# Subgradient of a strictly convex function.

I am beginning to study convex optimization and I found the following problem. Let $$E\subset \mathbb{R}^n$$ be a convex set and $$f:E\rightarrow \mathbb{R}$$ a convex, non-necessarily differentiable function. We defined the subgradient of $$f$$ at $$x_0$$ to be a vector $$v\in \mathbb{R}^n$$ such that $$f(x)\geq f(x_0)+v^t(x-x_0)$$, $$\forall x\in E$$. Later in a proof the professor uses that if $$f$$ is strictly convex, then the previous inequality is strict, which I am failing to see. Any ideas on how to prove this?

By Definition of the subgradient in the point $$x_0\in \mathbb{R}^n$$ it holds for all $$x\neq x_0$$ that $$f(x)\geq f(x_0)+v^t(x-x_0)$$. We assume that there is a $$x\neq x_0$$ with $$f(x)=f(x_0)+v^t(x-x_0)$$ and show a contradiction. Thus $$f(x)>f(x_0)+v^t(x-x_0)$$ has to hold.
By strict convex we get for all $$t\in]0,1[$$ $$f(tx_0+(1-t)x) For the point $$tx_0+(1-t)x\in E$$ we get by the subgradient condition in the point $$x_0$$ $$f(tx_0+(1-t)x)\geq f(x_0)+v^t(tx_0+(1-t)x-x_0)=f(x_0)+(1-t)v^t(x-x_0) .$$ This is a contradiction to the equation before.