We are given a function $f : \mathbb{R}^n \to \mathbb{R}$. We know that $f$ is convex and continuously differentiable on $\mathbb R^n\setminus \{0\}$ . Also $f$ is homogeneous of degree $1$, which means that $ \forall t \geq 0: f(tx)=tf(x). $
By using the properties of convex functions, prove that $$\forall x \in \mathbb{R}^n, \forall x' \in B(x,r) \subset \mathbb{R}^n \setminus \{0\}: f(x') \geq \nabla f(x) \cdot x' \, .$$
I tried to solve the problem by using the following property of convex functions: $f$ is a convex function $\iff \forall (x,y) \in \mathbb{R}^n \times \mathbb{R}^n$, $ f(x)-f(y) \geq \nabla f(x) \cdot (x-y) $. Since $f$ is linear I replace $y$ with $0$ and end up with $f(x) \geq \nabla f(x) \cdot x $.
Please help me find a way to go from this expression $f(x) \geq \nabla f(x) \cdot x $ to this one $f(x') \geq \nabla f(x) \cdot x'$.
There is a sign error in your characterization of convex functions via the gradient, it should be $$ \tag{*} f(y)-f(x) \ge \nabla f(x) \cdot (y-x) $$ for $x, y \in \Bbb R^n$. Setting $y=0$ then gives $f(x) \le \nabla f(x) \cdot x$, which is the opposite of what we need. But actually equality holds, and that can be seen as follows:
If $f$ homogeneous and $x \ne 0$ then $$ \nabla f(x) \cdot x = \lim_{h \to 0+} \frac{f(x+hx)-f(x)}{h} = \lim_{h \to 0+} \frac{(1+h)f(x)-f(x)}{h} = f(x) \, . $$
Then it follows from $(*)$ that $$ f(y) \ge f(x) + \nabla f(x) \cdot (y-x) = \nabla f(x) \cdot y + \underbrace{\bigl( f(x) - \nabla f(x) \cdot x \bigr)}_{=0} =\nabla f(x) \cdot y \, . $$
Note that his holds for all $x \in \Bbb R^n \setminus \{ 0 \}$ (where $f$ is differentiable), and all $y \in \Bbb R^n$.