Prove $\nabla^2 f(x) \preceq L I$ for convex $f$ with Lipschitz gradient I am given a convex and twice differentiable function $f$ who gradient is Lipschitz with constant "$L$". I am trying to prove
$$
\nabla^2 f(x) \preceq L\,I.
$$
I recognize that this is pretty easy, and suspect the proof will invoke the definition of the Hessian.
From $f$ having Lipschitz gradient, we know
$\|\nabla f(x) - \nabla f(y)\| \lt L\,\|x - y\|$.
And therefore
$$
\lim_{x \to y} \frac{\|\nabla f(x) - \nabla f(y)\|}{\|x - y\|} \lt 
\frac{L\,\|x - y\|}{\|x - y\|}
~~\Longrightarrow ~~\nabla^2 f(x) \preceq LI.
$$
But this isn't quite right since the Hessian is a matrix and the quantities inside the limit to the left of the implication arrow are scalars. What is the proper way to prove this?
Any help here would be helpful :-)
 A: @Brian-Borchers
Okay, here's another stab. Assume there exists $x_0$ such that $\nabla^2 f(x_0) \succ L\,I$. Then there exists a vector $v$ such that
$$
\frac{\|\,(\nabla^2 f(x_0))\,v\,\|}{\|v\|} > L.
$$
Let $x = x_0 + v$ and we assume $\|v\|$ is small enough such that $x \in D(x_0,\epsilon) \subseteq \text{Domain}(f)$.
By Taylor's theorem, we know$^*$ that
$$
\nabla f(x) = \nabla f(x_0) + (\nabla^2 f(x_0))(x - x_0) + o(\|x - x_0\|)(x-x_0). ~~~~~~~(1)
$$
Now, taking norms and taking the limit as $x \to x_0$ we have for some $x$ sufficiently close to $x_0$
$$
\lim_{x\to x_0} \Biggl\{\|\nabla f(x) - \nabla f(x_0) \| = \Bigl\| (\nabla^2 f(x_0))(x - x_0) + o(\|x - x_0\|)(x-x_0)
\Bigr\| > L \, \|x - x_0\|
\Biggr\}
$$
This contradicts our assumption that $\|\nabla f(x) - \nabla f(x_0) \| \le L \, \|x-x_0\|$  $~~~~\square$
$^*$However, I am not sure about my claim that equation (1) is true. I am not familiar with how to handle the remainder term in a multivariate Taylor expansion.
A: @Brian-Borchers
Okay, it looks like using a first order Taylor expansion of the gradient could work. Please let me know if this is right.
Using the first order Taylor expansion around a point $x_0$ for the gradient gives us
$$
\nabla f(x) = \nabla f(x_0) + (\nabla^2 f(\xi)) (x - x_0)
$$
where $\xi$ is some point between $x$ and $x_0$. This implies
$$
\nabla f(x) - \nabla f(x_0) = (\nabla^2 f(\xi)) (x - x_0).
$$
Taking norms we have
$$
\| \nabla f(x) - \nabla f(x_0) \| = \| (\nabla^2 f(\xi)) (x - x_0) \|.
$$
Since $\| \nabla f(x) - \nabla f(x_0) \| \le L\,\| x - x_0 \|$ by assumption, this implies
$$
\| (\nabla^2 f(\xi)) (x - x_0) \| \le L \, \|x - x_0\|.
$$
Therefore
$$
\frac{\| (\nabla^2 f(\xi)) (x - x_0) \|}{\|x - x_0\|} \le \frac{L \, \|x - x_0\|}{\|x - x_0\|}
$$
where the vector $x - x_0$ is arbitrary. Therefore
$$
\nabla^2 f(\xi) \preceq L \, I.
$$
Does that do it? Thanks again.
