# Proving that an embedding $G \hookrightarrow BC(G)$ is continuous.

I'm going over Professor Tao's presentation of the Birkhoff-Kakutani theorem and I don't see how it follows that $j = (g \mapsto \tau_gf)$ (between "Lemma 2" and "Remark 2") is continuous.

I don't even see how you'd apply that uniform continuity property of $f$ because to show continuity of $j$ at $x \in G$ requires that for all $\epsilon>0$ we can find a neighbourhood $U$ of $x$ such that

$$\|j(x) - j(y)\|_\infty = \sup_{g \in G} |f(x^{-1}g) - f(y^{-1}g)| < \epsilon \text{ for all }y \in U.$$

But the uniform continuity property of $f$ doesn't seem to apply to this at all.

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The definition of a uniformly continuous function in that reference goes like this: $f:G\to \mathbb R$ is uniformly continuous if for every $\varepsilon>0$ there exists a neighborhood $V$ of the identity for which $[a\in V, \;b\in G \Rightarrow |f(ab)-f(b)|\le \varepsilon].$ I call this right uniform continuity but more or less half of the people call it left uniform continuity (this is unfortunate, I know). Regardless of this, fix $x\in G$ and $\varepsilon>0.$ Fix $V$ as above, and define $U=xV.$ Then, if $y\in U$, define $a=x^{-1}y\in V;$ for every $g\in G,$ put $b=y^{-1}g$; since $x^{-1}g=ab$, we deduce that $|f(x^{-1}g)-f(y^{-1}g)|\le \varepsilon.$ I hope I got it right :)
Edit: The most compact and easiest-to-remember way of defining this (right) uniform continuity for me is the following: For every $\varepsilon>0$ there exists a neighborhood of the identity $U$ in $G$ such that $[xy^{-1}\in U\Rightarrow |f(x)-f(y)|\le \varepsilon.$