I am reading this paper: http://dl.acm.org/citation.cfm?id=1390696
Following optimization problem is defined in section 2: \begin{align} \max_{\mathbf{X}>0} \log det(\mathbf{X})-trace(\mathbf{SX})-\lambda||\mathbf{X}||_1 \end{align} where $\mathbf{X}$ and $\mathbf{S}$ are $p\times p$ matrix and $\mathbf{X}$ is constrained to be positive definite.
In the next subsection, the author creates a dual representation by writing \begin{align} ||\mathbf{X}||_1 = \max_{||\mathbf{U}||_\inf \leq 1} trace(\mathbf{XU}) \end{align}
This is fine, as $||\mathbf{X}||_1 = trace(\mathbf{XU})$, if all entries of U are 1, and that is the maximum value possible. But in the next step, this is substituted in the original optimization problem to form:
\begin{align}
\max_{\mathbf{X}>0} \min_{||\mathbf{U}||_\inf \leq \lambda}\log det(\mathbf{X})-trace(\mathbf{(S+U)X})
\end{align}
and the $\max_{||\mathbf{U}||_\inf \leq 1} $ is changed $\min_{||\mathbf{U}||_\inf \leq \lambda} $ . Changing 1 to $\lambda$ is obvious, but why is the max changed to min. Shouldn't a minus sign be placed before $\mathbf{U}$ in that case?
Is there some property of trace operator which allowed for this step?
$$a-\max_{x\in X} f(x)=\min_{x\in X}(a-f(x)),$$ this is just placing the minus sign inside the max which makes it a min.