Given a $C^2$ function $f$ which has a Lipschitz gradient, that is,
$ || \nabla f(x) - \nabla f(y) ||_2 \leq L || x - y ||_2$
Can we prove a bound on the largest eigenvalue of $\nabla^2 f$ ?
What does it mean to say -
$ \lim_{x \rightarrow y} \frac{|| \nabla f(x) - \nabla f(y) ||_2}{|| x - y ||_2} \leq L $
Fix vectors $x$ and $y$, and a constant $c>0$. Let $g(t) = \nabla f(x+tcy)$ and note $g'(t) = [\nabla^2 f(x+tcy)] (cy)$. By the mean value theorem, there exists some $t_c \in [0,1]$ such that $$\nabla f(x+cy) - \nabla f(x) = g(1)-g(0) = g'(t_c) = [\nabla^2 f(x+t_ccy)] (cy).$$ Taking the norm of both sides and applying the Lipschitz condition, we have $$\|[\nabla^2 f(x+t_ccy)]y\| \le L\|y\|.$$
By taking $c \to 0$ and using the fact that $f \in C^2$ and $t_c \in [0,1]$, we have $$\|[\nabla^2 f(x)]y\| \le L \|y\|.$$ Thus the absolute value of any eigenvalue of $\nabla^2 f(x)$ must be $\le L$.
Hint: you can use that fact that, if $A$ is a symmetric real matrix, then its maximal eigenvalue $\lambda_{max}(A)$ is characterized by $$ \lambda_{max}(A) = \max_{\|x\| = 1} \langle Ax, x \rangle. $$
