Let $f \colon O \to \mathbb R$ be continuously differentiable, where $O \subset \mathbb R^n$ is open and convex. Then, the following two assertions are equivalent:
- $\nabla f$ is Lipschitz-continuous with constant $1$, i.e., $$ \| \nabla f(y) - \nabla f(x) \| \le \| y - x \| \qquad \forall x,y \in O.$$
- we have $$ \bigl| (\nabla f(y) - \nabla f(x) )^\top (y - x) \bigr| \le \|y - x\|^2 \qquad\forall x,y \in O.$$
I am looking for a concise proof or a reference for this equivalence.
The direction $1 \Rightarrow 2$ is just an application of the Cauchy-Schwarz inequality.
The other direction is slightly harder. Let me outline a proof. If we assume additionally $f \in C^2$, then by using $y = x + t h$, with fixed $h \in \mathbb R^n$ in 2, we obtain $$ |h^\top \nabla^2f(x) h| \le \| h \|^2 \qquad\forall x \in O, h \in \mathbb R^n$$ by passing to the limit $t \to 0$. Since $\nabla^2 f(x)$ is symmetric, this yields $\| \nabla^2 f(x) \| \le 1$ and the Lipschitz continuity of $\nabla f$ follows. In the general case, one can apply this argument to the mollification $f_\epsilon = f \mathbin\star \psi_\epsilon$, where $\psi_\epsilon$ is a standard mollifier.
Is it possible give a direct proof of $2 \Rightarrow 1$ without the 'detour' to $C^2$-functions?
By the way: Although the statements 1 and 2 only depend on $\nabla f$, the equivalence no longer holds if $\nabla f$ is replaced by an arbitrary, continuous function $g \colon O \to \mathbb R^n$: Use, e.g., $g(x) = A x$, where $A \in \mathbb R^{n \times n}$ is skew symmetric.