# Smoothness and the eigenvalues of the Hessian

A continuously differentiable function $$f$$ is $$\beta$$-smooth if the gradient is $$\beta$$-Lipschitz: $$|| \nabla f(x) - \nabla f(y)|| \leq \beta ||x-y||,$$ where $$\nabla f(\cdot)$$ denotes the gradient vector of $$f$$ at a given point, $$\beta$$ is a scalar, and $$||\cdot||$$ is an $$\ell_2$$-norm.

In the book Convex Optimization: Algorithms and Complexity by Sebastien Bubeck it is said that:

The above Lipschitz condition implies (if the function is twice differentiable) the eigenvalues of the Hessians being smaller than $$\beta$$.

I cannot see how.

Consider $$y=x+v$$ with $$v \in \mathbb{R}^n$$ being any non-zero vector. Therefore, $$\frac{||\nabla f(x+v)-f(x)||_2}{||v||_2} \leq \beta$$ As the derivative is a linear approximation, we have $$\nabla f(x+v)=\nabla f(x)+\nabla^2f(x)v+r(v)$$ Here $$\lim_{v \to 0} \frac{r(v)}{||v||_2}=0$$. Rearranging, taking norms, dividing and the triangle inequality give $$\frac{||\nabla^2f(x)v||_2}{||v||_2} \leq\frac{||\nabla f(x+v)-f(x)||_2}{||v||_2}+ \frac{||r(v)||_2}{||v||^2}\leq \beta+\frac{||r(v)||_2}{||v||^2}$$
Note that the Hessian is symmetric. Assuming there exists a $$v \in \mathbb{R}^n$$ such that $$\nabla^2f(x) v=\lambda v$$ , we can simplify the leftmost term in the above inequality. Then, assuming $$|\lambda| >\beta$$ and plugging in $$tv$$ into $$v$$, for $$t\in \mathbb R \setminus \{ 0\}$$, in the right-hand side of the inequality, we obtain $$|\lambda| \leq \beta+\frac{||r(tv)||_2}{||tv||^2}.$$
Sending $$t \to 0$$ gives a contradiction.