Consider two Hilbert spaces $\mathcal{H}$ and $\mathcal{G}$ and an extended real valued function $f:\mathcal{H}\to\mathbb{R}\cup\{+\infty\}$ with $f\in\Gamma_0(\mathcal{H})$ the set of proper, lower semicontinuous, convex functions. We look at the following problem (we will refer to this as the primal),
$$\min f(x) : Ax=0$$
where $A:\mathcal{H}\to\mathcal{G}$ is a linear operator.
We can form the classical Lagrangian for our problem,
$$\mathcal{L}(x,\mu) = f(x) + \langle \mu, Ax\rangle$$
and state the dual problem,
$$\max_\mu\min_x\mathcal{L}(x,\mu)$$
Let us also assume that strong duality holds. My question, then, is the following:
Imagine you have a particular solution $\bar{\mu}$ to the dual problem and you also have a solution $x^*\in\arg\min\limits_{x}\mathcal{L}(x,\bar{\mu})$. Is it true that $Ax^*=0$, i.e. $x^*$ is feasible and thus $(x^*,\bar{\mu})$ is a saddle point/primal dual pair?
No, not necessarily. Take $\min_{x} \{ x : x = 0 \}$. The Lagrangian is $L(x,\mu)=x+\mu x$, the unique saddle point is $(x,\mu)=(0,-1)$, while any $x$ minimizes the Lagrangian.