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Let $X \subset \mathbb{R}^{n}$ be compact and convex. Moreover, let $f:X \rightarrow \mathbb{R}$ be a differentiable map with $\sup_{x \in X} \|\nabla f(x)\| = K < \infty$, where $\|\cdot\|$ denotes the Euclidean norm and $\nabla f$ the gradient.

Both in this question and in this webpage it is said that the best (minimal) Lipschitz constant of $f$ is $K$. If this is a basic result in real analysis, could someone please point me to a reference?

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  • $\begingroup$ Undoubtedly this is in Geometric Measure Theory by Federer. But I think this question belongs on math.stackexchange. $\endgroup$
    – Nik Weaver
    Commented Jan 11, 2019 at 14:49
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    $\begingroup$ It may be hard to find the proof in a textbook, but you can prove it yourself by taking a point $x$ with $\|\nabla f(x) \|\geq K-\varepsilon$ and finding a $y$ with $|f(y)-f(x)| \geq (K-2\varepsilon)|y-x|$ using the definition of differentiability at $x$. $\endgroup$
    – user126920
    Commented Jan 11, 2019 at 15:00
  • $\begingroup$ @StanleySnelson: you have to check the reverse inequality too, which is where convexity comes in: any two points are joined by a line segment and you can apply the one-dimensional mean value theorem. $\endgroup$
    – Nik Weaver
    Commented Jan 11, 2019 at 15:54
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    $\begingroup$ You know what, come to think of it, there's a proof of this in a book I just wrote ... $\endgroup$
    – Nik Weaver
    Commented Jan 11, 2019 at 15:59

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This is essentially Corollary 1.42 in my book Lipschitz Algebras (second edition).

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