On an abstraction of duality for an optimization problem Let $X,Y$ be finite-dimensional normed spaces over $\mathbb{R}$. Let $T:X\rightarrow Y$ be a (necessarily bounded) linear operator and $T^\ast:Y^\ast\rightarrow X^\ast$ its adjoint. Fix $a\in T
(X)$.
I am curious when the following equation holds:
$$\inf \{\|x\|_X:x\in X,\hspace{.2cm}Tx=a\} = \sup\{\phi(a):\phi\in Y^\ast,\hspace{.2cm}\|T^\ast\phi\|_{X^\ast}\leq 1\}.$$
Is this true in general? Or on some assumptions on $X,Y,T$?
The context is from optimal transport on discrete spaces. The L.H.S. arises as an optimal cost over flows between two fixed distributions on a network (represented by $a$, which in practice is the difference between the distributions); and the R.H.S. arises as its Lagrangian dual. The equation above can therefore be derived as an instance of strong duality for a certain optimization problem.
This in some sense is a more high-level form of said problem, and my interest is in generalizing the spaces in a certain way (omitted). Hence I'm wondering if others have encountered an equation of this type, or if it arises as a special artifact of a deeper notion of duality, etc..
Thanks a million for any input!
 A: Yes, the conjectured equation does hold.  In order to prove it, suppose that $\phi\in Y^*$ is such that $\|T^*\phi\|\leq 1$, and $x\in X$ is such that $T(x)=a$.  Then
$$
  \phi (a) \leq |\phi (a)|=|\phi (T(x))| = |(T^*(\phi ))(x)| \leq \|T^*(\phi )\|\|x\| \leq \|x\|,
  $$
thus proving that
$$
  \sup\{\phi(a):\phi\in Y^*,\ \|T^*\phi\|\leq 1\} \leq
  \inf \{\|x\|_X:x\in X,\ Tx=a\}.
  $$
In order to prove the reverse inequality, choose $x_0$ with minimum norm such that $T(x_0)=a$, and let $r=\|x_0\|$.
Denoting by $B_r$ the closed ball in $X$ with radius $r$ and centered at zero, observe that $a=T(x_0)\in T(B_r)$.
However we claim that, given any $\rho >1$, one has that $\rho a\notin T(B_r)$.  This is because otherwise there would
be some $y$ in $B_r$ such that $\rho a=T(y)$, or equivalently $a=T(\rho ^{-1}y)$.  But since
$$
  \|\rho ^{-1}y\|=\rho ^{-1}\|y\|\leq \rho ^{-1}r<r=\|x_0\|,
  $$
this would contradict the minimality of $\|x_0\|$.
Using the Hahn-Banach separation Theorem, we therefore obtain a linear functional $\phi $ on $Y$, and a positive scalar
$t$, such that
$$
  \phi (y)\leq t\leq \phi (\rho a), \quad\forall y\in T(B_r).
  $$
Multiplying $\phi $ by $r/t$, we may assume that $t=r$.  For every $x$ in $X$, with $\|x\|\leq 1$, we then have that
$rx\in B_r$, so
$$
  (T^*(\phi ))(rx) = \phi (T(rx)) \leq r,
  $$
whence $(T^*(\phi ))(x) \leq 1$, and we deduce that $\|T^*(\phi )\|\leq 1$.  Moreover
$$
  \rho \phi (a) = \phi (\rho a) \geq r = \|x_0\|,
  $$
so $\phi (a) \geq \rho ^{-1}\|x_0\|$.  This shows that
$$
  \sup\{\phi(a):\phi\in Y^*,\ \|T^*\phi\|\leq 1\} \geq \rho ^{-1}\|x_0\|,
  $$
and since $\rho $ is arbitrary, we get the desired conclusion.
