Let $f$ be a $\alpha$ strongly convex function and $\nabla f$ be $\beta $ Lipschitz then show that : $$ \int_{0}^{1} \langle \nabla f(x+t(y-x)), y-x \rangle \, dt \le \langle \nabla f(x), y-x\rangle + \frac{\beta}{2} \| x-y \|^{2} $$
Thanks and regards.
This is a basic result in many books or slides. And it is true that we do not need $\alpha$-strongly convex here.
First, we have \begin{equation} \begin{aligned} f(y) - f(x) &= \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. \end{aligned} \end{equation} Then, \begin{equation} \begin{aligned} |f(y) - f(x) - \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} |\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)\| \cdot \|y-x\| \, dt \\ & \leq \frac{\beta}{2}{\|y-x\|}^2. \end{aligned} \end{equation} Therefore, \begin{equation} f(y) - f(x) = \int_{0}^{1} \langle \nabla f(x+t(y-x)), y-x \rangle \, dt \leq \langle \nabla f(x), y-x \rangle + \frac{\beta}{2}{\|y-x\|}^2. \end{equation}