# Distributionally Robust Stochastic Programming - Help with derivation

I've been working through this book on robust optimization of electric energy systems, and in particular chapter 4 on distributionally robust optimization. In following the derivation of section 4.2.1.3, I cannot understand the last step. Here is the derivation I'm following and my understanding of each step, up until the point at which I get lost.

First, let $$x$$ be a decision vector and $$\xi$$ be a vector of random variables. A static distributionally robust stochastic optimization problem (DRSO) is

$$$$\underset{x\in X}{\text{min}} \hspace{0.2cm} \underset{\xi \in \mathbb{U}}{\text{max}} \hspace{0.05cm}\mathbb{E}_{\mathbb{P}}[f(x, \xi)]$$$$

where $$\mathbb{U}$$ is a support set defined as all $$\xi$$ such that $$(\xi - \xi_{0})\Lambda(\xi - \xi_{0}) \leq 1$$, with $$\Lambda$$ positive semi-definite. This uncertainty set defines an ellipsoid centered at nominal values of $$\xi_{0}$$. $$\mathbb{P}$$ is a probability distribution that lives in an ambiguity set $$\mathbf{P}$$.

The inner maximization is

$$\underset{\mu}{\text{max}} \int_{\mathbb{U}} f(x, \xi) d\mu(\xi) \\ \text{subject to} \\ \int_\mathbb{U}\xi d\mu(\xi) = \xi_{0} \\ \int_\mathbb{U}(\xi - \xi_{0})(\xi - \xi_{0}) d\mu(\xi) \preceq \gamma \Sigma_{0}\\ \int_\mathbb{U}d\mu(\xi)=1$$

which requires the covariance to lie in the cone $$\gamma \Sigma_{0}$$, where $$\Sigma_{0}$$ is the covariance matrix. Also, $$\mu$$ must integrate to 1 and the mean of $$\xi$$ must equal the notional mean $$\xi_{0}$$.

Then, if you form the Lagrangian and group terms, you get the following Langrangian dual problem:

$$L(\mu, \lambda, \mathbf{\beta}, Q) = \lambda + \xi_{0}\beta + \gamma \Sigma_{0} \bullet Q + \int_\mathbb{U}(f(x, \xi) - \lambda - \xi \beta - (\xi - \xi_{0})Q(\xi - \xi_{0}))d\mu(\xi) \\ \\ = \underset{\lambda, \beta, Q \succeq 0}{\text{min}} \hspace{0.05cm} \underset{\mu}{\text{max}} \hspace{0.05cm}\lambda + \xi_{0}\beta + \gamma \Sigma_{0} \bullet Q + \int_\mathbb{U}(f(x, \xi) - \lambda - \xi \beta - (\xi - \xi_{0})Q(\xi - \xi_{0}))d\mu(\xi) \\ =\underset{\lambda, \beta, Q \succeq0}{\text{min}} \lambda + \xi_{0} \beta +\gamma \Sigma_{0} \bullet Q \\\ \text{subject to} \\ f(x, \xi) - \lambda - \xi \beta - (\xi - \xi_{0})Q(\xi - \xi_{0}) \leq 0, \hspace{0.25cm} \forall \xi \in \mathbb{U}$$

where $$\bullet$$ is the Frobenius norm of two matrices, ie the component-wise sum.

The last step comes from the fact that the maximization over $$\mu$$ may be unbounded since the constraint that $$\mu$$ integrate to 1 has been relaxed in the Lagrangian, so requiring that the term in the integral be negative ensures the integral will converge and the maximization over $$\mu$$ will remain finite.

Now, let $$f(x,\xi)$$ = $$\underset{1 \leq k \leq K}{\text{max}} y_{k} \xi + l_{k}(x)$$. Substituting in for $$f(x, \xi)$$, we get

$$\underset{\lambda, \beta, Q \succeq0}{\text{min}} \lambda + \xi_{0} \beta +\gamma \Sigma_{0} \bullet Q \\\ \text{subject to} \\ \underset{1 \leq k \leq K}{\text{max}} y_{k} \xi + l_{k}(x) - \lambda - \xi \beta - (\xi - \xi_{0})Q(\xi - \xi_{0}) \leq 0, \hspace{0.25cm} \forall \xi \in \mathbb{U}$$

IF the above is true for the maximum over $$k$$, then it must be true for all $$k$$:

$$\underset{\lambda, \beta, Q \succeq0}{\text{min}} \lambda + \xi_{0} \beta +\gamma \Sigma_{0} \bullet Q \\\ \text{subject to} \\ y_{k} \xi + l_{k}(x) - \lambda - \xi \beta - (\xi - \xi_{0})Q(\xi - \xi_{0}) \leq 0, \hspace{0.25cm} \forall \xi \in \mathbb{U}, \hspace{0.25cm} \forall k \in K$$

Now putting a terms with $$\xi$$ on one side,

$$\underset{\lambda, \beta, Q \succeq0}{\text{min}} \lambda + \xi_{0} \beta +\gamma \Sigma_{0} \bullet Q \\\ \text{subject to} \\ y_{k} \xi - \xi \beta - (\xi - \xi_{0})Q(\xi - \xi_{0}) \leq \lambda - l_{k}(x), \hspace{0.25cm} \forall \xi \in \mathbb{U}, \hspace{0.25cm} \forall k \in K$$

Then, if we multiply the constraint by -1, we can eliminate the $$\forall \xi \in \mathbb{U}$$ by minimizing over $$\xi$$

$$\underset{\lambda, \beta, Q \succeq0}{\text{min}} \lambda + \xi_{0} \beta +\gamma \Sigma_{0} \bullet Q \\\ \text{subject to} \\ \underset{\xi : (\xi - \xi_{0})\Lambda(\xi - \xi_{0}) \leq 1}{\text{min}}(\xi - \xi_{0})Q(\xi - \xi_{0}) + (\beta - y_{k}) \xi \geq l_{k}(x) -\lambda, \hspace{0.25cm} \forall k \in K$$

Everything up to here I understand- it is the next step the authors derive (without explanation) that I don't follow. Somehow, the above problem can be converted into this form:

Unfortunately, there is no information in the book regarding how this last step is completed.