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In the final section of the section "General case", I am pretty sure that there is a substantial error. It is said that

${\displaystyle \langle \nabla j(v),\delta _{v}\rangle =d_{v}j(v;\delta _{v})=d_{v}{\mathcal {L}}(u_{v},v,\lambda _{v};\delta _{v}).}$

The last equality seems to be wrong. From the second equation from the Lagrange multiplier ansatz we instead find

${\displaystyle \langle \nabla j(v),\delta _{v}\rangle =d_{v}j(v;\delta _{v})=d_{v}J(u_{v},v;\delta _{v})=-\langle d_{v}D_{v}(u_{v};\delta _{v}),\lambda _{v}\rangle .}$

The example in the linear case should be changed accordingly. To be specific, instead of

${\displaystyle \langle \nabla j(v),\delta _{v}\rangle =\langle Au_{v},\delta _{v}\rangle +\langle \nabla _{v}B_{v}:\lambda _{v}\otimes u_{v},\delta _{v}\rangle ,}$

it should read

${\displaystyle \langle \nabla j(v),\delta _{v}\rangle =\langle Au_{v},\delta _{v}\rangle =-\langle \nabla _{v}B_{v}:u_{v}\otimes \delta _{v},\lambda _{v}\rangle .}$