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I'm slightly puzzled by the following: if $g(t)$ is a function in $L^q(X)$ then we can show that $g(t-x)$ is continuous function of $t$, i.e. for $\varepsilon > 0$ we can find $\delta$ such that $d(x,y)<\delta$ implies $\|g(t-x) - g(t-y)\|_q < \varepsilon$.

But $g$ is not necessarily continuous. Is this result stating something like $g$ is continuous with respect to norm $\|\cdot\|_q$? Because if $\tau_x$ is translation by $x$ then $g(t-x) = g \circ \tau_x$ is not necessarily continuous. When continuous I mean in topology on $X$ (e.g. $X \to \mathbb R$). Does this sort of continuity have a name?

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    @DavideGiraudo It is compact metric topological group with measure $1$.2012-08-05
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    @t.b. Please can you post this in answer?2012-08-05
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    "Continuous" is a property that depends on three things: a function $f : X \to Y$, a topology on $X$, and a topology on $Y$. Write down exactly what these things are for $g$ and for translation and your confusion should disappear.2012-08-05
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    "slightly puzzled" $\neq$ "confused".2012-08-15

2 Answers 2

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Let me write $G$ instead of $X$ since we're talking about compact groups and my groups refuse to be called $X$...

Silliness aside, the point here is that for each $x \in G$ the map $\tau_x\colon L^q(G) \to L^q(G)$ is an invertible (in fact isometric) linear map. You get a group homomorphism $\tau\colon G \to {\rm GL}(L^q(G))$ and as such $\tau$ is strongly continuous (provided $1 \leq q \lt \infty$). What this means is that for each $f \in L^q(G)$ the function $x \mapsto \tau_x f$ is continuous as a function $(G,d) \to (L^q(G),\lVert\cdot\rVert_q)$, which is precisely the continuity Thomas mentions in his answer.

The reason this is the case is that the space of continuous functions $C(G)$ is dense in $L^q(G)$ and that continuous functions on compact spaces are uniformly continuous. For (uniformly) continuous functions $f\colon G \to \mathbb{C}$ we have that $x \mapsto \lVert \tau_x f - f\rVert_\infty$ is continuous and since $\mu(G) = 1$ we have that $\lVert f \rVert_q \leq \lVert f\rVert_\infty$, so for continuous functions $f$ the map $x \mapsto \tau_x f$ is continuous with respect to all $L^q$-norms.

Now, for continuous functions it is clear that $\lVert \tau_x f\rVert_q = \lVert f\rVert_q$, so $\tau_x\colon (C(G),\lVert\cdot\rVert_q) \to (C(G),\lVert \cdot \rVert_q)$ is an isometry and thus it extends uniquely to an isometry of the completion $L^q(G)$ of $(C(G),\lVert \cdot \rVert_q)$. Since $\tau_{-x} \tau_x = {\rm id}_{C(G)} = \tau_{x} \tau_{-x}$ we see that each of those extensions is invertible. To see that the resulting map $\tau\colon G \to {\rm GL}(L^q(G))$ is strongly continuous, let $h \in L^q(G)$ and $\varepsilon \gt 0$. Then there is $f \in C(G)$ such that $\lVert f - h\rVert_q \lt \varepsilon$ and by continuity of $x \mapsto \lVert\tau_x f - f\rVert_\infty$ there is $\delta$ such that $d(x,e) \lt \delta$ implies that $\lVert\tau_x f - f\rVert_\infty \lt \varepsilon$. But then $$ \begin{align*} \lVert \tau_x h - h\rVert_q & \leq \lVert \tau_x h - \tau_x f\rVert_q + \lVert \tau_x f - f\rVert_q + \lVert f - h\rVert_q \\ &\leq \lVert \tau_x h - \tau_x f\rVert_q + \lVert \tau_x f - f\rVert_\infty + \lVert f - h\rVert_q \\ & \lt 3 \varepsilon \end{align*} $$ where we have used that $\lVert \tau_x h - \tau_x f\rVert_q = \lVert h - f\rVert_q$ by translation invariance of Haar measure.

To see that $q \lt \infty$ is essential, consider the characteristic function of a non-trivial segment $[0,\alpha]$ of the circle group $S^1$.


Added:

All I said here extends (with small modifications) to non-commutative locally compact groups, see the threads

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    But your elements in group don't refuse to be called $x$ : )2012-08-05
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    That is true but that's because functions $f$ and $h$ don't like a group element $g$ in between them :)2012-08-05
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    I will need some time to think about this. Thank you for this awesome answer.2012-08-05
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    @bananalyst: sure, take your time, there's no hurry... I'm aware that this is a bit heavy-going, so if you'd like to have some more details, feel free to ask. Thanks for the nice words!2012-08-05
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    For a map to be strongly continuous one would expect map $f: B(X,Y) \to Z$ or maybe $f: B(X,Y) \to B(X,Y)$. Then SOT is on domain or both domain and codomain. But $\tau$ is defined from group $G$ to $B(X,X)$ where $X=L^q$. Somewhat counter-intuitive. Sorry for taking a long time. And I'm not even done reading your answer.2012-08-13
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    We have a map $\tau\colon G \to GL(Y) \subset B(Y)$, where $G$ has its metric topology, $Y = L^q$ and where we equip $GL(Y)$ the relative strong topology from $B(Y) = B(Y,Y)$, so the topology generated by the semi-norms $\lvert T\rvert_y = \lVert Ty\rVert_Y$. Domain $G$ has metric topology, codomain $GL(Y)$ has strong topology.2012-08-13
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    Yes. Not intuitive. : )2012-08-13
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    In your last paragraph, did you mean to write "Now, for $L^q$ functions..." instead of "Now, for continuous functions..."?2012-08-15
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    No, I meant continuous functions: I'm talking about continuous functions right afterwards. The reason I'm phrasing it this way is that often it is much easier to prove the transformation formula for continuous functions than for measurable ones.2012-08-15
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    Ok. Now I'm almost finished with your answer, just need to read the two links, then I'll accept. I'll do that soon.2012-08-15
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You are confusing things. First we need to assume that $X$ is a subset of a vector space $V$ so we can form differences. Then we need to assume that $g\circ \tau_t$ is defined on $X$ if $g$ is, e.g. by assuming that $X = X - t$ (e.g. if $X=\mathbb{R}$) or by assuming that $g$ has compact support, whatever -- there are several possibilities. What is meant by the continuity of translation is that the map $t\mapsto g\circ\tau_t $ is continuous as a map from a neighbourhood of $0\in V$ to $L^q(X)$.

This does not imply in any sense that the translated function is continuous, but it means that if $|s-t|$ is small, then $||g\circ\tau_s - g\circ\tau_t||_q$ is small, too (in the $\varepsilon, \delta$ sense).

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    $X$ is topological group. Did not mention in question, I am sorry.2012-08-05
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    then "-" is the (inverse) group action. If $X$ is the full group you don't have to worry about $g\circ\tau_t$ being well defined (you do have to show that it maps $L^q$ to $L^q$, though).2012-08-05