$\newcommand{\loc}{\mathrm{loc}}$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?

  • 2
    $\begingroup$ 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.. $\endgroup$ Commented Jan 24, 2022 at 16:07
  • $\begingroup$ For step 2, if $E$ and $F$ are such Banach spaces, then is $\{e_n\otimes f_k\}_{n,k}$ a basis of $E\hat{\otimes}_{\pi} F$? $\endgroup$
    – ABIM
    Commented Jan 24, 2022 at 16:49

1 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.

  • $\begingroup$ 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? $\endgroup$
    – ABIM
    Commented Jan 24, 2022 at 13:50
  • $\begingroup$ @JohanessHahn any luck on the full detailed version? $\endgroup$
    – ABIM
    Commented Jan 25, 2022 at 23:53
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jan 26, 2022 at 22:00
  • $\begingroup$ @JohanessHahn I thought it wasn't true (but maybe there is some assumption implicitly working where which I'm not aware of). $\endgroup$
    – ABIM
    Commented Jan 27, 2022 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.