Consider the following convex program:
\begin{aligned} \textrm{minimize} & \quad f_0(x)=\sum_{i=1}^n f_i(x_i) \\ \textrm{subject to} & \quad a^Tx=b \end{aligned}
where $a\in \mathbb{R}^n, b\in \mathbb R$, and $f_i:\mathbb R \to \mathbb R$ are differentiable and strictly convex. It is stated in Boyd & Vandenberghe's book (Example 5.4) that if there exists an $x_0 \in \mathrm{dom}\:f_0$ such that $a^Tx_0=b$, then the convex program has a unique optimal point. I don't quite see why. Does someone know the reason?
It appear to me that a counter-example could be: $f_0(x_1, x_2)=e^{x_1}+e^{x_2}$, subject to $x_1-x_2=0$. Then $\mathrm{dom}\:f_0=\mathbb R^2$ and includes the line $x_1-x_2=0$. And the optimal value of this program would be $0$, but is never attained, isn't it? I'd appreciate it, if someone can point out where I missed.