{B_{r}(x)}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy.$$

2. Now fix $x \in \mathbb{R}^{n}$ and using 1. we have $$\begin{equation}\begin{aligned}|u(x)|\leq \int_{B_{r}(x)}|u(y)-u(x)|dy + \int_{B_{r}(x)}|u(y)|dy \\ \leq C\int_{B_{r}(x)}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy + \|u\|_{L^{1}(B_{1}(x))}\end{aligned}\end{equation}$$

Since $L^{\infty}(B_{1}(x)) \hookrightarrow L^{1}(B_{1}(x))$ (I'm not sure in this affirmation) we have $$\begin{equation}\begin{aligned}|u(x)|\leq C\int_{B_{r}(x)}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy + \|u\|_{L^{\infty}(B_{1}(x))}\end{aligned}\end{equation}$$ 3. I'm stuck here, I want to Prove that (using the argument in Evans) $$\int_{B_{r}(x)}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy<C_{2}\|\nabla u\|_{L^{\infty}(B_{1}(x))} $$

and so conclude that $$\|u\|_{0}\leq C_{3}\|u\|_{W^{1,\infty}(\mathbb{R}^{n})}$$

Answer 1

The Hölder inequality does also hold for $p=\infty$, $q=1$: $$\Vert fg\Vert_{L^1}\leq\Vert f\Vert_{L^\infty}\Vert g\Vert_{L^1}.$$ That is because $\vert fg\vert=\vert f\vert\vert g\vert\leq\Vert f\Vert_{L^\infty}\vert g\vert$ holds by definition of the $L^\infty$-norm almost everywhere. Applying this to the equation in 3. we get $$\int_{B_1(x)}\frac{\vert\nabla u(y)\vert}{\vert x-y\vert^{n-1}}\mathrm{d}y\leq\Vert\nabla u\Vert_{L^\infty}\int_{B_1(x)}\frac{1}{\vert x-y\vert^{n-1}}\mathrm{d}y=\Vert\nabla u\Vert_{L^\infty}\underbrace{\int_{B_1(0)}\frac{1}{\vert y\vert^{n-1}}\mathrm{d}y}_{=:C_2<\infty}.$$

The continuous embedding $L^\infty\hookrightarrow L^1$ on sets of finite measure is also a direct consequence of the Hölder inequality: Just choose $g\equiv 1$.