Extension of mean value theorem for convex functions

Question

I'm trying to solve this exercise that claims to prove an extension of mean value theorem for convex functions. The exercise is as follows:

Let $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ be a convex function, $x, y \in \mathbb{R}^n$ two distinct points and $\phi:[0,1] \longrightarrow \mathbb{R}$ defined by $$\phi(t)=f(x+t(y-x))-f(x)+t(f(x)-f(y))$$

1. Prove that $\phi$ is a convex function and it get its minimum in a point $t_0 \in ]0,1[$
2. Obtain the value of $\phi'(t_0;d)$ for any $d \in \mathbb{R}$
3. Prove that if $z_0=x+t_0(y-x)$ then $f'(z_0; y-x)=f(y)-f(x)$
4. Prove that there exists a point $u_0 \in \partial f(z_0)$ such that $$f(y)-f(x)=\langle u_0,y-x \rangle$$

*In the book, the directional derivative for a function $g$ is defined by $$g'(x;d)=\lim_{\lambda \rightarrow 0^+} \dfrac{g(x+\lambda d)-g(x)}{\lambda}$$

I know how to solve 1, it follows from the definition of convexity and, for the minimum, by showing that $\phi(0)=\phi(1)=0$, $\phi(t) \leq 0, \forall t \in [0,1]$ and noticing that $\phi$ is continuous (because $f$ is continuous for being convex in $\mathbb{R}^n$).

For 2. I've proven that $$\phi'(t_0;d)=d(f(x)-f(y))+f'(x+t_0(y-x);d(y-x))$$ and, I suspect that the value must be zero buy I don't know how to prove it.

For 3, if in fact $\phi'(t_0;d)=0$, from the expression I give above the required formula follows by taking $d=1$.

Lastly, in 4, knowing that $$f'(z_0;y-x)=\max_{u \in \partial f(z_0)} \langle u, y-x \rangle$$ and that $\partial f(z_0)$ is a compact set, it also follows the extension formula of the mean value theorem we want to prove.

So, can someone give me any ideas of how to solve 2, and if the rest of my reasoning is correct?

Thank you.

Answer 1

I've find that in fact 3. is not true as it can be shown taking $f(u,v)=|u+v|$ and the points $x=(0,-1)$ and $y=(2,-1)$.