How to prove Campanato space is a Banach space

Let $p\ge1$, $\mu\ge0$ and $\Omega$ be a bounded open set in $\mathbb{R}^n$. Campanato space embraces all $u$'s which

$$[u]_{p,\mu}=[u]_{p,\mu;\Omega}=\sup_{\substack{x\in\Omega\\0<\rho<\mathrm{diam}\Omega}}\left(\rho^{-\mu}\int_{\Omega_\rho(x)}\left|u(y)-u_{x,\rho}\right|^p\,dy\right)^{\frac{1}{p}}<+\infty,$$ where $\Omega_\rho(x)=\Omega\cap B_\rho(x)$ ($B_\rho(x)$ denotes a ball centered at $x$ with a radium $\rho$) and $$u_{x,\rho}=\frac{1}{\left|\Omega_\rho\right|}\int_{\Omega_\rho(x)}u(y)\,dy,$$

equipped with a norm $$\|u\|_{L^{p,\mu}}=\|u\|_{L^{p,\mu}(\Omega)}=[u]_{p,\mu;\Omega}+\|u\|_{L^p(\Omega)}.$$

Let $\{u_k\}$ be a Cauchy sequence in Campanato space, one can determine a $u$ because of the completeness of $L^p$ space. What are the next steps to prove that $u$ is also the right limit of $\{u_k\}$ in the sense of $\|\cdot\|_{L^{p,\mu}(\Omega)}$ and therefore Campanato space is complete? Thank you~

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I do not know anything about these spaces - so no much hope to help you. But quick google search led me to the book Kufner A, John O, Fucik S - Function Spaces. They claim that the proof should be almost the same as the proof for Marcinkiewicz space in Theorem 4.2.2. books.google.com/… – Martin Sleziak Jun 25 '11 at 8:01
Thank you, @Martin Sleziak. I did google before, using "Campanato space Banach" as keywords, but didn't get much information. – ziyuang Jun 25 '11 at 13:16