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.

Davide Giraudo
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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 05 '12 at 02:57
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    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 05 '12 at 05:11
  • I added some details to the answer. Let me know if you need more elaborations. – t.b. May 05 '12 at 16:11

2 Answers2


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](http://math.stackexchange.com/questions/74447/) – t.b. May 05 '12 at 06:46
  • 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 05:42
  • @Jeff: Point (1) is nothing but the fact that a function with compact support is uniformly continuous. See Pedersen, [Analysis Now](http://books.google.com/books?id=a1R0livwR9AC), 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 02:04
  • After working through this (1 year later) as it applied to a problem I'm working on, just wanted to say thanks very much! – roo May 14 '13 at 21:29
  • Is it possible to say something about net convergence if the function is a Fourier transform of measure? – Canjioh Nov 18 '19 at 15:50

Here is a nice theorem that might interest you ("‎Methods of Modern Mathematical Physics‎: ‎Functional analysis‎", ‎Volume 1‎, ‎Michael Reed‎, ‎Barry Simon)‎:

‎Let $\mu$ be a regular Borel measure on a compact Hausdorff space $X$‎. ‎Let $(f_\alpha)$ be an increasing net of continuous functions. Then $f=\lim_\alpha f_\alpha \in L^1(X,\mu )$ if and only if $\sup_\alpha \| f_\alpha \|_1 < \infty$, and in this case $\lim_\alpha \| f-f_\alpha\|_1 = 0$‎.

Alex M.
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Ramin Faal
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  • Hard to understand how this could be useful: in the question, $G$ is not supposed compact, and the net of translates of $f$ is not increasing and not made of continuous functions either (they are only in $L^1$). This post seems to have been made quite at random. – Alex M. Jun 27 '19 at 21:02