We know that if a locally integrable function $f:\mathbb{R}\to\mathbb{R}$ is bounded, then the function $F(t)=\int_0^tf(x)dx$ is Lipschitz. In fact $$|F(t)-F(s)|=\left|\int_s^tf(x)dx\right|\leq|f|_\infty|t-s|.$$ I don't know what we can say if we replace the boundedness of $f$ by the weaker condition $$M:=\sup_{t\in\mathbb{R}}\int_t^{t+1}\left|f(x)\right|dx<+\infty.$$ I tried $$|F(t)-F(s)|\leq\int_{\lfloor s \rfloor}^{\lfloor t \rfloor +1}|f(x)|dx=\sum_{k= \lfloor s \rfloor}^{\lfloor t \rfloor}\int_{k}^{k+1}|f(x)|dx\leq M\left| \lfloor t \rfloor - \lfloor s \rfloor +1 \right|,$$ where $\lfloor t \rfloor$ denotes the largest integer not greater than $t$.

I am also interested if we can still have the uniform continuity of $F$.


Nope. Let $f$ be the function

$$f(x) = \left\{\begin{array}{cc} \frac{1}{2\sqrt x} &: (0,1) \\ 0 &: \text{ else}\end{array}\right.$$

This function is integrable ($\|f\|_1 = 1$), but the corresponding $F$ is not Lispchitz; one can directly compute that

$$F(x) = \Big(\sqrt x \Big) \chi_{[0,1]} + \chi_{(1, \infty)}$$

which is not Lipschitz at $0$.

  • $\begingroup$ Thanks !!! I don't know if we still have the uniform continuity of $F$ or not. $\endgroup$ – user50618 Jul 15 '14 at 1:04

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