For a convex optimization problem, if primal and dual optimal variables exist, does strong duality hold? Is there a convex optimization problem such that a primal optimal solution exists, and a dual optimal solution exists, but the primal optimal value is not equal to the dual optimal value?
 A: I found an example here, which I'll summarize below.
Consider the optimization problem
\begin{align}
\operatorname{minimize} & \quad e^{-x} \\
\text{subject to} & \quad \frac{x^2}{y} \leq 0.
\end{align}
The domain for the problem is $D = \{ (x,y) \mid y > 0 \}$. It can be shown that this problem is convex. Clearly, $x = 0, y = 1$ is a primal optimal solution. 
The Lagrangian is
$$
L(x,y,z) = e^{-x} + \frac{z x^2}{y}.
$$
The dual function is
\begin{align}
G(z) &= \inf \, \left\{ e^{-x} + \frac{z x^2}{y} \mid y > 0 \right\}.
\end{align}
You can check that $G(z) = 0$ for all $z \geq 0$.
Thus, any $z \geq 0$ is a dual optimal solution.
So, this is a problem for which primal and dual optimal variables exist, but strong duality fails to hold.
A: No.  Here is a standard counter-example (perhaps also found elsewhere on this site?)  
Define $\mathcal{X} = [0,1]$.  Define the convex function $f:\mathcal{X}\rightarrow\mathbb{R}$ by
$$f(x) = \left\{ \begin{array}{ll}
1 &\mbox{ if $x =0$} \\
0  & \mbox{ if $x \in (0,1]$} 
\end{array}
\right.$$
Then consider the convex program: 
\begin{align}
\mbox{Minimize:} & \quad f(x) \\
\mbox{Subject to:} & \quad x \leq 0 \\
& \quad x \in \mathcal{X}
\end{align}
The optimal solution to the primal is $x^*=0$ and $f(x^*)=1$. 
The dual function (defined for $\mu\geq 0$) is: 
$$ d(\mu) = \inf_{x \in \mathcal{X}} [f(x) + \mu x] =0 \quad \forall \mu \geq 0 $$
So the dual function is constant, in particular $\mu^*=0$ maximizes the dual function and $d(\mu^*)=0$.  So $d(\mu^*)<f(x^*)$ and we have a duality gap of 1.  
