How can this equality of sup and maximum be proven? Let me first introduce some notation. $\Delta_n$ denotes the probability simplex, i.e. $\Delta_n :=\{(p_0,\dots,p_n):\sum_{i=0}^{n-1} p_i= 1, p_i\ge 0, i=0,\dots,n-1\}$. The confidence region for a Pearson's $\chi^2$ test is defined as 
$$G := \{p\in \Delta_n: \sum_{i=0}^{n-1}\frac{(p_i-\hat{p}_i)^2}{2 p_i}\le \frac{1}{2N}\chi^2_{n-1,1-\alpha}\}$$
where $\chi^2_{n-1,1-\alpha}$ is the $1-\alpha$ quantile of a $\chi^2$ distribution with $n-1$ degrees of freedom. 
For a given vectors $ a_0,\dots,a_{n-1}$ with $a_i\in \mathbb{R}^n$ and $\epsilon > 0$ we define 
$$H := \{x\in \mathbb{R}^n: x = \sum_{i=0}^{n-1}q_i a_i,q\in\Delta_n,q\le\frac{1}{\epsilon}p,p\in G\}$$
and one more set is needed. For a fixed $p\in\Delta_n$
$$I(p) := \{x\in \mathbb{R}^n: x = \sum_{i=0}^{n-1}q_i a_i,q\in\Delta_n,q\le\frac{1}{\epsilon}p\}$$
For my question the probabilistic background doesn't play a role since it's a pure analysis question to establish a certain equality. My notes say based on the definition of the sets $G,I(p)$ and $H$ the following should be true for a $v\in \mathbb{R}^n$
$$\sup_{p\in G}\max_{x\in I(p)} \langle v,x\rangle = \max_{x\in H}\langle v, x\rangle$$
 where $\langle v,x\rangle$ denotes the inner product in $\mathbb{R}^n$.
My attempt so far: The set $H$ uses all $p\in G$ in its definition. Therefore we clearly have for any $p\in G$, $\max_{x\in I(p)} \langle v,x\rangle \ge \max_{x\in H}\langle v, x\rangle$ leading to $\sup_{p\in G}\max_{x\in I(p)} \langle v,x\rangle \ge \max_{x\in H}\langle v, x\rangle$. I'm sturggeling to prove the opposite direction. Any help would be appreciated.
This question arises in the following paper about robust optimization. It's about one step in the proof of theorem 5. The proof can be found in the appendix.
 A: The key observation is that
$$
H=\bigcup_{p\in G}I(p).
$$
LHS$\ \leq\, $ RHS.
For all $p\in G$, we have that $I(p)\subseteq H$, and therefore $\max_{x\in I(p)}\langle v,x\rangle\leq \max_{x\in H}\langle v,x\rangle.$ Taking the supremum over $p\in G$ on both sides yields that
$$
\sup_{p\in G}\max_{x\in I(p)}\langle v,x\rangle\leq \max_{x\in H}\langle v,x\rangle.
$$
LHS$\ \geq\, $ RHS. Let $x^*\in H$ be a point at which $\langle v,x\rangle$ attains its maximum over $x\in H$. (Such a point exists since $H$ is compact.) Let $p^*\in G$ be a point such that $x^*\in I(p^*)$. Then
$$
\sup_{p\in G}\max_{x\in I(p)}\langle v,x\rangle\geq \max_{x\in I(p^*)}\langle v,x\rangle\geq \langle v,x^*\rangle=\max_{x\in H}\langle v,x\rangle.
$$

Note. It is easy to see that not only $H$ but also each $I(p)$ is compact, since they are closed subsets of the probability simplex (which is itself compact). The authors are implicitly using this in the notation, by writing $\max\, \langle v,x\rangle$ rather than $\sup\, \langle v,x\rangle$. (Typically one writes $\max$ only when the optimum is attained.)
