Trivial Embeddings in Morrey & Campanato spaces

Question

I'm taking a Nonlinear PDEs course this semester and the last time our professor introduced us to Morrey & Campanato Spaces. We have for $\lambda \gt 0$ that:

• The Morrey space $L^{2,\lambda}(\Omega)$ consists of all functions $f\in L^2(\Omega)$ for which the seminorm

$[f]_{L^{2,\lambda}}=\sup_{x \in \Omega} \sup_{r \lt diam\Omega} \frac{1}{r^\lambda} \int_{B(x,r)\cap \Omega} |f|^2 dy \lt \infty$

• The Campanato space $\mathcal L^{2,\lambda}(\Omega)$ consists of all functions $f\in L^2(\Omega)$ for which the seminorm

$[f]_{\mathcal L^{2,\lambda}}=\sup_{x \in \Omega} \sup_{r \lt diam\Omega} \frac{1}{r^\lambda} \int_{B(x,r)\cap \Omega} |f-f_{x,r}|^2 dy \lt \infty$

where $f_{x,r}:=\frac{1}{\mathcal L^n(B(x,r)\cap\Omega)} \int_{B(x,r)\cap \Omega} f dy$ and $\Omega \subset \mathbb R^n$

The first examples that the professor gave are:

1. $f\in L^{\infty} \Rightarrow f \in L^{2,\lambda} \;\;\forall \lambda \in (0,n]$
2. $f \in W^{1,\infty} \Rightarrow f \in \mathcal L^{2,n+2}$
3. $f \in C^{0,\alpha} \Rightarrow f \in \mathcal L^{2,n+2\alpha}$

Although they seem to be quite trivial since there is no special proof of the above nowhere, I have trouble understanding them. I think this double $\sup$ in the definition confuses me a lot because I don't know how to handle them. Why do these 3 examples hold?

I only have some thoughts about 2.:

If $f \in W^{1,\infty}(\Omega)$ then $f$ is a Lipschitz function. So we write

$|f(x)-f_{x,r}(x)|=|\frac{1}{\mathcal L^n(B(x,r)\cap \Omega)} \int_{B(x,r)\cap \Omega} f(x)-f(y) dy| \le \frac{1}{\mathcal L^n(B(x,r)\cap \Omega)} \int_{B(x,r)\cap \Omega} |f(x)-f(y)| dy \le \frac{1}{\mathcal L^n(B(x,r)\cap \Omega)} \int_{B(x,r)\cap \Omega} M|x-y| dy \le \frac{1}{\mathcal L^n(B(x,r)\cap \Omega)} \int_{B(x,r)\cap \Omega} Mr\; dy=Mr$

Hence $\sup_{x \in \Omega} \sup_{r \lt diam\Omega} \frac{1}{r^\lambda} \int_{B(x,r)\cap \Omega} |f-f_{x,r}|^2 dy \le \sup_{x \in \Omega} \sup_{r \lt diam\Omega} \frac{1}{r^\lambda} \int_{B(x,r)\cap \Omega} M^2 r^2 dy=\sup_{x \in \Omega} \sup_{r \lt diam\Omega} \frac{1}{r^{\lambda-2}} M^2 \mathcal L^n(B(x,r)\cap \Omega)$

At this point I've been stuck. How do I proceed? Could somebody provide me some hints in order to prove the rest too?

Any help is much appreciated. Thanks in advance!

Answer 1

For the third example one might argue

\begin{eqnarray*} &&\frac{1}{r^{n+2\lambda}}\int_{B(x,r)\cap \Omega} |f - f_{x,r}|^2dy \leq \frac{1}{r^{n+2\lambda}}\int_{B(x,r)\cap \Omega} |f - f(z)|^2dy \\ & = & \frac{1}{r^n}\int_{B(x,r) \cap \Omega} \left( \frac{|f - f(z)|}{r^\lambda} \right)^2dy \leq \frac{1}{r^n}\mathcal{L}^n(B(x,r)) \sup\limits_{y \in B(x,r)}\left( \frac{|f(y) - f(z)|}{r^\lambda}\right)^2 \\ & \leq & c [f]_{C^{0,\lambda}}^2 < \infty \end{eqnarray*}

for $z \in B(x,r)$ arbitrary but fixed. The second one should then follow by the Morrey embedding. For the first one I'm not sure right now but I think when we know that $f$ is essentially bounded then the statement follows as $\frac{1}{r^\lambda}\int_B1$ is bounded by some constant (at least for $\Omega$ bounded so $r$ can not become too large.)