Im studying an optimization method and I am at a point where I am given these two following assumptions ($x_1$ is the starting point of the method's algorithm and f the objective function we're trying to minimize).
1) The level set $\mathcal{L} = \{ x: f(x) \leq f(x_1) \}$ is bounded.
2) In some neighborhood $\mathcal{N}$ of $\mathcal{L}$, the objective function $f$ is continuously differentiable, and it's gradient is Lipschitz continuous, i.e., there exists a constant $L>0$ such that $$\| \nabla f(x) - \nabla f(\bar x)\| \leq L\| x- \bar x\|$$ for all $x, \bar x \in \mathcal{N}$.
Now, the book proceeds to say that these assumptions imply that there is a constant $\gamma$ such that $$ \| \nabla f(x) \| \leq \gamma, \forall x \in \mathcal{L}$$
I cant understand how this inequality is implied from these two assumptions. Any help would be appreciated, thank you!
EDIT: My tags are probably horrible but I dont know if there are more suitable ones. Sorry in advance.
Hint: By definition $\mathcal{L}$ is closed. Since it is also bounded, it is a compact set. Also by the given assumptions, $\|\nabla f(x)\|$ is continuous on $\mathcal{L}$. By a theorem of Weierstrass, a continuous function defined on compact set achieves maximum and minimum. (See here for details and generalization.) It suffices to take $\gamma$ to be the maximum.
Remark: One can also use the Lipschitz condition together with triangle ineqality to give a different proof: $$\| \nabla f(x)\| -\|\nabla f(x_1)\|\leq \|\nabla f(x)-\nabla f(x_1)\|\leq L\|x-x_1\|$$ $$\Rightarrow \|\nabla f(x)\|\leq \|\nabla f(x_1)\|+L\|x-x_1\|\leq \|\nabla f(x_1)\|+Ld,$$ where $d$ can be taken to be $\max_{x\in\mathcal{L}}\|x-x_1\|$ by Weierstrass's theorem or any upper bound of $\|x-x_1\|$ using the given assumption that $\mathcal{L}$ is bounded.