$\newcommand{\loc}{\mathrm{loc}}$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](https://en.wikipedia.org/wiki/Haar_wavelet) *(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?*



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


