How to show two different definitions of $\alpha$-strongly convex are equivalent?

In my optimization lecture notes I have the following definition:

Definition: Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ be differentiable. The $$f$$ is strongly convex it $$\exists$$ a positive constant $$\alpha > 0$$ such that $$\langle \nabla f(y) - \nabla f(x),y-x \rangle \geq \alpha||y-x||^2 \,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n \tag{1}$$ However, in the Online Linear Optimization via Smoothing on page 2, it says the function is $$\beta$$-strongly convex if

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

How can we show that (1) is equivalent to (2) with appropriate choice of $$\beta$$?

Let’s take $$(1)$$ and write $$g_1=f-\frac{\alpha}{2}\|\cdot\|_2^2$$. Then $$(1)$$ is equivalent to $$\langle \nabla g_1(y)-\nabla g_1(x),\,y-x \rangle \geq 0$$, ie $$g_1$$ convex.

Do the same for $$(2)$$ with $$g_2= f-\frac{\beta}{2}\|\cdot\|_2^2$$.

• In (1) there is no $f$, we have $\nabla f$, could you write what you mean. I do not think it is as easy as that your saying. – Saeed Jan 11 at 13:42

I will post the following derivation which is expressed in Mindlak's answer just to have better understanding:

From (1) we have:

$$\langle \nabla f(y) -\nabla f(x),y-x \rangle \geq \alpha||y-x||^2= \alpha \langle y - x,y-x \rangle\,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n$$ So, $$\langle \nabla f(y) - \alpha y -(\nabla f(x)-\alpha x),y-x \rangle \geq 0\,\,\,\,\,,\,\,\,\forall x,y \in \mathbb{R}^n \tag{3}$$ (3) means $$g_1(x)=f(x)-\frac{\alpha}{2} \|x\|_2^2$$ is convex, because it is the monotonicity property of convex function. From $$g_1(s) \geq g_1(t) + \langle \nabla g_1(t),s-t \rangle \,\,\,\,\,,\,\,\,\forall s,t \in \mathbb{R}^n$$ we have

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

$$f(y)-f(x)+ \langle \nabla f(x),y-x \rangle \geq -\frac{\alpha}{2} \|x\|_2^2 +\frac{\alpha}{2} \|y\|_2^2 +\langle -\alpha x,y-x \rangle =\frac{\alpha}{2} \|y-x\|_2^2 \\ \forall \,\,\,\,\,y,x \in \mathbb{R}^n$$ and $$\alpha = \beta$$