Schauder basis of $L^1_{\mathrm{loc}}(\mathbb{R}^n,H)$

Question

$\newcommand{\loc}{\mathrm{loc}}$Let $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),\mu)$ denote the Euclidean space $\mathbb{R}^n$ with its Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$ equipped with the Lebesgue measure $\mu$ and let $H$ be a separable Hilbert space. Let $L^1_{\loc}(\mathbb{R}^n,H)$ denote the space of all (equivalence classes of) measurable vector-valued functions from $\mathbb{R}^n$ to $H$ such that, given any $x\in \mathbb{R}^n$ and any $\epsilon>0$ $$ \int I_{B(x,\epsilon)}(u) \,\|f(u)\|_Hd\mu(u)<\infty. $$ We recall that $L^1_{\loc}(\mathbb{R}^n,H)$ can be made into a Fréchet space when equipping it with the metric $$ d(f,g):=\sum_{n=1}^{\infty}\frac1{2^n} \frac{\int I_{B(x,n)}(u) \,\|f(u)-g(u)\|_Hd\mu(u)}{ 1+ \int I_{B(x,n)}(u) \,\|f(u)-g(u)\|_Hd\mu(u)} . $$

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I expect that $L^1_{\loc}(\mathbb{R}^n,H)$ admits a Schauder basis however, is it true that $$ \left\{ \psi_{i,j}\cdot h_k \right\}_{i,j,k} $$ is a Schauder basis of $L^1_{\loc}(\mathbb{R}^n,H)$ where $\{\psi_{i,j}\}_{i,j=1}^{\infty}$ the Haar-system (defined by $\psi_{i,j}(t)\triangleq 2^{i/2}\psi(2^it -j)$ and $\psi(t)=I_{[0,1/2)}(t) -I_{[1/2,1)}(t)$) and where $\{h_k\}_{k=1}^{\infty}$ is a fixed orthonormal basis of $H$.

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Reasoning: My reasoning is the following.

The subset $L^1(\mathbb{R})\subset L^1_{loc}(\mathbb{R})$ is dense and therefore, for each $f\in L^1_{loc}(\mathbb{R})$ there exists a sequence $\{f_n\}_n\in L^1(\mathbb{R})$ satisfying $$ \lim\limits_{n\to\infty}\, d(f_n,f)=0. $$ Now, the Haar-system is a Schauder basis of $L^1(\mathbb{R})$, for its norm topology, and the norm topology is strictly stronger than the topology the subspace topology on the set $L^1_{}(\mathbb{R})$ inherited from $L^1_{loc}(\mathbb{R})$. Therefore, there exist unique $(F^{i,j})_{i,j=1}^{\infty} \in L^1(\mathbb{R})'$ such that $$ \lim\limits_{s\to\infty}\, \|f_n-\sum_{i',j'=1}^{s} F^{i,j}(f_n)\psi_{i',j'}\|_{L^1}=0. $$ Combining both expressions, we find that for every $f\in L^1_{loc}(\mathbb{R})$ we have $$ \lim\limits_{n\to\infty} \,d(f,\sum_{i',j'=1}^{n} F^{i,j}(f_n)\psi_{i',j'}) =0. $$ I suspect that there is an issue with the argument when discussing uniqueness, but I can't put my finger on it... am I missing something?

Answer 1

Instead of the balls, you can equivalently look at convergence on cubes $Q_k:=k+[0,1)^n$ for $k\in\mathbb{Z}^n$. More precisely: Convergence in your metric is equivalent to convergence to w.r.t. the set of seminorms $$f\mapsto\int_{B_0(r)} \|f(x)\|_H \,dx$$ for all $r\in\mathbb{N}$ which is equivalent to convergence w.r.t. the set of seminorms $$f\mapsto\int_{Q_k} \|f(x)\|_H \,dx$$ for all $k\in\mathbb{Z}^n$. Now since the $Q_k$ are pairwise disjoint, it is sufficient to find a Schauerbasis for $L^1(Q_k,H)=L^1([0,1]^n, H)$.

I think I can prove that if $(\psi_m)_m$ is a Schauder basis of $L^1([0,1]^n)$ and $h_k$ an orthonormal basis of $H$, then there is Schauder basis consisting of products $\psi_m\cdot h_k$ of $L^1([0,1]^n,H)$.

I' being a bit vague, because $L^1$ does not have an unconditional Schauder basis, so the precise ordering matters and all relevant series will be conditionally convergent at most. I have to get back to work now, but tonight I'll write my proof down in detail.