Testwiki:Reference desk/Archives/Mathematics/2021 February 4

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < February 3 ! width="25%" align="center"|<< Jan | February | Mar >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

February 4

Continuously differentiable implies Holder continuous

On Holder condition it is mentioned that continuously differentiable implies Holder continuous. Where can the proof be found? ThanksAbdul Muhsy (talk) 13:46, 4 February 2021 (UTC)

Note that this only holds (in general) on a closed and bounded non-trivial interval of the real line. Since Lipschitz continuous means the same as 1-Hölder continuous, all we need to prove is that continuous differentiability implies Lipschitz continuity. Let $f$ be a function that is continuously differentiable on some closed and bounded interval $I$. Then it has a derivative $f^{\prime }$ that is continuous on that interval, and so is the absolute value of the derivative. By the extreme value theorem, the latter attains some maximum $K$ on the interval, so for all $x\in I$, $|f^{\prime }(x)|\le K.$ Lipschitz continuity now follows (with that constant $K$) from the mean value theorem.  --Lambiam 14:33, 4 February 2021 (UTC)