Equivalent definitions of $\beta$-smoothness Suppose that the gradient of $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is $\beta$-Lipschitz, for some $\beta \geq 0$, i.e.
\begin{equation}
      \|\nabla f(x)-\nabla f(y)\|_2\leq \beta\|x-y\|_2,
\end{equation}
for all $x,y\in\mathbb{R}^n$.
Show that this is equivalent to
\begin{align*}
        f(\theta x+(1-\theta)y)&amp;\geq \theta f(x)+(1-\theta)f(y)-\tfrac{\beta}{2}\theta(1-\theta)\|x-y\|^2,\\
        f(\theta x+(1-\theta)y)&amp;\leq \theta f(x)+(1-\theta)f(y)+\tfrac{\beta}{2}\theta(1-\theta)\|x-y\|^2.
\end{align*}
for all $x,y\in\mathbb{R}^n$ and all $\theta \in [0,1]$,
Moreover, show that it is equivalent to
\begin{align*}
        f(y)&amp;\geq f(x)+ \nabla f(x)^{T}(y-x) - \tfrac{\beta}{2}\|x-y\|^2,\\
        f(y)&amp;\leq f(x) + \nabla f(x)^{T}(y-x) + \tfrac{\beta}{2}\|x-y\|^2,
\end{align*}
for all $x,y\in\mathbb{R}^n$.
Since the gradient of $f$ is $\beta$-smooth over $\mathbb{R}^n$, we have that by the Descent Lemma $$f(y)\leq f(x)+\nabla f(x)^T(y-x)+\frac{L}{2}\|x-y\|^2$$ for all $x,y\in\mathbb{R}^n$. Then can further write
$$f(x)\leq f(\theta x+(1-\theta) y)+\nabla f(\theta x+(1-\theta) y)^T(x-(\theta x+(1-\theta) y))+\frac{L}{2}\|x-(\theta x+(1-\theta) y)\|^2 \\
f(y)\leq f(\theta x+(1-\theta) y)+\nabla f(\theta x+(1-\theta) y)^T(y-(\theta x+(1-\theta) y))+\frac{L}{2}\|y-(\theta x+(1-\theta) y)\|^2$$
which is the same as
$$f(x)\leq f(\theta x+(1-\theta) y)+(1-\theta)\nabla f(\theta x+(1-\theta) y))^T(x-y))+\frac{L(1-\theta)^2}{2}\|x-y\|^2 \\
f(y)\leq f(\theta x+(1-\theta) y)+\theta\nabla f(\theta x+(1-\theta) y))^T(y-x)+\frac{L\theta^2}{2}\|x-y\|^2$$
Now multiply the first inequality by $\theta$ and the second by $(1-\theta)$ and add them together to get the desired result for the first inequality you're trying to prove. The second inequality proof is similar.
Since the gradient of $f$ is $\beta$-smooth over $\mathbb{R}^n$, we have that by the Descent Lemma $$f(y)\leq f(x)+\nabla f(x)^T(y-x)+\frac{L}{2}\|x-y\|^2$$ for all $x,y\in\mathbb{R}^n$. Then can further write
$$f(x)\leq f(\theta x+(1-\theta) y)+\nabla f(\theta x+(1-\theta) y)^T(x-(\theta x+(1-\theta) y))+\frac{L}{2}\|x-(\theta x+(1-\theta) y)\|^2 \\
f(y)\leq f(\theta x+(1-\theta) y)+\nabla f(\theta x+(1-\theta) y)^T(y-(\theta x+(1-\theta) y))+\frac{L}{2}\|y-(\theta x+(1-\theta) y)\|^2$$
which is the same as
$$f(x)\leq f(\theta x+(1-\theta) y)+(1-\theta)\nabla f(\theta x+(1-\theta) y))^T(x-y))+\frac{L(1-\theta)^2}{2}\|x-y\|^2 \\
f(y)\leq f(\theta x+(1-\theta) y)+\theta\nabla f(\theta x+(1-\theta) y))^T(y-x)+\frac{L\theta^2}{2}\|x-y\|^2$$
Now multiply the first inequality by $\theta$ and the second by $(1-\theta)$ and add them together to get the desired result for the first inequality you're trying to prove. The second inequality proof is similar.