1

Let $f(x)=\frac{1}{2}\|x-y\|^2$ and $\|\cdot\|$ be 2-norm.

$x,y \in R^n$

Let the fenchel conjugate be defined as $f(z)=\sup_x(z^Tx -\frac{1}{2}\|x-y\|^2)$.

Taking the derivative w.r.t. $x$ and setting it $0$:

$z - (x-y)=0$ and $x=z+y$

Then the fenchel conjugate is $f^*(z)=\frac{1}{2}\|z\|^2+z^Ty$.

Could somebody please check my calculations?

Emin Ozkan
  • 187

1 Answers1

3

Yes this is correct. There is a calculus of convex conjugates that allows you to compute the convex conjugate of a translated function so you could also derive the result as follows: let $h(x) = \frac{1}{2}\|x\|^2$. Then $f(x) = h(x-y)$ and we obtain using the translation rule: $$f^*(z) = h^*(z) + z^{\intercal}y$$ and it is an easy exercise to show that $h^*(z)= h(z)$.

E G
  • 991