# Continuity under the integral sign?

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.

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

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 $e$.
• 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. – Vladimir Sotirov May 23 '14 at 0:54
• 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. – InTransit May 23 '14 at 8:04