Convolution of a function and a measure.

Consider a locally compact group $\mathrm{G}$ and a left-invariant Haar measure $\lambda$ on it. Let $\mu$ be a probability measure. Suppose $f$ is a function continuous and bounded. Denote by $\Delta$ the modulus of the group $\mathrm{G}.$ I want to show that the function $$(f \ast \mu)(x) = \int\limits_\mathrm{G} f(xs^{-1}) \Delta(s^{-1})\ d\mu(s)$$ is (1) defined everywhere, (2) continuous and (3) bounded.

EDIT: I am looking at the integral on a locally compact space, the usual way they handle this is by means of the Daniell integral. So, I guess the measures here are called Radon measures. In particular, they are regular (both inner and outer) and finite on every compact set. Also, if $\mathscr{A}_\mu$ denotes the $\mu$-integrable sets, then $\mathscr{A}_\mu$ contains the topology of $\mathrm{X}$ (remark $\mu$ is a probability measure, so there are no issues of infinite measure here).

In the main text the proof of the same statement is given for the convolution $\mu \ast f$ which is given by $$(\mu \ast f)(x) = \int\limits_\mathrm{G} f(s^{-1} x)\ d\mu(s).$$

I tried to immitate the proof, but the are problems that arise. The easiest way to illustrate this is when they show boundedness. For the case $\mu \ast f$ the author uses a well-known inequality and boundedness of $f$ $$|(\mu \ast f)(x)| \leq \int\limits_\mathrm{G} |f(s^{-1} x)|\ d\mu(s) \leq \|f\|,$$ since $\mu$ is a probability measure. Obviously, the same proof can't proceed for the convolution $f \ast \mu$ since one would get $$|(\mu \ast f)(x)| \leq \|f\| \int\limits_\mathrm{G} \Delta(s^{-1})\ d\mu(s)$$ and I don't think the modulus function is integrable.

Any suggestions is greatly appreciated.

• By the way, it seems to be the case that all there is to this exercise is to show that the modulus is integrable with respect to every probability measure. However, this is my first time tackling Haar measure and I lack any intuition regarding the modulus. – Will M. Mar 2 '18 at 17:23
• I don't know what definition of measure on a locally compact topological space you have but I am using the one I encounter everywhere: a measure $\mu$ on a locally compact topological space $\mathrm{X}$ is a linear form on the set $\mathscr{K}(\mathrm{X})$ of (complex-valued) continuous functions with compact support such that for every compact subset $\mathrm{K}$ of $\mathscr{X}$ there exists a constant $a_\mathrm{K} > 0$ such that $\mu(f) \leq a_\mathrm{K} \| f\|.$ – Will M. Mar 5 '18 at 4:54
• And of course, every measure $\mu$ is identified with all its unique extension to the set $\mathscr{L}^1$ of $\mu$-integrable functions (those that are (a) defined $\mu$-almost everywhere, (b) $\mu$-measurable and (c) the upper integral of the absolute value is finite.) – Will M. Mar 5 '18 at 4:56
• What is the text you are using? – Dominique R.F. Mar 5 '18 at 20:50
• I am following "Treatise on analysis" by Jean Dieudonne, chapter XIV, p. 284, paragraph just before (14.9.4) he says he shall leave to the reader to prove analogous propositions for the convolution $f \ast \mu$ which I now think are not true. – Will M. Mar 5 '18 at 20:56

Too long for a comment:

You need some more assumptions on $f$ or on the measure $\mu$.

For example, for the function $f$ take the constant function $f \equiv 1$. Of course, this is continuous and bounded. For this choice of $f$ your question amounts to showing that the function $s \mapsto \Delta(s^{-1})$ is integrable with respect to any (suitable) probability measure.

The map $s \mapsto \Delta(s^{-1})$ is a group morphism to the multiplicative group of positive reals. Hence it is either constant (in this case the question is trivial) or unbounded. But for any unbounded function $h$ one can always find a probability measure on $G$ such that $h$ is not integrable with respect to this measure. The idea is to take a sequence $x_n \in G$ such that $|h(x_n)| > 2^n$, and define the measure $\mu = \sum_{n=1}^{\infty} 2^{-n} \delta_{x_n}$, where $\delta_x$ denotes the Dirac measure at $x$ (see e.g. here for some more details). Hence for such a probability measure the function $f * \mu$ as you defined it is identically infinite.

Maybe you wanted $f$ to be compactly supported? Or integrable with respect to $\lambda$?

• Thanks, I did notice that by taking $f = 1$ I just needed to prove $\Delta$ was integrable with respect to every probability measure (which sounded unlikely, but I am afraid I am not familiar with non-unimodular groups). And the author wrote, I quote, "We shall leave to the reader the task of stating the corresponding propositions for the convolution $f \ast \mu$." The problem is that two chapters later the author uses both convolutions $f \ast \mu$ and $\mu \ast f$ and that was bugging me. I think, that, indeed, the "corresponding propositions" are false. – Will M. Mar 5 '18 at 20:53
• Anyway, I am only interested in $\mathbf{R}^d$ and $\mathbf{Z}^b,$ so I can say for sure that the result is true for unimodular groups. – Will M. Mar 5 '18 at 20:54
• What are other conditions on $f$ and $\lambda$ that would imply the boundedness? It seems like you could contrive some functions such as $f(xs^{-1}) = \Delta(s^{-1})^{-1}$, or $f(xs^{-1}) = g(x)\Delta(s^{-1})^{-1}$ for some continuous function $g(x)$ to imply continuity, but I'm curious if there is something more general – Ryan Warnick Mar 5 '18 at 21:53