Considering a convex function $f:\mathbb{R}^n \rightarrow \mathbb{R}$, if it is a $\beta$-smooth function, namely $$\forall x,y\in \mathbb{R}^n, \quad f(y) \leq f(x) + \nabla f(x)^\mathrm{T}(y-x)+\frac{\beta}{2}\lVert y-x \rVert^2 $$
then we have $$\lVert \nabla f(x)\rVert^2 \leq \beta f(x) $$
This conclusion comes from a lecture note from Dimitris Papailiopoulos's course ECE 901 but without proof.
I just wanna figure out how to prove this conclusion? Thx.
You copied incorrectly the definition of $\beta$-smooth. In the lecture notes you linked to, $\beta$-smooth is defined by $$ \| \nabla f(x) - \nabla f(y) \| \leq \beta \| x - y\|. $$ (What you copied down is almost the definition of $\beta$-strong convex; you have the wrong inequality.)
That said, the claim you are asking about is false (at least without additional assumptions). Any linear function $f(x) = Lx$ is convex and $\beta$-smooth for any $\beta \geq 0$. But $\| \nabla f(0)\|^2 = \|L\|^2 \not\leq \beta f(0) = 0$ whenever $L \neq 0$.