I am trying to follow a proof for the fact that convex envelope of the rank function of a matrix $\mathbf{X} \in \mathbf{R}^{m \times n}$ (with unit spectral norm) is the trace norm or nuclear norm given in a paper.
The proof is based on finding the conjugate of the rank function. It proceeds as follows,
$$ f^*(y) = \sup_\limits{\|\mathbf{X}\| \leq 1} \left( \operatorname{Tr} \left( \mathbf{Y}^T \mathbf{X} \right) - \operatorname{rank} (\mathbf{X}) \right) $$
and makes use of the von Neumann's trace theorem
$$ \operatorname{Tr} \left({\bf Y}^T {\bf X} \right) \leq \sum_{i=1}^{q} \mathbf{\sigma_i(Y) \, \sigma_i(X)} \\ \text{where} \qquad q = \min(m,n) $$
and the fact that eigenvalues of $\mathbf{X}$ and $\mathbf{Y}$ does not depend on the rank function.
The paper says, Let $\mathbf{X = U_X\Sigma_X V_X}$ and $\mathbf{Y = U_Y\Sigma_Y V_Y}$ be the singular value decomposition of $\mathbf{X}$ and $\mathbf{Y}$. Since the rank function is independent of $\mathbf{U_X}$ and $\mathbf{Y_X}$, we pick $\mathbf{U_X} = \mathbf{U_Y}$ and $\mathbf{V_X} = \mathbf{V_Y}$ to maximize the term $\operatorname{Tr}\left(\mathbf{Y^TX}\right)$. Then it follow that
$$ f^*(y) = \sup_\limits{\|\mathbf{X}\| \leq 1}\left(\sum_{i=1}^q \mathbf{\sigma_i(X)\sigma_i(Y)} - \operatorname{rank}(\mathbf{X})\right) $$ I do not understand the last equality. It should be an inequality according to me. It should be, $$ f^*(y) \leq \sup_\limits{\|\mathbf{X}\| \leq 1}\left(\sum_{i=1}^q \mathbf{\sigma_i(X)\sigma_i(Y)} - \operatorname{rank}(\mathbf{X})\right) $$
Why it is an equality instead of inequality ? Link to the paper https://web.stanford.edu/~boyd/papers/pdf/rank_min_heur_sys_approx.pdf