# How to show $\frac{1}{L}\|f(y)-f(x)\|_2^2 \leq \langle \nabla f(y) - \nabla f(x) ,y-x \rangle$ for strongly convex function?

Suppose $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ is a strongly convex function, i.e.,

$$\langle \nabla f(y) - \nabla f(x) ,y-x \rangle \geq \theta \|y-x\|_2^2 \,\,\,\,\, \forall x,y \in \mathbb{R}^n$$

or

$$z^T(\nabla^2 f(x) -\theta I)z\geq 0 \,\,\,\,\forall z \neq 0$$

Also, suppose $$\nabla f(x)$$ is a Lipschitz with Lipschitz constant $$L>0$$, i.e., $$\|\nabla f(y)- \nabla f(x)\|_2 \leq L \|y-x\|_2$$.

Show $$\frac{1}{L}\|\nabla f(y)-\nabla f(x)\|_2^2 \leq \langle \nabla f(y) - \nabla f(x) ,y-x \rangle$$

• Your statement is definitely wrong. Just suppose $f(x)=\frac{1}{2}x^2$, then $L=1$ and the right side should be $(x-y)^2$. However, the left side is $\frac{1}{4}|y^4-x^4|$. It is clear that this inequality can't hold. – foxell May 6 at 6:37
• To know about the proof of some properties of strongly convex function, I recommend the book "Lecture on Convex Optimization" of Nesterov. – foxell May 6 at 6:38
• @foxell: Thank you for your careful reading. I had dropped $\nabla$ in the left hand side. By the way, your left hand side becomes $(y^2-x^2)$ . – Saeed May 6 at 17:59

In order to get the conclusion we only need the convexity of $$f(x)$$ rather than the strong convexity. It is easy to show that if $$\nabla f(x)$$ is Lipschitz continuous with constant $$L$$ then we have:$$f(y)-f(x)-\langle\nabla f(x),y-x\rangle\leq\frac{L}{2}\| x-y\|^2$$ To show this inequality, we have:\begin{align} f(y)-f(x)-\langle\nabla f(x),y-x\rangle&=\int_{0}^{1}\langle \nabla f(x+t(y-x)),y-x \rangle dt-\langle\nabla f(x),y-x\rangle \\ &=\int_{0}^{1}\langle \nabla f(x+t(y-x))-\nabla f(x),y-x \rangle dt \\ &\leq\int_0^1\|\nabla f(x+t(y-x))-\nabla f(x)\|\|y-x\| dt\\ &\leq\int_0^1Lt\|y-x \|^2dt=\frac{L}{2}\|y-x\|^2 \end{align}

In order to get the result we need to consider the next function $$h(x)=f(x)-\langle \nabla f(y),x\rangle$$ It is clear that $$h(x)$$ is convex and also $$\nabla h(x)$$ is Lipschitz continuous with constant $$L$$, and hence $$y\in\arg\min_{x}h(x)$$ because $$\nabla h(y)=0$$. We have:$$h(x-\frac{1}{L}\nabla h(x))\geq h(y)$$ Use the first inequality: $$h(x-\frac{1}{L}\nabla h(x))\leq h(x)-\frac{1}{2L}\|\nabla h(x)\|^2$$ We have: $$h(y)\leq h(x)-\frac{1}{2L}\|\nabla h(x) \|^2$$ $$f(y)+\langle\nabla f(y),x-y\rangle\leq f(x)-\frac{1}{2L}\| \nabla f(x)-\nabla f(y)\|^2$$ $$f(y)-f(x)+\langle\nabla f(y),x-y\rangle\leq -\frac{1}{2L}\| \nabla f(x)-\nabla f(y)\|^2$$ Exchange $$x,y$$: $$f(x)-f(y)+\langle\nabla f(x),y-x\rangle\leq -\frac{1}{2L}\| \nabla f(x)-\nabla f(y)\|^2$$ Add them together then you can get the conclusion.

• Could you show me how you get the first line? Also, why $h(x-\frac{1}{L}\nabla h(x)) \geq h(y)$? – Saeed May 7 at 18:16
• Ok, I have given the proof of the first line now. And the answer to your second question is that as I said $y$ is the global minimum of $h(x)$ so $h(y)\leq h(z),\forall z\in R^n$ – foxell May 8 at 1:37
• It was beautiful! – Saeed May 9 at 0:53

By definition of strong convexity we have the following:

$$\begin{equation} f(x) \geq f(y) + <\nabla f(y), x-y> + \frac{\sigma}{2} \left\lVert x-y\right\rVert^2 \end{equation}$$

Similarly we have:

$$\begin{equation} f(y) \geq f(x) + <\nabla f(x), y-x> + \frac{\sigma}{2} \left\lVert x-y\right\rVert^2 \end{equation}$$

If you sum up these two inequalities and organize them you will end up with the following:

$$\begin{equation} <\nabla f(y) - \nabla f(x), y-x> \geq \sigma.\left\lVert x-y\right\rVert^2 \end{equation}$$

Which is what you have. We call this function $$\sigma$$-strongly convex. To give an example, "negative entropy" is a 1-strongly convex (also known as 1-SC) function wrt L1-norm which is used as regularization term for making a non-smooth function smooth in L1-setting (in learning with experts setup)--Please note that in L1-setting everything is a simplex.

Additionally, you can use "Hölder's inequality" to prove the general case shown below (not just norm 2):

$$\begin{equation} \left\lVert \nabla f(y) - \nabla f(x)\right\rVert_{\infty} \left\lVert x-y \right\rVert_{1} \geq <\nabla f(x) - \nabla f(y), x-y> \geq \sigma.\left\lVert x-y\right\rVert_{1}^2 \end{equation}$$ • What I have is not what you have. My lower bound is in terms of $\nabla f(x)$ and $\nabla f(y)$ while yours is in terms of $\|x-y\|^2$. Please correct your derivation. – Saeed May 7 at 18:13