$$\begin{array}{ll} \text{minimize} & f(\mathcal{X})= \| \mathcal{X} - \mathcal{A} \|_F^2\\ \text{subject to} & \mathcal{X}_\Omega = \mathcal{T}_\Omega \end{array}$$
where $\mathcal{X}$, $\mathcal{A}$ and $\mathcal{T}$ are all third-order tensor, and $\Omega$ is the observed element. $\mathcal{X}_\Omega = \mathcal{T}_\Omega$ means that their observed elements should be the same.
An solution for above is from the paper (its equation (24) ):https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6979248
I place its solution here:
By introducing Lagrangian multiplier $\mathcal{Q}$ for the constraint $\mathcal{X}_\Omega = \mathcal{T}_\Omega$
we have the Lagrangian function:
$F(\mathcal{X},\mathcal{Q}) = \|\mathcal{X} - \mathcal{A} \|_F^2 + <\mathcal{Q}, \mathcal{X}_\Omega - \mathcal{T}_\Omega>$
take derivative to zero, we have the KKT conditions
$2(\mathcal{X} - \mathcal{A}) + \mathcal{Q}_\Omega =0$
$ \mathcal{X}_\Omega - \mathcal{T}_\Omega = 0$
they then get the solution
$\mathcal{X} = \mathcal{T}_\Omega +\mathcal{A}_{\Omega^C}$
where ${\Omega^C}$ is the complement of ${\Omega}$
My question is that: How to get the solution from above KKT condition? it seems that the $\mathcal{Q}_\Omega$ is gone.
Thanks.