Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f$ is a convex, differentiable and $\|\nabla f(x)-\nabla f(y)\|\leq L\|x-y\|$. The minimum of $f$ is $0$. ($f$ may not be twice differentiable.)

How to show $f(x)\geq\frac{1}{2L}\|\nabla f(x)\|^2$, $\forall x$?

share|cite|improve this question
up vote 2 down vote accepted

I'll assume $f$ is defined on all of $\mathbb{R}^n$. Let $f^* = \inf_{y \in \mathbb{R}^n} f(y)$. (You have told us that $f^* = 0$, but I prefer just to call it $f^*$.)

Because $\nabla f(x)$ is Lipschitz continuous with parameter $L$, we have the following quadratic upper bound on $f$:

\begin{equation} f(y) \leq f(x) + \langle \nabla f(x),y - x \rangle + \frac{L}{2} \| y - x \|^2 \end{equation} for all $x,y \in \mathbb{R}^n$.

It follows that \begin{equation} \inf_y \, f(y) \leq \inf_y \, f(x) + \langle \nabla f(x),y - x \rangle + \frac{L}{2} \| y - x \|^2 \end{equation} for all $x \in \mathbb{R}^n$.

The infimum on the left hand side is just $f^*$. The infimum on the right hand side is $f(x) -\frac{1}{2L} \| \nabla f(x) \|^2$. Thus we find that \begin{align*} & f^* \leq f(x) - \frac{1}{2L} \| \nabla f(x) \|^2 \\ \implies& f(x) - f^* \geq \frac{1}{2L} \| \nabla f(x) \|^2. \end{align*}

This result appears in chapter 1 ("Gradient method") of Vandenberghe's 236c lecture notes (see slide 1-14 entitled "Quadratic upper bound").

share|cite|improve this answer
Thank you, littleO, especially for pointing out the reference. – mining Nov 10 '12 at 19:32

I started this last night, but got stuck with the transition from the one dimensional example back to $f$. (A problem related to the Pinot Grigio technique I was using, I think.)

Let $\phi:\mathbb{R}\to [0,\infty)$ be convex, differentiable, $\phi'$ Lipschitz continuous with rank $L$. Then Taylor gives: $\phi(y) = \phi(x)+\phi'(x)(y-x) + \int_x^y (\phi'(t)-\phi'(x) )dt \leq \phi(x)+\phi'(x)(y-x) + L\int_x^y |t-x|dt$. Integrating gives the bound $\phi(y) \leq \phi(x)+\phi'(x)(y-x) + \frac{L}{2} (y-x)^2$. By assumption $\phi(t) \geq 0$, so we have $\phi(x)+\phi'(x)(y-x) + \frac{L}{2} (y-x)^2 \geq 0$ for all $x,y$. Minimizing over $y$ gives $\phi(x) \geq \frac{1}{2L} (\phi'(x))^2$, which is basically the desired result.

To connect with $f$ above, choose $x\neq y$, and let $\phi(t) = \frac{f(y+t(x-y))}{\|x-y\|^2} $, then $\phi'(t) = \frac{\partial f(y+t(x-y))}{\partial x}\frac{(x-y)}{\|x-y\|^2}$, and a quick computation shows $|\phi'(t)-\phi'(s)|\leq L|t-s|$. This gives $\phi(1) \geq \frac{1}{2L} (\phi'(1))^2$, or equivalently, $f(x) \geq \frac{1}{2L} |\frac{\partial f(x)}{\partial x}\frac{(x-y)}{\|x-y\|}|^2$ (for all $x\neq y$). Since $\|v\| = \max_{\|u\|=1} |v^T u|$, this gives the desired result $f(x) \geq \frac{1}{2L} \|\frac{\partial f(x)}{\partial x}\|^2$.

share|cite|improve this answer
Thank you, copper.hat. – mining Nov 10 '12 at 19:32
You are welcome, sorry I left you hanging last night. Late night brain fuzz... – copper.hat Nov 10 '12 at 19:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.