It is classical that $L^1(G, m)$ is a Banach algebra when $G$ is a locally compact group with Haar measure $m$ by using the operation of convolution via the integral


I'm interested in knowing if there is a natural Banach algebra structure on $L^1(X, \mathcal{B},\mu)$ for a locally compact, Hausdorff measure space $X$, with a Borel sigma algebra, $\mathcal{B}$ and radon measure, $\mu$ with the added assumption that the measure is induced by a (say normalized) Haar measure on a compact abelian group, $G$.

In particular, the situation I'm most interested in is

$$X=\coprod_{n\in\mathbb{N}} X_n$$

with each $X_n$ a compact space topologically a homogeneous $G$-space, i.e. a quotient space $G/H_n$ with $H_n$ closed subgroups of $G$. With these assumptions I can define just such a measure on $X$ by first taking $f\in C(X_n)$ and selecting any point $x_n\in X_n$ and using the normalized Haar measure $m$ on $G$, setting

$$I_n(F)=\int_G f(\tau x_n)\,dm(\tau)$$

Since $G$ acts transitively on its coset spaces, this integral is independent of choice of $x_n$. Then, on $X$, I define for $F\in C_c(X)$

$$I(F)=\sum_n I_n(F\cdot 1_{X_n})$$

Since $F$ has compact support, this sum is well-defined (being finite). I know that, since $I(F)$ is a linear functional, it is represented by integration against a Radon measure, $\lambda$ on $X$ and by construction and the transitivity on the pieces, this measure is invariant under the action of $G$, that is to say, there is a radon measure $\mu$ on $X$ and a $\sigma$-algebra, $\Sigma$, containing the Borel $\sigma$-algebra so that for every $F\in L^1(X,\Sigma,\mu)$

$$I(F)=\int_X F\,d\mu.$$

Moreover it is clear from the definition that this has the property that

$$\int_{X_n}F(x)\,d\mu(x)=\int_{X_n}F(\tau x)\,d\mu(x)$$

for every $\tau\in G$ and $F\in L^1(X)$. By writing

$$F=\sum_n F\cdot 1_{X_n}$$

we can see now that

$$\int_X F\,d\mu=\sum_n F(x)1_{X_n}\,d\mu(x)$$ $$=\sum_n\int_{X_n}F(x)\,d\mu=\sum_n \int_{X_n}F(\tau x)\,d\mu(x)$$ $$=\int_X F(\tau x)\,d\mu(x).$$

This establishes that the measure $\mu$ is invariant under $G$.

Now, $X$ itself lacks a group structure, but can we still somehow define a Banach algebra structure on $L^1(X)$ by using the fact that the measure $\mu$ is essentially pushing forward the measure $m$ on $G$ and $G$ does have the structure of a compact group?

Edit (Adam): I've added the fact about the $X_n$ being homogeneous spaces of $G$ by subgroups and clarified the invariance of $\mu$ under the $G$-action.

  • $\begingroup$ Since you are working with an homogeneus $G$-space $X$ the action of $G$ in $X$ will not induce an algebra structure on $L_1(X)$ but a natural Banach $L_1(G)$-module structure (right or left depending on your choice of actions). I do not know if that is a satisfactory answer. $\endgroup$ Nov 28, 2013 at 7:08

2 Answers 2


If the $H$ are compact and $G$ unimodular, there is indeed a algebra structure on $L^1(G/ H)$ simply by convolution over $G$. It does however not contain more structure then the subalgebra $A=L_1( G//H) \subset L^1(G)$ of $H_n$-biinvariant functions.

You can drop modularity of $G$ an will obtain a slightly different $H$-invariance from the left.

This actually works fine still if $H$ is containing and compact modulo the center $Z$ of $G$. You obtain as subalgebras $$L^1(G//H) \subset L^1(G/Z).$$

This naive approach fails if $G//H$ is not nicely defined, say $H$ being the subgroup of upper triangular matrices in $G=GL_2(\mathbb{R})$. Then $G//H$ has cardinality two, and one of the double coset carries the full measure, the other zero measure.

This does of course not rule out any different Banach algebra structure on $L^1(G/H)$, but $G/H$ is pretty well-known (i.e. projective space). So perhaps you should start with that specific example?

  • $\begingroup$ It seems you've addressed the compact case. For the situation described, does one just do the natural thing and define the action of convolution on $X$ by writing integrals as sums over the component pieces to get the structure on a union of such $G$-spaces then, or is there some subtlety I'm potentially not seeing? $\endgroup$ Nov 29, 2013 at 9:49
  • $\begingroup$ Yes, you can do the usual integration on $G$ and ignore invariance. $\endgroup$
    – Marc Palm
    Nov 29, 2013 at 14:40

I am not sure that this will be useful for you, but Yulia Kuznetsova proved not long ago that for a locally compact group $G$ and a measurable function $w:G\to{\mathbb R}_+$ the weighted space $$ L_1^w(G)=\{f:\quad fw\in L_1(G)\} $$ is an algebra (with respect to the convolution) if and only if the weight $w$ is equivalent in some sense to a continuous submultiplicative function: $$ w(st)\le w(s)w(t), \qquad s,t\in G. $$ She also considers $L_p^w(G)$, $p>1$.


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