The Morrey space $L^{2, \nu} (\Omega)$ for an open set $\Omega \subset \mathbb{R}^n$ and for $1 \leq \nu \leq n$ is defined as $$L^{2, n} (\Omega) = \{ f \in L^2(\Omega); \hspace{2mm} [f]^2_{L^{2, \nu}} = \sup_{x_0 \in \Omega, \hspace{1mm} 0<r<1} \left( r^{-\nu} \int_{\Omega_r(x_0)}|f|^2 dx \right) < \infty \}$$ where $\Omega_r(x_0) = \Omega \cap B_r(x_0)$ with $B_r(x_0) \subset \mathbb{R}^n$ the ball with radius $r$.
I've read that $L^{2, n} (\Omega) \cong L^{\infty} \cap L^2 (\Omega)$.
I can see that we can embed $L^{\infty} \cap L^2 (\Omega)$ in the space $L^{2, n} (\Omega)$.
But how can one see the other embedding?
Thanks in advance for any help!
Take $f\in L^{2,n}(\Omega)$. Almost every point of $\Omega$ is a Lebesgue point of $|f|^2$. At such points $x$, $$ |f(x)|^2 = \lim_{r\to 0} \left( r^{-n} \int_{\Omega_r(x_0)}|f|^2\, dx \right) \le [f]^2_{L^{2,n}}$$ Therefore, $\|f\|_{L^\infty}\le [f]_{L^{2,n}}$.
When the measure of $\Omega$ is finite, the $L^\infty$ norm controls the $L^2$ norm and we get the continuous embedding $L^{2,n}\to L^\infty \cap L^2 = L^\infty$.
When the measure of $\Omega$ is infinite, we still have set-theoretical inclusion $L^{2,n}\subset L^2\cap L^\infty$, since $L^{2,n}\subset L^2$ by definition. However, this is not a continuous embedding. Indeed, let $\Omega=\mathbb R^n$ and define $f_n=R^{-n/2}\chi_{B(0,R)}$. Then $\|f_n\|_{L^2}=1$ while $[f_n]_{L^{2,n}}\to 0$.
I'd say that the definition of Morrey space that you stated is meant for domains with finite measure. For general domains, one should probably add $L^2$ norm to the Morrey norm.