Let $u:\Omega \rightarrow \mathbb{R}$ a twice differential function, with $\Omega$ a subset of $\mathbb{R}^n$.
Suppose that we have the following: $$D^2u\geq - \dfrac{(1+K^2)^{1/2}}{\epsilon}I$$
where the inequality is in the sense of quadratic form, so it means $D^2u+\dfrac{(1+K^2)^{1/2}}{\epsilon}I$ is positive semidefinite.
Now if $A$ is a matrix then define the norm of $|A|=\sqrt{\sum a_{ij}^2}$.
I want to show that $$\epsilon|D^2u|\geq 1$$.
My attempt:
First off i think that one can considerate some different definition of "norm" of a matrix, since all norms on the space of matrices are equivalent.
I tried to use the fact that all the eigenvalues of $D^2u$ are in the set $[-\frac{(1+K^2)^{1/2}}{\epsilon},+\infty)$. I tried with fact that sum of element of the Hessian is greather than $n\frac{(1+K^2)^{1/2}}{\epsilon}$. Nothing arise.
The claim is false. Take $u = 0$ (the zero function), then $D^2 u = 0$ with $$D^2 u + \frac{(1+K^2)^{1/2}}{\epsilon} I \succeq 0$$ but $\epsilon\lVert D^2 u \rVert_F = 0 \ngeq 1$.