(Folland 8.5) What do they mean by continuity of translation? For each $y\in \mathbb R^n$, it is defined a translation operator $(\tau_y f)(x) = f(x-y)$. Folland's Proposition 8.5 says

If $1\leq p< \infty$, translation is continuous in the $L^p$ norm;
that is, if $f\in L^p$ and $z\in \mathbb R^n$, then $\lim_{y\to 0}
 ||\tau_{y+z} f - \tau_z f||_p =0$.

I'm a bit confused by what they mean by "continuous" here. Clearly they aren't saying that each $\tau_y$ is continuous from $L^p$ to itself, right? Is this just saying that for each $f\in L^p$ the function $\phi_f(x) : \mathbb R^n \to L^p$ given by $\phi_f(x) = \tau_xf$ is continuous?
 A: Yes, that condition means that $\tau_x f$ is continuous in $x\in\mathbb R^n$ for fixed $f$. (The continuity in $f$ for fixed $x$ is obvious).
WARNING. In a previous version of this answer I wrote "joint continuity in $(f, x)$ is false". That's not true. Indeed, if $x_n\to x_0$ and $f_n\to f$ then
$$
\begin{split}
\lVert \tau_{x_n}f_n-\tau_{x_0}f\rVert_p\le &\ \lVert \tau_{x_n}f_n-\tau_{x_n}f\rVert_p +\lVert \tau_{x_n}f-\tau_{x_0}f\rVert_p \\ 
=&\ \lVert f_n-f\rVert_p +\lVert \tau_{x_n}f-\tau_{x_0}f\rVert_p\to 0.
\end{split}$$
This proves that $(f, x)\to \tau_x f$ is a continuous map of $L^p\times \mathbb R^d$ into $L^p$.
MORE REMARKS. There are several forms of continuity for this kind of families of operators (the keyword to look for is "operator semigroup"). The one mentioned in this question, that is, continuity in $x$ for fixed $f$ is called strong continuity. There is another important one, which is actually stronger than the previous despite the name: uniform continuity. This is the statement that
$$\tag{FALSE}
\sup_{\lVert f\rVert_p=1} \lVert\tau_{x_n} f-\tau_{x_0} f\rVert_p\to 0,\quad \text{if }x_n\to x_0.$$
It is this one that does not hold for the translation group $\tau_x$, not the joint continuity as I mistakenly claimed.
