# Why taking integral of both sides of matrix inequality is allowed?

How to show if $$\nabla^2 f(x) \succeq \alpha I$$, then the function is $$\alpha$$-strongly convex?

In my optimization notes I have $$\nabla^2 f(x) \succeq \alpha I \rightarrow \alpha\text{-strongly convex} \,\,\,\,\,\,\forall x$$ where $$x \in \mathbb{R}^n$$ and $$A\succeq B$$ means $$A-B$$ is positive semi-definite.

For the proof we use mean value theorem $$\nabla f(y)-\nabla f(y) = \int_0^1 \nabla^2 f(x+t(y-x))(y-x)dt$$ where $$x, y \in \mathbb{R}^n$$ and $$0 \leq t \leq 1$$. $$\langle \nabla f(y)-\nabla f(y) ,y-x \rangle= \langle \int_0^1 \nabla^2 f(x+t(y-x))(y-x)dt,y-x \rangle$$ Since $$\nabla^2 f(x+t(y-x)) \succeq \alpha I \tag{1}$$ and the fact that $$\langle Ad,d \rangle \geq c\|d\|^2 \leftrightarrow A \succeq cI \tag{2}$$

$$\langle \int_0^1 \nabla^2 f(x+t(y-x))(y-x)dt,y-x \rangle \geq \alpha \|y-x\|^2 \tag{3}$$ I do not understand why how we can use $$(1)$$ to get $$(2)$$? Because I do not know why we can use $$(1)$$ and write $$A=\int_0^1 \nabla^2 f(x+t(y-x))dt \succeq \int_0^1 \alpha I dt= \alpha I$$ and then use $$(2)$$.

Is it possible to take integral from any matrix inequality?

I think it would be better to state the integral equality as $$\langle \nabla f(y) - \nabla f(x), y-x \rangle = \int_0^1 \langle \nabla^2 f(x + t(y-x)) (y-x), y-x \rangle \, dt.$$ (This is simply the fundamental theorem of calculus $$g(1) - g(0) = \int_0^1 g'(t) \, dt$$ applied to $$g(t) := \langle \nabla f(x + t(y-x)), y-x \rangle$$.)
From here, you can immediately use $$\nabla^2 f(x + t(y-x)) (y-x), y-x \rangle \ge \alpha \|y-x\|^2$$ from (2), as desired.