## Continuity of the translation mapping in L^1

A problem from class. Fix an $L^1 (\mathbb{R})$ function $f(x)$. Define the translation mapping $T:\mathbb{R} \rightarrow L^1$ sending $h \rightarrow f(x+h)$. How does one prove that $T$ is continuous at $h=0$?

We begin by proving it for a dense set. One can either use compactly supported continuous functions, or instead 'very simple functions' (my own terminology). Let us use the latter, but I recommend you to try and prove it using compactly supported continuous function as well. By a 'very simple function', I mean a simple function $f=\sum a_i \chi _{E_i}$ where the 'very' means that each $E_i$ is an open interval. Given such a function $f$, we notice that by the triangle inequality,

However, it's not hard to see that this integral (for a fixed $i$) is the measure of the symmetric difference of $E_i$ and $E_i-h$. One can easily compute this integral, and show that this symmetric difference is either twice the measure of $E_i$ (whenever $|h|$ is larger than the measure of $E_i$) or is 2|h| (when $|h|$ is smaller than the measure of $E_i$). Note that this calculations hold because we assumed $E_i$ to be an open interval. Thus, if $h$ is small enough (small enough means smaller than the measure of all the $E_i$'s), we have

and therefore

This proves the statement for $f$ being a 'very simple function'. Let us now prove it in the general case. Let $f$ be any $L^1$ function. Fix $\varepsilon>0$ and take a 'very simple function' $g$ such that $\|f-g\|_{L^1} < \varepsilon$. By the triangle inequality,

where the last line holds because of the translation invariance of the Lebesgue integral. Now since $g$ is a 'very simple function' for which we already proven the translation to be continuous, when taking the limit $h\rightarrow 0$ we get

This is of course true for all $\varepsilon>0$, and therefore

as we wanted. Yay math.