A closed convex program with a positive duality gap when Slater fails

The epigraph of $\varphi$ is exactly $E=\{(x,y,t):y\ge 0,t\ge 0,x^2\le yt\}$. For $y>0$ this is the inequality $t\ge x^2\mathrm{/}y$; when $y=0$ it forces $x=0$ and $t\ge 0$; for $y<0$ it is empty. Thus $E$ is closed, being defined by continuous non-strict inequalities. It is convex: if $\mathrm{\mid }{x}_j\mathrm{\mid }\le \sqrt{y_j{t}_j}$ for $j=1,2$, then for $0\le \theta \le 1$, $\mathrm{\mid }\theta x_1+(1-\theta )x_2\mathrm{\mid }\le \theta \sqrt{y_1{t}_1}+(1-\theta )\sqrt{y_2{t}_2}\le \sqrt{(\theta y_1+(1-\theta )y_2)(\theta t_1+(1-\theta )t_2)}$ by Cauchy–Schwarz. Therefore $\varphi$ is closed convex. The objective is continuous and has second derivative $e^{-x}>0$ in $x$, so it is closed convex and proper as well.

On its domain $\varphi \ge 0$. If $y>0$, $\varphi (x,y)\le 0$ forces $x=0$; if $y=0$, finiteness already forces $x=0$. Hence the feasible set is $\{(0,y):y\ge 0\}$ and $f_0=1$ throughout it, so $p^{\ast }=1$ and the primal minimum is attained, for example at $(0,0)$. No point has $\varphi (x,y)<0$, so Slater fails.

For $\lambda \ge 0$, every point in the Lagrangian domain has $e^{-x}+\lambda \varphi (x,y)\ge 0$, so $q(\lambda )\ge 0$. At $\lambda =0$, taking $x\to +\mathrm{\infty }$ gives $q(0)=0$. For $\lambda >0$, take $(x_k,y_k)=(k,k^3)$; then $e^{-x_k}+\lambda \varphi (x_k,y_k)=e^{-k}+\lambda \mathrm{/}k\to 0$, so again $q(\lambda )=0$. Therefore $d^{\ast }={\sup }_{\lambda \ge 0}q(\lambda )=0$ and the positive duality gap is $p^{\ast }-d^{\ast }=1$. This does not contradict [F2], because its Slater hypothesis fails.

The calculation gives a closed convex primal problem with attained finite optimum and a strict gap; the missing strict-feasibility condition is exactly the boundary that blocks the Slater theorem's certificate.