Conditions for integral function to be Lipschitz

Question

I’m looking for conditions for the function $$F(x) = \int_{a(x)}^{b(x)} f(x,t)dt$$ To be Lipschitz. I tried to take $F(x_1) - F(x_2)$ and add and subtract things but the limits make it not provable. Any suggestions please?

Answer 1

If $f$ is bounded, and $g=a-b$ is Lipschitz, then $F$ is Lipschitz.

Edit: I only just noticed that $f$ also depends on $x$. The conditions then would be that $g$ is Lipschitz and $|f(x,t)-f(y,u)|<M|t-u|$ for some $M>0$ and all $x,y,t,u$. I don't know if this property has a name. It is not as nice as the case when $f$ depends only on $t$.

Answer 2

Assume $f$ is bounded, measurable, and uniformly Lipschitz in $x$, i.e. there is $c$ such that $|f(x,t) - f(y,t)| \le c |x - y|$ for all $t$, and $a(t)$ and $b(t)$ are Lipschitz and $b - a$ bounded. We have

$$ \eqalign{|F(x) - F(y)| &= \left|\int_{a(x)}^{b(x)} f(x,t)\; dt - \int_{a(y)}^{b(y)} f(y,t)\; dt \right|\cr &\le \left| \int_{a(x)}^{b(x)} f(x,t)\; dt - \int_{a(y)}^{b(y)} f(x,t)\; dt \right| + \left|\int_{a(y)}^{b(y)} f(x,t)\; dt - \int_{a(y)}^{b(y)} f(y,t)\; dt \right| } $$ If $G(t)$ is an antiderivative of $f(x,t)$, then $G$ is Lipschitz because $f$ is bounded, and we have $$ \eqalign{\left| \int_{a(x)}^{b(x)} f(x,t)\; dt - \int_{a(y)}^{b(y)} f(x,t)\; dt \right| &= \left| G(b(x)) - G(a(x)) - G(b(y)) + G(a(y))\right| \cr &\le |G(b(x)) - G(b(y))| + |G(a(y)) - G(a(x))|\cr &\le const \cdot |x - y|} $$ and since $f$ is uniformly Lipschitz in $x$, $$ \left|\int_{a(y)}^{b(y)} f(x,t) \; dt - \int_{a(y)}^{b(y)}\; f(y,t)\; dt \right| \le |b(y) - a(y)| c |x - y| $$