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(Sorry for asking so many questions of the same type. There is an underlying issue that I think once resolved will allow me to understand them all at once.)

Let $G$ be a locally compact group, and $f\in L^{1}(G)$. I'm trying to verify that $\|f^{*}\|_{p} = \|f\|_{p}$, where the involution $*$ is defined by $f^{*}(x) = \delta(x)^{-1}\overline{f(x^{-1})}$.

So my computations so far look like:

\begin{eqnarray*} \int_{G}f^{*}(y)\mu(dy) &=& \int_{G}\left|\delta(y^{-1})\overline{f(y^{-1})}\right|^{p}\mu(dy)\\ &=& \int_{G}\delta(y^{-1})^{p}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy)\\ \end{eqnarray*}

Here I am stuck at the central issue I am facing where the calculations stop making sense and I feel like I'm just throwing things around hoping it's true.

How do I use the definition of the $\delta$ map:
$$\text{the unique group homomorphism }\delta:G\to(0,\infty)\text{ such that }\\\delta(x)\mu(E) = \mu(Ex)\text{ for all Borel sets }E$$

when its argument is not independant of the variable of integration?

Ted has already confirmed in his answer to:

How to use the modular function of a locally compact group?

that the definition of $\delta$ implies:

$$\int_{G}\delta(x)f(y)\mu(dy) = \int_{G}f(y)\mu(dy\cdot x)$$

If I try to use that step in my current problem, I end up with

\begin{eqnarray*} \int_{G}\delta(y^{-1})^{p}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy) &=& \int_{G}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy\cdot (y^{-1})^{p}) \end{eqnarray*} which I don't have any clue how to interpret, and I suspect is wrong. What is the correct way to handle this function? I dont know what measure to associate with the notation $\mu(dy\cdot (y^{-1})^{p})$.

I think the answer to this question is what I really need to be able to finish the verifications I'm working on.

Thanks to whoever can help me!

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roo
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1 Answers1

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I think you would find the last chapter of Real Analysis by Folland helpful (it covers the fundamentals of Haar measures). The way I think of it, $\delta^{-1}$ is a factor attached to $\mu$ (a left Haar measure), not to $f$. To begin with, the measure $\delta^{-1}\mu$ is right-invariant: $$\int_{E\,y}\delta(x)^{-1}\,d\mu(x)=\int_{E}\delta(xy)^{-1}\,\delta(y)\,d\mu(x)= \int_{E}\delta(x)^{-1} \,d\mu(x) $$ And here is another way to get a right invariant measure from $\mu$. Let $\iota:G\to G$ be the inversion, $\iota(x)=x^{-1}$. This is an anti-homomorphism, it changes the order of group multiplication. Therefore, the pushforward $\iota^*\mu$ is a right Haar measure.

Fact: $\delta^{-1}\mu = \iota^*\mu$.

Proof. They are both right Haar measures, therefore one is a constant multiple of each other. To see that the multiple must be $1$, consider a symmetric neighborhood of identity, called $U$, which is small enough so that $|\delta(x)^{-1}-1|<\epsilon$ for $x\in U$. (Such $U$ exists because the modular function is continuous.) Now, $\iota^*\mu(U)=\mu(U)$ by symmetry, and also $|\delta^{-1}\mu(U)-\mu(U)|<\epsilon \,\mu(U) $. Since the ratio of $\iota^*\mu(U)$ to $\delta^{-1}\mu(U)$ can be made arbitrarily close to $1$, but is also a constant (independent ot $U$), the ratio is in fact $1$. $\Box$

The rest is a tautology from abstract measure theory: $$\int (f\circ \iota) \,d\mu = \int f \,d (\iota^*\mu) \tag1$$ hence $$\int f(x^{-1})\,d\mu(x) = \int f(x)\,\delta(x)^{-1} \,d\mu(x) \tag2$$ Actually, you want to put $f\circ \iota$ instead of $f$ here, so that the result is exactly as required: $$\int f(x )\,d\mu(x) = \int f(x^{-1})\,\delta(x)^{-1} \,d\mu(x) \tag3$$

But I don't think your construction produces an isometry on $L^p$, at least not as stated. The identity (3) immediately generalizes to $$\int |f(x )|^p\,d\mu(x) = \int |f(x^{-1})|^{p}\delta(x)^{-1} \,d\mu(x) \tag4$$ but if you rewrite (4) by attaching $\delta^{-1}$ to $f$, it has to be raised to power $1/p$.

75064
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  • Thanks so much for the thorough answer. I will read it tomorrow morning with a fresh pair of eyes. Just one clarification:

    When you say the measure $\lambda:=\delta^{-1}\mu$ is right invariant. I get confused because $\delta$ is not a set function, so I don't know how to interpret this. What is the definition of $\lambda(E)$ for Borel subsets $E$ of $X$?

     – roo May 13 '13 at 07:50
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    @Kyle One can multiply a measure $\mu$ by a nonnegative measurable function $f$, and obtain a new measure $\lambda$, namely $\lambda(E)=\int_E f,d\mu$. This is often denoted $f\mu$, or $f,d\mu$. – 75064 May 13 '13 at 14:47
  • I think that last fact may have what I needed to understand all of my questions. so $\delta^{-1}\mu(E) = \int_{G}\delta(y^{-1})\mu(dy)$? (using my favourite notation..) – roo May 13 '13 at 18:32
  • As for the issue, I made a mistake, it's only an isometry on $L^{1}(G)$, turning $L^{1}(G)$ into an involutive Banach Algebra. – roo May 13 '13 at 18:35
  • @Kyle Yes, it's the same thing. In everything that involves both functions and measures, mathematical notation is remarkably inconsistent. – 75064 May 13 '13 at 18:46
  • Does the equation (1) hold for any homeomorphism $\iota:G\to G$? – roo May 15 '13 at 00:33
  • @Kyle Yes, and for continuous map too (even measurability is enough). It's pretty much the definition of pushforward measure. – 75064 May 15 '13 at 00:39
  • So just to confirm: for general measurable $E$, measurable $\iota:G\to G$, $\int_{E}f\circ \iota d\mu = \int_{\iota(E)}fd(\iota^{*}(\mu))$? – roo May 15 '13 at 00:59
  • @Kyle No, you have to be careful with images of sets. For example, the sets $[-1,1]$ and $[0,1]$ have the same image under the map $x\mapsto x^2$. But it is true that $\int_{\iota^{-1}(E)}f\circ i,d\mu = \int_E f,d(\iota^*\mu)$. – 75064 May 15 '13 at 01:03
  • Thank you very much for your help. That was amazingly helpful! – roo May 15 '13 at 01:04
  • Sorry, I'm stuck once again. In the right hand side of (3) $\delta(x)^{-1}$ attached to $\mu$. But in the definition of $|f^{*}|_{1}$, the $\delta(x)^{-1}$ will be attached to $f(x)$. I suspect the solution is that it doesn't matter and both interpretations yield the same result. Assuming this is true, my last question (written in my favourite notation) is:

    How do we know that $\int_{G}\delta(x)^{-1}f(x)\mu(dx) = \int_{G}f(x)\delta^{-1}\mu(dx)$.

     – roo May 15 '13 at 21:57
  • @Kyle Multiplication is commutative and associative. My remark about attaching a factor was to describe the way I think of it. – 75064 May 15 '13 at 22:00