Let $x \mapsto f(x) \in \mathcal{C}^2$ be convex, i.e. $\forall x \in \mathbb{R}^n$, $\nabla^2f(x) \succeq 0$. Let $A \in \mathbb{R}^{m \times n}$ and suppose we have $M I_n \succeq \nabla^2f(x) + A^\top A \succeq m I_n$, where $M \geq m > 0$. Is it then true that \begin{equation} M I_n \succeq \int_0^1\nabla^2f(x^\ast + \tau(x-x^\ast))d\tau + A^\top A \succeq m I_n, \end{equation} for all $x,x^\ast \in \mathbb{R}^n$ and $\tau \in [0,1]$ (element-wise integration)?
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Yes, it is true.
Since $M I_n \succeq \nabla^2f(x) + A^\top A \succeq m I_n$ is true for all $x\in \mathbb{R}^n$ and $x^*+\tau(x-x^*)$ is such a point, you can say that:
$$ M I_n \succeq \nabla^2f(x^*+\tau(x-x^*)) + A^\top A \succeq m I_n $$ Now, multiply each element by $d\tau$ and integrate to get the result.
Mehdi Jafarnia Jahromi
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