- 1

- 0

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

Prove that f*(y,x) = f(x,y)

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[tex]\epsilon[/tex]R[tex]^{N}[/tex] {<x,y> - F(x)}

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

Information I know:

Skew-symmetric : A*=-A

f* computation: f*(y) = sup(x[tex]\epsilon[/tex]R[tex]^{N}[/tex] {<x,y> - F(x)}

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