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Let $G$ be a locally compact Hausdorff Abelian topological group. Let $\mu$ be a Haar measure on $G$, i.e. a regular translation invariant measure. Let $f$ be fixed in $\mathcal{L}^1(G, \mu)$. Define the function from $G$ to $\mathcal{L}^1(G)$ that assigns to each $y$ in $G$ the function that takes $x$ in $G$ to $f(xy^{-1})$. I.e. it assigns the $y$-translate of $f$. Is this a continuous function of $y$ into $\mathcal{L}^1$?

On p.94 3.10 of these notes it is asserted that it is. The usual proof of this sort of thing in cases where $G=\mathbb{R}$, say, is you use density of compactly supported continuous functions, and then dominated convergence. The former survives, but dominated convergence does not work on nets. In fact, a limit of a net of measurable functions need not be measurable.

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The map is continuous. Try checking this first when $f$ is continuous with compact support. That is a dense subset of $L^1(G)$, so you can try to extend the result to that larger space, once it is established on the smaller space, by an approximation argument. –  KCd May 5 '12 at 2:57
    
It is the part where I need to prove it for f continuous with compact support that I am having trouble with. There is no dominated convergence to use, as there is in the R case, because in the R case continuity is wrt metrics, whereas here it comes out of a topological group, hence nets are required. –  Jeff May 5 '12 at 5:11
    
I added some details to the answer. Let me know if you need more elaborations. –  t.b. May 5 '12 at 16:11

1 Answer 1

up vote 7 down vote accepted

There simply is no version of dominated convergence which holds for nets, not even if all the functions involved are continuous and compactly supported.

Here's a cautionary example on the unit interval which you can modify to concoct all sorts of counterexamples:

Let $\Lambda$ be the set of finite subsets of $[0,1]$ ordered by inclusion. For each $\lambda \in \Lambda$ choose a continuous function $f_{\lambda}: [0,1] \to [0,1]$ such that $f_{\lambda}(x) = 1$ for all $x \in \lambda$ and $\int f_{\lambda} \leq \frac{1}{2}$. Then $f_{\lambda} \to 1$ everywhere on $[0,1]$, while $\frac{1}{2} \leq \int (1 - f_{\lambda}) \nrightarrow 0$…

In particular, (possible failure of) measurability of the limit is not really the issue.


That said, try the following approach instead (as suggested by KCd):

  1. Show that for $g \in C_c(G)$ the map $y \mapsto g_y$ (where $g_{y}(x) = g(xy^{-1})$) is uniformly continuous as a map $G \to C_c(G)$ where the latter is equipped with the sup-norm. More precisely: for every $\varepsilon \gt 0$ there is a neighborhood $U$ of the identity such that $\lVert g_y - g\rVert_{\infty} \lt \varepsilon$ for all $y \in U$.

  2. If $g \in C_{c}(G)$ then there is a compact set $K$ outside of which $g$ vanishes. Take a symmetric compact neighborhood $K'$ of the identity. Then $KK'$ is compact and thus has finite Haar measure. Apply 1. to find a compact neighborhood of the identity with $C \subset K'$ such that $\lVert g_y - g\rVert_\infty \leq \varepsilon/ \mu(KK')$ for all $y \in C$. Note that $g_{y} - g$ vanishes outside $KK'$ so that $$ \lVert g_y - g\rVert_1 = \int \lvert g_y - g\rvert \leq \int_{KK'} \lVert g_y - g\rVert_\infty \leq \varepsilon \quad \text{for all }y \in C. $$

  3. Now use that $C_{c}(G)$ is dense in $L^1(G)$. Given $\varepsilon \gt 0$ choose $g \in C_c(G)$ such that $\lVert f-g\rVert_1 \lt \varepsilon$, and note that invariance of Haar measure gives $\lVert f_y - g_y\rVert_1 \lt \varepsilon$. Choose $C$ for $g$ and $\varepsilon \gt 0$ as in point 2.

    For every net $y_\lambda \to 1$ in $G$ we have $y_{\lambda} \in C$ eventually so that $$ \lVert f_{y_\lambda} - f\rVert_1 \leq \lVert f_{y_\lambda} - g_{y_\lambda}\rVert_1 + \lVert g_{y_\lambda} - g\rVert_1 + \lVert g-f\rVert \lt 3\varepsilon $$ and we conclude that $y \mapsto f_y$ is (uniformly) continuous for all $f \in L^1(G)$.

  4. Generalize to $L^p(G)$ for $1 \leq p \lt \infty$.

  5. Added: To see that the action of $G$ on $L^\infty(G) = L^1(G)^\ast$ is not strongly continuous, consider $G = S^1$ and let $I = [0,t]$ with $t$ be a non-trivial interval. Then the characteristic function $f$ of $I$ has the property that $\lVert gf - f\rVert_\infty = 1$ whenever $g \neq 1$. In particular, for $g_n \to 1 \in G$ we can't have $g_n f \to f$. In fact, it is essentially the definition of right uniform continuity of $f$ that $\lVert gf - f\rVert_\infty \to 0$ whenever $g \to 0$ in $G$.

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There's no need to assume $G$ to be abelian. The result you ask about is proved in any text on abstract harmonic analysis, e.g. the ones suggested in my answer here –  t.b. May 5 '12 at 6:46
    
I have not replied to this yet because I have been unable to prove, and also unable to secure a text in which (1) from your comment, t.b., is proven. I have not given much thought to the things after that because I'm still stuck there. –  Jeff May 11 '12 at 5:42
    
@Jeff: Point (1) is nothing but the fact that a function with compact support is uniformly continuous. See Pedersen, Analysis Now, Lemma 6.6.10. Relevant to your question are also Lemma 6.6.11, as well as Proposition 6.6.19 (and some of the other facts in that section). As I mentioned in my previous comment, all that's covered in essentially all the texts I mention in the linked answer. –  t.b. May 14 '12 at 2:04
    
After working through this (1 year later) as it applied to a problem I'm working on, just wanted to say thanks very much! –  Kyle Schlitt May 14 '13 at 21:29

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