Is $e^{-\int_0^t f(u)du} $ a Lipschitz function?

Question

If $f$ is a Lipschitz function and non-negative, do we have the following function Lipschitz $$ e^{-\int_0^t f(u)du} $$

It is enough to show that $$ |e^{-\int_0^x f(u)du}-e^{-\int_0^y f(u)du}|\le L|x-y|. $$

Answer 1

As an immediate consequence of the mean value theorem, if $h$ is a differentiable function on $(a, b)$, $$|h(b)- h(a) |\le \sup_{(a,b)} |h^\prime(t)||b-a|$$ Now $$\frac{d}{dt} \exp\left(-\int_0^t f(u)\, du\right) = -\exp\left(-\int_0^t f(u)\, du\right)f(t) $$ For any given interval $[a,b]$ this expression is bounded by a constant $C = C(f,a,b)$ - for this conclusion weaker conditions (like continuity of $f$) would suffice.

By combining these two results you get Lipshitz continuity of $\exp\left(-\int_0^t f(u)\, du\right) $ with Lipshitz constant $C$.

As a potentially quite poor bound you could take $C= \exp (\frac{d^2L}{2} + f(0)d)(Ld+f(0))$ if $L$ is the Lipshitz constant of $f$ and $d=\max\{a,b\}$, since

$$|f(t)|\le L |t|+ f(0)$$

and

$$|\int_0^t f(u) du| \le\left| \int_0^t |f(u)| du \right|\le\left|\int_0^t L|u| +f(0) du\right|\le \frac{L}2 |t|^2 + f(0)|t| $$