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:
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.
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.