Is it true that if $\varphi$ is a $C^2$ functional on a Hilbert space $X$ and $||\nabla \varphi(u)||, ||\nabla \varphi(v)|| \geq 2\varepsilon$ then $$ \left| \frac{||\nabla \varphi(v)||}{||\nabla \varphi(u)||} - \frac{||\nabla \varphi(u)||}{||\nabla \varphi(v)||}\right| \leq C ||\nabla \varphi(u) - \nabla \varphi(v)||? $$
How to prove this?
Context: We are trying to show that the field $f$ (the right-hand side of the Cauchy problem) in the proof of Willem's Quantitative Deformation Lemma is locally Lipschitz.
Any hints will be the most appreciated.
Thanks in advance and kind regards
The left hand side of what you have is $$ \frac{|\nabla \phi(v)|^2 - |\nabla \phi(u)|^2}{|\nabla\phi(u)||\nabla \phi(v)|} = \frac{|\nabla \phi(u)| + |\nabla \phi(v)|}{|\nabla \phi(u)||\nabla \phi(v)|} \cdot (|\nabla \phi(u)| - |\nabla \phi(v)|). $$ Use the (reverse) triangle inequality for the parenthetical term. To upper bound the denominator, use your assumed lower bound. To get a constant $C$ you'll need to assume that $\nabla \phi$ is bounded, or restrict to a compact domain.