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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

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