A problem, asked by Boyd and Vandenberghe (3.36f), is:
Compute the conjugate function $f^*(y,u) = \sup_{x,t} \{ y^Tx + ut - f(x) \}$ of the negative generalized logarithm for second-order cone: $f(x,t) = −\log{(t^2 −x^Tx)}$ on $\{(x,t)\in \mathbb{R}^n \times \mathbb{R} \mid \|x\|_2 < t \}$
The solution manual states $dom f^* = \{(y,u) \mid ||y|| < -u \}$.
Well, I disagree, and I don't see why my disagreement is false.
Let $(y,u) \in dom f^*$.
If $u > 0$, then for $x=0$, $y^Tx + ut - f(x) = ut + \log{t^2} \rightarrow_{t+\infty} +\infty$. So $(y,u) \notin dom f^*$ Therefore, $u \leq 0$.
We consider now $u \leq 0$. If $\exists k, y_k > 0$, then for $z>0$, $x=(\delta_{ki}z)_i$, $t=-2z$, $y^Tx + ut - f(x) = z(y_k - 2u) + \log{3z^2} \rightarrow_{z+\infty} +\infty$. So $(y,u) \notin dom f^*$ Therefore, $\forall k, y_k \leq 0$.
If $\exists k, y_k > 0$, then for $z>0$, $x=(-\delta_{ki}z)_i$, $t=-2z$, $y^Tx + ut - f(x) = z(-y_k - 2u) + \log{3z^2} \rightarrow_{z+\infty} +\infty$. So $(y,u) \notin dom f^*$ Therefore, $\forall k, y_k = 0$.
This is quite different. What is wrong?