Let $$ F(r, x) = \frac{1}{2}\| r\|^2_2 + \lambda \| x\|_1 + \langle y, Ax - b - r\rangle, $$ where $A \in \mathbb R^{m\times n}$, $x \in \mathbb R^n$, $b, r, y \in \mathbb R^m$ and $\lambda > 0$
Considering that $x$ is constant, find the convex conjugate of $F$
This is my work so far: \begin{align*} F^*(z) &= \sup_{u} (\langle z, u\rangle - F(u))\\ &= \sup_{u}(\langle z, u\rangle - \frac{1}{2}\| u\|^2_2 - \langle y, Ax - b - u\rangle - \lambda \| x\|_1)\\ &= \sup_{u}(\langle z, u\rangle - \frac{1}{2}\langle u, u\rangle - \langle y, Ax - b\rangle + \langle y, u\rangle - \lambda \| x\|_1)\\ &= \sup_{u}(\langle z, u\rangle) - \frac{1}{2}\langle u, u\rangle + \langle y, u\rangle) - \langle y, Ax - b\rangle - \lambda \| x\|_1\\ &= \sup_{u}(\langle z + y , u\rangle - \frac{1}{2}\langle u, u\rangle) - \langle y, Ax - b\rangle - \lambda \| x\|_1\\ &= \sup_{u}\left( \frac{1}{2}\| z + y \|_2^2 -\frac{1}{2}(\| u - (z + y)\|_2^2\right) - \langle y, Ax - b\rangle - \lambda \| x\|_1\\ &=\frac{1}{2}\|z+y \|^2_2 - \langle y, Ax - b\rangle - \lambda \| x\|_1, \end{align*} but I'm not sure sure because of the supremum and $x$ being constant.