I am stuck with this problem from my textbook and I cannot see the solution. I'm certain the solution is fairly simple and I am just missing the mark somehow. Any help would be appreciated.
Suppose the function $f: B_1(0) \rightarrow \mathbb{R}$ has continuous bounded partial derivatives on the unit Euclidean ball $B_1(0)$ in $\mathbb{R}^n$. Prove that $f$ is uniformly continuous on $B_1(0)$
Edit: Here is my take using Lipschitz and multivariable MVT, can someone please let me know if I am on the right track? Is it as simple as this?
let $x,y \in B_1(0)$ then by MVT we have $|f(x) – f(y)| = \nabla f(c)|x-y| \leq M|x-y|$ for some c on the line segment connecting x and y, thus Lipschitz thus uniformly cts?
Best Answer
Let us prove that $f$ is Lipschitz continuous, with Lipschitz constant $M := \sup_{x\in B_1} |\nabla f(x)|$.
Given two points $x,y\in B_1$, let $$ g(t) := f((1-t)x + ty), \qquad t\in [0,1]. $$ Since $f$ is of class $C^1$, then also $g$ is, and by the chain rule formula $$ g'(t) = \nabla f((1-t)x + t y) \cdot (y-x). $$ By the mean value theorem there exists $\tau \in (0,1)$ such that $g(1) - g(0) = g'(\tau)$, hence $$ |f(y) - f(x)| = |g(1) - g(0)| = |g'(\tau)| \leq |\nabla f((1-t)x + ty)| \, |x-y| \leq M |x-y|. $$ (The first inequality follows from the Cauchy-Schwarz inequality.)