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

$$\newcommand{\loc}{\mathrm{loc}}$$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 $$d(f,g):=\sum_{n=1}^{\infty}\frac1{2^n} \frac{\int I_{B(x,n)}(u) \,\|f(u)-g(u)\|_Hd\mu(u)}{ 1+ \int I_{B(x,n)}(u) \,\|f(u)-g(u)\|_Hd\mu(u)} .$$

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 $$H$$.

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?

• 1. $L^1([0,1])$ has a basis. 2. The projective tensor product of two Banach spaces with a basis also has one. 3. The dierct sum (in the sense of l.c.s's) of a sewuence of Banach spaces with bases is a Fréchet space with a basis. The rest follows from an identification of the the way your space is constructed from its components.. Commented Jan 24, 2022 at 16:07
• For step 2, if $E$ and $F$ are such Banach spaces, then is $\{e_n\otimes f_k\}_{n,k}$ a basis of $E\hat{\otimes}_{\pi} F$?
– ABIM
Commented Jan 24, 2022 at 16:49

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.

• I guess, since $L^1(Q_k,H)\cong L^1(Q_k)\hat{\otimes} H \cong L^1([0,1]^n)\otimes H$ and then, the density of $\{\psi_{\cdot}\}$ in $L^1([0,1]^n)$ implies that $\{\psi_{\cdot}\cdot h_{\cdot}\}$ is dense in $L^1(Q_k,H)$. Thus, by your semi-norm formulation of the topology in $L^1_{loc}(\mathbb{R},H)$ we would have our conclusion; no?
– ABIM
Commented Jan 24, 2022 at 13:50
• @JohanessHahn any luck on the full detailed version?
– ABIM
Commented Jan 25, 2022 at 23:53
• Honestly, no. I found a flaw in my original argument and haven't had the time to try to fix it yet. At least I found some articles that claim your more general conjecture is true that the tensor product of two Schauder basis is a Schauder basis of the (projective) tensor product, but alas it was stated without a reference. Is that because it is a well known fact to people in functional analysis? I'm hopeful. Commented Jan 26, 2022 at 22:00
• @JohanessHahn I thought it wasn't true (but maybe there is some assumption implicitly working where which I'm not aware of).
– ABIM
Commented Jan 27, 2022 at 12:55