Consider $$C([0,T] ; L^1(\mathbb{R}^{N}) \cap L^{\infty}(\mathbb{R}^{N})) := \{ u:(0,T) \rightarrow L^1(\mathbb{R}^{N}) \cap L^{\infty}(\mathbb{R}^{N}) \ | \ \text{u continuous on }[0,T] \}$$ with $T>0$, $N \in \mathbb{N}$. I am reading a paper which uses the Banach Fixed Point Theorem for a contraction mapping $\phi$ on the closed subset
$$B := \{ u \in C([0,T] ; L^1(\mathbb{R}^{N}) \cap L^{\infty}(\mathbb{R}^{N})) \ | \ ||u|| \leq R \}$$ with $R \in \mathbb{R}$ being some constant. The norm $||u||$ is defined as follows:
$$||u|| := \sup_{0 < t < T} \left(||u(t)||_{L^{1}(\mathbb{R^{N}})} + ||u(t)||_{L^{\infty}(\mathbb{R^{N}})}\right)$$
Upon looking at the statement of the Banach Fixed Point Theorem, I noticed that the metric space in question must be complete. I have had a hard time showing that $C([0,T] ; L^1(\mathbb{R}^{N}) \cap L^{\infty}(\mathbb{R}^{N}))$ is complete under the norm $||\cdot||$, so I am wondering if it even is a complete space?
If not, are we allowed to use a closed (and thus cauchy) subset of a metric space for the purpose of the Banach Fixed Point Theorem?
Yes, it is complete.
The first step would be to show that $L^1 \cap L^\infty$ is complete in the given norm. To see this, note that if $\{f_n\}$ is Cauchy in this norm, then it is Cauchy in both the $L^1$ and $L^\infty$ norms. Since $L^1$ and $L^\infty$ are each complete, this means that there are functions $f,g$ such that $f_n \to f$ in $L^1$ norm, and $f_n \to g$ in $L^\infty$. Now verify that in fact $f = g$, and that $f_n \to f$ in the $\|\cdot\|_{L^1} + \|\cdot\|_{L^\infty}$ norm.
To finish, it is a fact that for any complete metric space $X$, the space $C([0,1],X)$ is complete in the uniform metric. Just look back at the proof that $C([0,1], \mathbb{R})$ is complete, and note that it goes through (with the obvious changes) replacing $\mathbb{R}$ by $X$.