2
$\begingroup$

I've been reading Folland's Harmonic analysis book, in which he claims the following on page 56:

  • Suppose $G$ is a locally compact (and of course Hausdorff) topological group G, $H$ a (closed) subgroup of $G$, and $\displaystyle f\mapsto\int_Hf(h)dh$ a continuous linear functional on $C_c(H)$, the compactly supported functions on $H$. Then for any $f\in C_c(G)$, the function on $G$ given by $\displaystyle g\mapsto \int_Hf(gh)dh$ is obviously continuous.

How is this obvious? The only remotely close results of this sort I have seen are:

  1. If $f\in C_c(X\times Y)$ and $\mu$ is a Radon measure on $Y$, then $\displaystyle x\mapsto \int f(x,y)d\mu(y)$ is continuous.

  2. If $X$ is non-empty subset of a metric space, and $(Y,\Sigma,\mu)$ is a measure space, then $\displaystyle x\mapsto\int_Y f_x(y)d\mu(y)$ is continuous if $f\colon X\times Y\to\mathbb R$ is such that

    • each $f_x$ (given by $f_x(y)=f(x,y)$) is measurable

    • $x\mapsto f(x,y)$ is continuous for almost all $y$

    • there is an integrable $g\colon Y\to\mathbb R$ so that $|f(x,y)|\leq g(y)$

Neither of these shows what Folland claims since $(g,h)\mapsto f(g\cdot h)$ does not have to be compactly supported, and locally compact groups don't have to be metrizable.

$\endgroup$
  • $\begingroup$ Just a guess: can you uniformly approximate $f$ by a sum of things that look like delta functions? $\endgroup$ – Qiaochu Yuan May 20 '14 at 4:59
1
$\begingroup$

OK, this should work. Since $f$ is $C_c(G)$, so is $L_{g^{-1}}f$. Let $K,K_g$ be the respective supports. Now apply the fact that $f$ is left uniformly continuous (cf. Prop 2.6, p.34) and that the Haar measures of compact sets are finite (Folland, $Real \ Analysis$, 2nd ed., p.341, Prop.11.4): If $F(g):= \int\limits_Hf(gh)\,dh$, one could prove the continuity of $F$ at $e$ as follows. Given $\epsilon>0$ there is an open $U\subset G$, a nbd. of $e$ such that for each $g\in U$, $||L_{g^{-1}}f - f||_\infty < \epsilon$. Hence, on $U$ $|F(g) - F(e)| = \left|\int\limits_H f(gh)\,dh - \int\limits_H f(h)\,dh\right| \le \int\limits_{K\cup K_g}|L_{g^{-1}}f(h) - f(h)|\,dh \le \epsilon\,\mu(K\cup K_g)$ where $\mu$ is the Haar measure on $H$. Now using the local compactness of $G$ and the usual compactness arguments, the RHS can be made independent of $g$ in a nbd of $e$.

$\endgroup$
  • $\begingroup$ Yes. Do you have a reference for this argument? Unless I'm misreading, Nachbin's book seems to mistakenly quote result #1 in my question (which leads me to believe Folland may have that incorrect argument in mind). In any case, I figured out that this proof can be unraveled and modified to show that for $f\in C_c(G/H)$, the association $gH\mapsto f_{gH}$ where $f_{gH}(\gamma)=f(\gamma gH)$ is a continuous map $G/H\to C_c(\Gamma)$ as long as one of $\Gamma$ or $H$ is compact (but both are closed). Any reference to this stronger fact would be appreciated as well. $\endgroup$ – Vladimir Sotirov May 23 '14 at 0:54
  • $\begingroup$ The refs for the "left uniformly continuous" bit are Folland (AHA and RA). The compactness argument is of a routine sort. Btw, the dominated convergence theorem can also be used if instead one shows using compactness that the integrand is majorised by an $L^1$ function on some nbd of $e$. The first result quoted in your OP is proved as 7.34 Lemma, p.227, in Folland's RA, 2ed, provided both $\mu$ and $\nu$ are Radon. I don't have a reference to hand for the continuity of the map you mention. $\endgroup$ – InTransit May 23 '14 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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