it is well known that if a function $f:[0,T]\to\mathbb{R}$ satisfies the inequality $$\vert f(t)-f(s)\vert\leq \int_s^t{m(r) dr},$$ for $s<t$ and some $m\in L^1([0,T])$ then $f$ is absolutely continuous. On the other hand, if the function satisfies the inequality $$\vert f(t)-f(s)\vert\leq (g(s)+g(t))\vert t-s\vert,$$ for some $g\in L^1([0,T])$, then $f\in W^{1,1}([0,T])$, indeed both conclusions are the same.

My question is: if the inequality $$\vert f(t)-f(s)\vert\leq (g(s)+g(t))\vert t-s\vert+\int_s^t{m(r) dr}$$ is valid, do we have the same conclusion? I think that it is true, but I don't know how to show it.


I think the answer to this question is ``yes" if in the last inequality you assume that $g\in L^1$ and $m\in L^1$. There are two simple ways to see this:

  1. In one dimensional case, $f\in W^{1,1}$ if and only if $f$ is absolutely continuous. Both the first inequality and the second inequality implies $f$ is absolutely continuous, essentially just by definition using absolutely continuity of integral (when you fix a division of the interval, the right-hand side of the second inequality can be seen as a discrete approximation of the integral over the interval).

  2. The second convenient way is to use the “differential quotient", namely for $h\in (0,T)$, consider \begin{equation*} \Delta_hf(t)=\frac{|f(t+h)-f(t)|}{h}, \quad t\in [0,T-h]. \end{equation*} Then the Sobolev space $W^{1‚1}$ can be characterized as $$W^{1,1}([0,T])=\Big\{f\in L^1:\sup_{0<h<T}\int_0^{T-h}\Delta_hf(t)dt<\infty\Big\}.$$ It is straightforward to verify this from the last inequality you have.

  • $\begingroup$ Thanks, I had not remembered the characterization of $W^{1,1}$. $\endgroup$ – Julio Valencia Nov 26 '14 at 20:01
  • $\begingroup$ The fact that the second inequality implies absolute continuity is not so obvious although your sketch is correct. I did not know the second characterization of $W^{1,1}$ can you add references? $\endgroup$ – Piotr Hajlasz Apr 5 '18 at 0:11

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