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Consider the function $$f(x,y) = \big(b-x^T A y\big)^2$$ where $x \in \mathbb{R}^{p\times 1}$, $y \in \mathbb{R}^{n\times 1}$, $A \in \mathbb{R}^{p \times n}$ and $b \in \mathbb{R}$.

What is the Fenchel conjugate of $f$, that is $$ f^*(u,v) := \sup_{x,y} \big[ u^T x + v^T y - f(x,y)\big] = \;? $$

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This is just a partial answer, but maybe it's still of some use to you. First, I would simplify the system a bit by reducing it to just one variable $\tilde x\in \mathbb R^{n+p}$ and $\tilde u\in \mathbb R^{n+p}$ by defining $\tilde x=\begin{bmatrix}x\\y\end{bmatrix}$, $\tilde u=\begin{bmatrix}u\\v\end{bmatrix}$ and $\tilde A=\frac12 \begin{bmatrix}0 & A \\ A^T & 0\end{bmatrix}$. Note that $\tilde A$ is symmetric. Then $$\tilde f(\tilde x)=(b-\tilde x^T\tilde A \tilde x)^2$$ and the Fenchel transform becomes $$\tilde f^*(\tilde u)=\sup_{\tilde x} [\tilde u^T \tilde x - \tilde f(\tilde x)]$$ Since $\tilde f$ is differentiable, we get as necessary condition $$\partial_{\tilde x} (\tilde u^T \tilde x - \tilde f(\tilde x)) =\tilde u - 4(b-\tilde x^T \tilde A\tilde x)\tilde A x=0$$ If $\tilde g$, defined by $\tilde g(\tilde x)=4(b-\tilde x^T\tilde A\tilde x)\tilde A\tilde x$, is invertible, you get $$\tilde f^*(\tilde u)=\tilde u\tilde g^{-1}(\tilde u)-\tilde f(\tilde g^{-1}(\tilde u))$$ However, I cannot see how one could come up with an analytical expression for $\tilde g^{-1}$, nor how one could get rid of the $\tilde x$ terms in the expression for $\tilde f^*$ otherwise. (I tried a bit around with premultiplying the equation for $\tilde u$ by $\tilde x^T$ and then trying some substitutions in $\tilde f^*$, but that didn't work out.).

NB: If you need to do the transform numerically, then of course you can use stuff like e.g. Newton's method or Brent's method to find $\tilde g^{-1}$.

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  • $\begingroup$ Thanks for the effort. I think it is easier to separately maximize first over, say, $x$ and then $y$. It seems to me that the $\sup$ is going to be $\infty$ for most $(u,v)$ pairs. $\endgroup$
    – passerby51
    Feb 22, 2013 at 2:04
  • $\begingroup$ I tried that too with $x$ and $y$ separately, but than you get a couple system which doesn't make things easier IMO. $\endgroup$ Feb 23, 2013 at 8:50
  • $\begingroup$ I think you're right with the second part: just take some $\tilde x\in \ker(\tilde A)$ and $\tilde u^T \tilde x - \tilde f(\tilde x)=\tilde u^T \tilde x - b^2$. Since for any scalar factor $\alpha$ you also have $\alpha \tilde x\in\ker(\tilde A)$, this cannot be bounded. Except, of course, if $n=p$ and $A$ is non-singular. $\endgroup$ Feb 23, 2013 at 8:58
  • $\begingroup$ ... and not $\tilde u\in\ker(\tilde A)^\bot$. $\endgroup$ Feb 23, 2013 at 9:06

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