Variational calculus proof

[jeremyfischer]

1. Suppose f:R$$^{NxN}$$ is defined by f(x,y) = $$\varphi$$(x) + $$\varphi$$*(Ax+y) where $$\varphi$$$$\epsilon$$$$\Gamma$$(R$$^{N}$$) and A$$\epsilon$$R$$^{NxN}$$ is skew-symmetric.
Prove that f*(y,x) = f(x,y)


3. The Attempt at a Solution
Information I know:
Skew-symmetric : A*=-A
f* computation: f*(y) = sup(x$$\epsilon$$R$$^{N}$$ {<x,y> - F(x)}

Would really appreciate it if someone would help me start this proof

 

[lippyka]

! Let f*(y,x) = F*(y) + \varphi*(Ay+x). Since A is skew-symmetric, A*=-A. Therefore, f*(y,x) = F*(y) + \varphi*(-Ax+x) = F*(y) + \varphi*(x-Ax) = F*(y) + \varphi*(Ax+x-Ax) = F*(y) + \varphi*(Ax+x) = f(x,y). Therefore, f*(y,x) = f(x,y).