Lipschitz condition and Leibniz rule

[phoenix2014]

Hi,
I am reading a paper that states: "We note that if an integrable function satisfies the Lipschitz condition of order one, then differentiation and integration can be interchanged. This provides a more compact way to take the derivative. Consequently, in our proofs, if an integrable function satisfies the Lipschitz condition of order one, then we interchange the differentiation and integration when we take the derivative of the function. Otherwise, we use the Leibniz’ rule to take the derivative".

Do they mean that if a function is not Lipschitz continuous (of order 1) then Leibniz rule can be used?

Thanks!

 

[jvicens]

Hello,

Thank you for your question. Yes, that is correct. The paper is stating that if a function satisfies the Lipschitz condition of order one, then differentiation and integration can be interchanged, which provides a more compact way to take the derivative. However, if the function does not satisfy this condition, then the Leibniz rule can be used to take the derivative. This is because the Lipschitz condition of order one guarantees that the function is differentiable almost everywhere, which allows for the interchange of differentiation and integration. If the function does not satisfy this condition, then the Leibniz rule, which allows for differentiation under the integral sign, can be used. I hope this clarifies the statement in the paper for you. Let me know if you have any further questions.