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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 $$ |e^{-\int_0^x f(u)du}-e^{-\int_0^y f(u)du}|\le L|x-y|. $$

RobPratt
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Hermi
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    You should provide more information about $f$, starting from its domain and its regularity. You may also enjoy the fact that a differentiable function is Lipschitz if and only if its derivative is bounded. – blamethelag May 07 '22 at 17:13
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    Is $f$ assumed to be non-negative? Hint: on which interval the function exp is Lipschitz? – Christophe Leuridan May 07 '22 at 18:26
  • @ChristopheLeuridan Yes, $f$ should be non-negative. https://math.stackexchange.com/questions/3848634/the-exponential-function-is-locally-lipschitz-continuous-with-the-lipschitz-cons shows that $\exp$ is locally Lip? – Hermi May 08 '22 at 02:01
  • You need to restrict your domain for your function to be lipschitz. Consider $f=1$, then the resulting function $e^{-t}$ is not lipschitz. However, if you restrict to $[C,\infty)$ it is lipschitz (here $C$ is any constant). – Severin Schraven May 10 '22 at 01:36
  • @SeverinSchraven Is it Lipschitz on a closed interval? say $[0,a]$? – Hermi May 10 '22 at 03:17
  • If $f$ is continuous and non-negative then $$ \left| {\frac{d}{{dt}}e^{ - \int_0^t {f(u)du} } } \right| = e^{ - \int_0^t {f(u)du} } f(t) \le f(t) $$ is bounded on any compact interval in the domain of $f$, and hence your $t\mapsto e^{ - \int_0^t {f(u)du} }$ is Lipschitz. – Gary May 10 '22 at 03:20
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    @Gary in your last inequality you seem to be asssuming that the exponent is nonpositive. This need not be correct for $t<0$. – Thomas May 10 '22 at 04:55
  • @Thomas Oh, you are right. My bad. – Gary May 10 '22 at 05:54

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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

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

Thomas
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  • Why $|f(t)|\leq L|t|$ and not $|f(t)-f(0)|\leq L|t|$? Why $(L/2)|t|^3$ and not $(L/2)|t|^2$? – Gary May 10 '22 at 06:09
  • @Gary The example estimate I wrote in a hurry and I guess I have to fix that, my bad. Will not be able to do that before late noon, though... – Thomas May 10 '22 at 06:39
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    Could you clarify $|f(t)|<L|t|$? – Gary May 10 '22 at 23:02
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    @Gary I can't ;-) Somehow I was under the impression that $f(0)=0$. Not my day, yesteday - I hope things are finally fixed now. Thanks for pointing that out. – Thomas May 11 '22 at 05:49
  • No just fix the spelling of Lipschitz and your solution is complete. :) – Gary May 11 '22 at 05:51
  • But we need to prove that $ |e^{-\int_0^x f(u)du}-e^{-\int_0^y f(u)du}|\le L|x-y|. $? – Hermi May 12 '22 at 00:05
  • @Thomas Is this function $L-$Lipschitz and $L<1$? How to get a smaller $0<L<1$? – Hermi May 12 '22 at 02:11
  • @Hermi 'we need to prove...' - that follows from the combination of the first equation with the bound on the derivative for the exponential. From what I have written the function is $C$ Lipschitz with a constant $C$ which depends on $f$ and the boundary of the interval, which I denoted by $C(a,b,f)$. I do not know whether that $C$ might be less than $1$. There was also no such information about the $L$ for the $f$ in your question. If you want to apply, e.g., the contraction theorem, you have to provide more information (on $L$ and $f$). – Thomas May 12 '22 at 04:59
  • @Thomas Yes, I try to prove contraction. It seems that we need the $f$ is $L$-Lip with $L<1$? – Hermi May 12 '22 at 06:20
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    @Hermi you should formulate your question in such a way that people don't have to guess what you are really after. If you look at how I defined $C$ you will note that the length of the intervall enters there as a factor. If you happen to know that $f(0) = 0$ (or $f(0) < 1$ you could conclude that $C<1$ in a neighbourhood of $0$. – Thomas May 12 '22 at 08:10