# Shifting a function is continuous

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?

• @DavideGiraudo It is compact metric topological group with measure $1$. – Rudy the Reindeer Aug 5 '12 at 15:39
• @t.b. Please can you post this in answer? – Rudy the Reindeer Aug 5 '12 at 15:56
• "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. – Qiaochu Yuan Aug 5 '12 at 16:41
• "slightly puzzled" $\neq$ "confused". – Rudy the Reindeer Aug 15 '12 at 19:55

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 $S^1$.

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

• But your elements in group don't refuse to be called $x$ : ) – Rudy the Reindeer Aug 5 '12 at 16:31
• That is true but that's because functions $f$ and $h$ don't like a group element $g$ in between them :) – t.b. Aug 5 '12 at 16:32
• I will need some time to think about this. Thank you for this awesome answer. – Rudy the Reindeer Aug 5 '12 at 18:35
• @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! – t.b. Aug 5 '12 at 18:59
• 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. – t.b. Aug 13 '12 at 19:48

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).

• $X$ is topological group. Did not mention in question, I am sorry. – Rudy the Reindeer Aug 5 '12 at 15:40
• 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). – user20266 Aug 5 '12 at 15:51