Boundedness of a sequence in $L^\infty(I,H^1(M))\cap\mbox{Lip}(I,L^2(M))$ implies that its temporal derivative is bounded as well

Question

I asked my question in mathoverflow, but it seems to be inappropriate there, so I try my luck here.

https://mathoverflow.net/questions/203258/boundedness-of-a-sequence-in-l-inftyi-h1m-cap-lipi-l2m-implies

"Hi I have the next claim which I would like to find a proof of it.

I have a sequence of functions $u_\epsilon(t,x) \in H^1(M)$ where $M$ is a compact manifold, and $u_\epsilon \in L^\infty(I,H^1(M))\cap \mbox{Lip}(I,L^2(M))$ where $I$ is some interval that includes zero in it.

The claim is that $\partial_t u_\epsilon$ is bounded in $L^\infty(I,L^2(M))$, why is that necessarily true?

Thanks in advance." I forgot to mention that $u_\epsilon$ is bounded.

Answer 1

Forget about the unnecessary detail that complicates things. A Lipschitz continuous functions, if it is differentiable$^{[1]}$, has a bounded derivative. The bound on the derivative is the Lipschitz constant. This follows trivially from the definition and from your favorite version of the mean value theorem.

In this case we have a sequence which is uniformly Lipschitz. So its first derivative is uniformly bounded. It remains only to unwrap this idea and apply to the fancy spaces you mentioned, but that's all.

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$^{[1]}$ And this happens almost everywhere, according to Rademacher's theorem.