We are given a function ${f \colon \mathbb{R^n} \rightarrow \mathbb{R}\\}$. $ f $ is a convex function ,also homogeneous of degree $1$ . So for every $t \geq 0 $, $ f(tx)= tf(x)$. Now we want to prove that : $\forall x \in \mathbb{R^n}, f(-x ) \geq -f(x) $
The hint for this exercise is that i should use the property of convex functions that says : $ \forall t \in [0,1]\hspace{0.5cm} \forall (x,y) \in \mathbb{R^n}\times\mathbb{R^n}, \hspace{0.5cm} f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) $
From this point i don't know how to end the demonstration since i can't use the homogeneous of degree 1 property because $-1$ is negative. Please give me a hint on how i can from the previous expression to $f(-x ) \geq -f(x) $ . Thank You.
As mentioned in the comments, we need that $f(0) = 0$, which follows from the homogeneity. Then $$ 0 = f(0) = f\left( \frac{(-x) + x}{2}\right)\le \frac{f(-x) + f(x)}{2} $$ implies $f(-x ) \geq -f(x)$.