How do I prove monotonicity of risk measures?

Question

Assume I have a risk measure, that is a functional $R:L^\infty(\Omega, F,P)\to \mathbb{R}$, that applies to random variables $X:\Omega\to\mathbb{R}$. Assume the functional satisfies the following properties

1. It is monotonic: $X\geq 0 \implies R(X)\leq 0$
2. It is convex: $R(\lambda X+(1-\lambda)Y)\leq \lambda R(X)+(1-\lambda)R(Y)$ for $\lambda\in[0,1]$.
3. It is positively homogeneous: $R(\lambda X)=\lambda R(X)$ for $\lambda\geq 0$.

Given these three properties, I should be able to derive the following. For any two random variables $X$ and $Y$, with $X\geq Y$, then the risk measure satisfied $R(X)\leq R(Y)$. That is, greater values imply smaller risk.

This fact should be a mere consequence of the three properties above. However, I am not able to complete the proof.

My attempt

This is how I am starting: \begin{align*} R(\lambda X + (1-\lambda)Y)&=R(\lambda X + Y -\lambda Y)\\ &\leq \lambda R(X)+(1-\lambda)R(Y)\\ &=\lambda R(X)+ R(Y)-\lambda R(Y) \end{align*} I was hoping the fact that $X-Y\geq 0$ and $R(X-Y)\leq 0$ could turn out useful, but I cannot see where to use it.

My question

Any hint as to how to continue?

Answer 1

By convexity and positive homogeneity, $$\begin{align}R(X)&=R\left(\frac{2Y+2(X-Y)}2\right)\\ &\le\frac{R(2Y)+R(2(X-Y))}2\\ &=R(Y)+R(X-Y), \end{align}$$ hence by monotonicity, if $X-Y\ge0$ then $R(X)\le R(Y).$