         \int_{0}^{T} \dual{A \dot\bSigma^\tau + B^\star \dot\bu^\tau}{\bT - \bSigma^\tau} \, \d t         \le         \int_{0}^{T} \dual{A \dot{\widetilde\bSigma}      + B^\star \dot{\widetilde\bu}     }{\bT - \widetilde\bSigma     } \, \d t