I'd like to check 2 proofs I'm using. Both were made after some discussions I read here, but I have doubts about whether I'm making right interpretations and whether I'm formulating correctly both statements.
Theorem 1-. Let $X \subset \mathbb{R}^{N}$ be a convex, compact set. Let $f \colon X \longrightarrow \mathbb{R}$ be a continuous, differentiable function with continuous gradient. Function $f$ is Lipschitz-continuous then.
Proof:
Let $x, y \in X$, consider $[x, y]$ segment and its parametrization $\gamma(t) = (1 - t)x + ty, \ \forall t \in [0, 1]$ (with $\gamma'(t) = y - x$). Then $$|f(y) - f(x)| = \left| \int_{0}^{1} \frac{d \big(f\circ \gamma \big) (t)}{dt} dt \right| = \left| \int_{0}^{1} \nabla f \big( \gamma(t) \big) \cdot \gamma'(t) \ dt \right| \leq \int_{0}^{1} \big| \nabla f \big( \gamma(t) \big) \cdot (y - x) \big| dt$$ As $\nabla f$ is continuous, $||\nabla f|| \colon X \longrightarrow \mathbb{R}$ reachs an absolute maximum, so $\exists K > 0$ such that $||\nabla f(x)|| \leq K, \forall x \in X$. By Cauchy-Schwartz inequality, can conclude that $$|f(y) - f(x)| \leq \int_{0}^{1} \big| \big\langle \nabla f\big( \gamma (t) \big), \gamma'(t) \big\rangle \big| \leq \int_{0}^{1} ||\nabla f \big( \gamma(t) \big)|| \ ||y - x|| \leq K ||y - x||, \ \forall x, y \in X$$
Theorem 2-. Let $A \subseteq \mathbb{R}^{N}$ be an open set and let $f \colon A \longrightarrow \mathbb{R}$ be a differentiable function. Function $f$ is locally Lipschitz-continuous then.
Proof:
Since $f$ is differentiable over set $A$, given $x_{0} \in A$ it is followed that $$\forall \varepsilon > 0, \ \exists \delta > 0 \ / \ \text{si } ||x - x_{0}|| < \delta \ \Longrightarrow \ \frac{|f(x) - f(x_{0}) - \nabla f(x) \cdot (x - x_{0})|}{||x - x_{0}||} < \varepsilon$$ At the same time, settled $\varepsilon > 0$ it is followed that $$\begin{array}{rcl} |f(x) - f(x_{0})| & \leq & \displaystyle |f(x) - f(x_{0}) - \nabla f(x) \cdot (x - x_{0})| + |\nabla f(x) \cdot (x - x_{0})| < \\ & < & \varepsilon ||x - x_{0}|| + ||\nabla f(x)|| \ ||x - x_{0}|| = \big( \varepsilon + ||\nabla f(x)|| \big) ||x - x_{0}|| \end{array}$$ Just remain taking $k = \varepsilon + ||\nabla f(x)||$ with associated $\delta$ to conclude $f$ is locally Lipschitz-continuous.
My doubt here is, are these really ok? In both cases I'm using norms related to gradient in different ways; as in first one $||\nabla f|| \colon X \longrightarrow \mathbb{R}$ and in second one I think $||\nabla f(x_{0})||$ is intended to be the maximum over unit circunference (at least in its original post for $f\colon \mathbb{R}^{N} \longrightarrow \mathbb{R}^{M}$). I'm getting confused with interpretations (I've been a long while without use them so I can be failing with basics).
Thanks for your help and time.
