I've been trying to prove that the space $BV[0, 1]$ of elements of $L^1 [0, 1]$ with bounded variation, is compactly embedded in $L^1[0, 1]$, using the Fréchet-Kolmogorov theorem.
For context, a function $f : [0, 1] \to \mathbb C$ is said to have bounded variation if its total variation, defined by $$ V_0^1 (f) = \sup_P \sum_{j = 1}^n |f(x_j) - f(x_{j-1})| $$ is finite (where the supremum I taken over the set of partitions $P = \{ x_0, x_1, ..., x_n \}$ with $0 = x_0 < x_1 < ... < x_n = 1$). If $f$ is instead an equivalence class of functions equal a.e. (e.g., an element of $L^1$), $f$ is said to be of bounded variation if it has a version that has bounded variation, and we set $V_0^1 (f) = \inf\{ V_0^1(g) : g \in f \}$.
The space $BV[0, 1]$ is endowed with the norm $\|f\|_{BV} = \|f\|_1 + V_0^1(f)$. To prove that the embedding $BV \hookrightarrow L^1$ is compact, it suffices to show that $B = \{ f \in BV[0, 1] : \|f\|_{BV} \le 1 \}$ is relatively compact in $L^1$.
Since we're working over $[0, 1]$ which is a bounded subset of $\mathbb R$, we have equitightness of $B$. All that remains to check is the equicontinuity property: $$ \| \tau_h f - f \|_1 \underset{h \to 0}{\longrightarrow} 0, \text{ uniformly for } f\in B $$ where $\tau_h f (x) = f(x + h)$.
Note: functions defined on $[0, 1]$ are extended as usual by the value $0$ outside of $[0, 1]$.
My attempt: let $\epsilon > 0$ and consider a version $g_\epsilon$ of $f \in B$ s.t. $V_0^1(g_\epsilon) \le V_0^1(f) + \epsilon$ (such a $g_\epsilon$ exists by definition of the infimum).
I tried using the inequality $$ |g_\epsilon (a) - g_\epsilon (b)| \le V_a^b (g_\epsilon) \label{1}\tag{$*$} $$ with $0\le a\le b \le 1$, in the hope of being able to use the condition that $\|f\|_{BV} \le 1$ to get something in the likes of $\| \tau_h f - f \|_1 \le \epsilon + R(h)$, where $R(h)$ would be something independent of $f$ and $R(h) \underset{h \to 0}{\longrightarrow} 0$. However, I am unsure how to proceed from \eqref{1} to obtain such an inequality. Does anyone have any idea?
Let $\delta>0$ be small and take $0<\varepsilon<\delta$. If you play with the fundamental theorem of calculus you should get $$\int_0^{1-\delta}|f_\varepsilon(x+h)-f_\varepsilon(x)|dx\le h\int_0^1|f’_\varepsilon(x)|\, dx=h V^1_0(f_\varepsilon).$$ Then you use the fact that the variation of the mollifiers is less than the variation of the function (see variation of mollifier) so you get$$\int_0^{1-\delta}|f_\varepsilon(x+h)-f_\varepsilon(x)|dx\le h V^1_0(f).$$ Then you let $\varepsilon $ goes to zero and use Fatous’s lemma on the left hand side. To use mollifiers, you should probably extend your function a bit outside $(0,1)$, that is define $f(x)=f(1)$ for $x>1$ and $f(x)=f(0)$ for $x<0$. I am skipping a lot of details.
Then you have $$\int_{1-\delta}^1|f(x+h)-f(x)|\, dx\le 2\delta\sup|f|.$$ I guess you could take $\delta=h$.
EDIT A simpler proof that does not use mollifiers. Extend $f$ to be $f(1)$ for $x\geq1$. Let $m=\lfloor1/h\rfloor$, so that $mh\leq1\leq(m+1)h$. By the change of variables $x=kh+y$, \begin{align*} \int_{0}^{1}|f(x+h)-f(x)|\,dx & \leq\sum_{k=0}^{m}\int_{kh}^{(k+1)h}% |f(x+h)-f(x)|\,dx\\ & =\sum_{k=0}^{m}\int_{kh}^{(k+1)h}|f(x+h)-f(x)|\,dx\\ & =\int_{0}^{h}\sum_{k=0}^{m}|f((k+1)h+y)-f(kh+y)|\,dy\\ & \leq h\operatorname*{Var}f \end{align*}