 Solution

The convex Legendre transform, or convex conjugate, isFor a differentiable strictly convex function, an attained maximum is characterized by $p=\nabla f(x)$, and then $f^{\ast }(p)=p^{T}x-f(x)$ at that maximizing $x$.

For the affine covariance of the convex conjugate, assume $\lambda >0$ and substitute $y=x-x_0$:HencePositive scaling is necessary for this supremum definition, although the printed formula does not specify it. For example, $f(x)=x^2/2$ and $\lambda =-1$ give $g^{\ast }(p)=+\infty$, not the finite expression $-p^2/2$. If the transform is instead defined as a stationary value using a locally invertible gradient, the same algebraic identity holds for any $\lambda \ne 0$: the critical-point relation is $p=\lambda \nabla f(y)$, and substitution of that stationary $y$ gives the displayed formula. That extension does not assert a supremum for negative scaling.

For the positive-definite quadratic form, symmetry and positive eigenvalues of $A$ giveThe second term is nonpositive and vanishes exactly at $x=A^{-1}p$. Therefore