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Let $Q \subset \mathbb{E} \times \mathbb{R}$ be a set, where $\mathbb{E}$ represents the Euclidean space. I have the following:

We know that $E_Q(x)=\inf\{r:(x,r) \in Q\}$ by definition. Furthermore, $f$ is convex if for all $x \in \mathbb{E}$, $y \in \mathbb{E}$, and $\lambda \in [0, 1]$, $f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$. Now, we will show that for any $x \in \mathbb{E}$, $y \in \mathbb{E}$, and $\lambda \in [0, 1]$, $E_Q(\lambda x+(1-\lambda)y) \leq \lambda E_Q(x)+(1-\lambda)E_Q(y)$.

I am trying to start this proof, but I don't know how to do this with $\overline{\mathbb{R}}$. I would like to know how to apply the definitions for this proof. I start by letting $x \in \mathbb{E}$, $y \in \mathbb{E}$, and $\lambda \in [0, 1]$ be chosen arbitrarily. I also don't know how to really apply the definitions of convexity and envelope or prove what I have so far. I would appreciate some help.

Sid Meka
  • 65

1 Answers1

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I got it:

For any $\epsilon > 0$, by the definition of infimum, there exist $r_x \in \mathbb{R} \text{ and } r_y \in \mathbb{R}$ such that $$(x, r_x) \in Q\,, \quad r_x < E_Q(x) + \epsilon$$ $$(y, r_y) \in Q\,, \quad r_y < E_Q(y) + \epsilon$$ Since $Q$ is convex, the convex combination $$\left( \lambda x + (1-\lambda)y,\, \lambda r_x + (1-\lambda) r_y \right) \in Q$$ By the definition of $E_Q$, $$E_Q\left(\lambda x + (1-\lambda)y\right) \leq \lambda r_x + (1-\lambda) r_y$$ But $r_x < E_Q(x) + \epsilon$ and $r_y < E_Q(y) + \epsilon$, so $$\lambda r_x + (1-\lambda) r_y < \lambda E_Q(x) + (1-\lambda) E_Q(y) + \epsilon$$ Therefore, $$E_Q\left(\lambda x + (1-\lambda)y\right) < \lambda E_Q(x) + (1-\lambda) E_Q(y) + \epsilon$$ Since $\epsilon > 0$ was arbitrary, it follows that $$E_Q\left(\lambda x + (1-\lambda)y\right) \leq \lambda E_Q(x) + (1-\lambda) E_Q(y)$$ Thus, $E_Q$ is convex.

Let me know if this is correct or not. Also, I forgot to put this earlier, but we should show that if $Q$ is convex, then the envelope $E_Q:\mathbb{E}\rightarrow \overline{\mathbb{R}}$ is a convex function.

Sid Meka
  • 65