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Let $G$ be a locally compact group with right-invariant Haar measure $\mu$. I know that the space of compactly supported continuous functions $C_c(G)$ is dense in $L^1(G,\mu)$. If $G$ is 1st-countable, I can show that whenever $y_n\to y$ in $G$, we have that for any $\varphi\in C_c(G)$, if we let $\varphi_{y_n}(x)=\varphi(x y_n)$, then $$\varphi_{y_n}\xrightarrow[L^1(G,\mu)]{n\to\infty}\varphi_y$$ by dominated convergence, so that the map $$G\to L^1(G,\mu),\quad \phi\to\phi_y$$ is continuous. But is this still true if $G$ is not 1st countable?

Dominic Wynter
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  • How does your proof depend on countability? – Mariano Suárez-Álvarez Mar 10 '18 at 02:46
  • It requires countability because I need to use dominated convergence. – Dominic Wynter Mar 10 '18 at 02:50
  • I don't understand. What version of the dominated convergence theorem requires first countability? The theorem does not even involve a topology, no? – Mariano Suárez-Álvarez Mar 10 '18 at 03:08
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    Possible duplicate of A net version of dominated convergence? – Eric Wofsey Mar 10 '18 at 03:10
  • @MarianoSuárez-Álvarez: We are trying to prove a certain map $G\to L^1(G,\mu)$ is continuous. We can use dominated convergence to prove the map preserves convergence of sequences. But if $G$ is not first countable, that is not enough to conclude our map is continuous. – Eric Wofsey Mar 10 '18 at 03:12
  • That's what I thought was the problem, but I could not see why it was needed to be able to use the DCT, because we don't ;-) – Mariano Suárez-Álvarez Mar 10 '18 at 03:16
  • @EricWofsey I don’t understand the first step of the approved answer. How do we show uniform continuity is implied by compact support? – Dominic Wynter Mar 10 '18 at 03:27
  • In the notation of the answer, that is essentially just the statement that $g$ itself is a uniformly continuous function $G\to\mathbb{C}$. – Eric Wofsey Mar 10 '18 at 03:48
  • That was precisely my question — how do we show uniform continuity of $g$ from the fact that it has compact support and is continuous? This seems like a Lebesgue-number type argument to me, but I can’t get it to work. – Dominic Wynter Mar 10 '18 at 03:52
  • It's the same as the usual proof for metric spaces; just with the $\delta/2$s replaced by small enough neighborhoods of the identity. As a first step, given $\epsilon>0$, for each $x\in G$ choose a symmetric neighborhood $U_x$ of the identity such that $|g(y)-g(x)|<\epsilon/2$ whenever $yx^{-1}\in U_xU_x$ (the doubling of $U_x$ here is playing the role of dividing "$\delta$" by $2$). Now choose a finite open subcover of ${U_xx}_{x\in G}$. – Eric Wofsey Mar 10 '18 at 05:11
  • However, we require the neighborhood to be translates of a single neighborhood, such that for any $x\in K$ where $K$ is our compact set, $y\in Ux$ implies $|g(x)-g(y)|<\varepsilon$. We don’t have a single notion of size of the neighborhoods that we can minimize over $K$. – Dominic Wynter Mar 10 '18 at 06:32
  • Just intersect all the $U_x$'s once you have a finite subcover. – Eric Wofsey Mar 10 '18 at 07:59

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