Continuity of the translation mapping in L^1

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,

\|T(h) - T(0)\|_{L^1} \leq \sum_i |a_i| \int_{\mathbb{R}} |\chi_{E_i} (x+h) - \chi_{E_i} (x)| dx

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

\|T(h) - T(0)\|_{L^1} \leq 2|h| \sum_i |a_i|

and therefore

 T(h)\xrightarrow[h\rightarrow 0]{L^1} T(0)

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,

\begin{array}{cl}<br />
\|T(h)-T(0)\|_{L^1} & \leq\|f(x+h) - g(x+h)\|_{L^1} + \|g(x+h) - g(x)\|_{L^1} + \| g(x) - f(x) \|_{L^1} \\&\leq 2\varepsilon +\|g(x+h) - g(x)\|_{L^1}\end{array}

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

\lim _{h\rightarrow 0}\|T(h)-T(0)\|_{L^1} \leq 2\varepsilon

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

 T(h)\xrightarrow[h\rightarrow 0]{L^1} T(0)

 as we wanted. Yay math.