For given $D \in \mathbb{R}^{N\times M}$, $\psi\in \mathbb{R}^M$ and $s^0 \in \mathbb{R}^N$, I have the following relationship for all $h\in\mathbb{R}^N$: $$ h^T s^0 = h^TD\psi$$
Does it imply that: $$ s^0 = D\psi$$
I want to say yes, but can't seem to write a valid proof.
On
Take $h = e^i$, the $i$th standard basis vector, which has $1$ as its $i$th component and all other components zero. Then for any vector $x \in \mathbb R^n$, $h^T x = x_i$, the $i$th component of $x$.
Since for all $h \in \mathbb R^N$, we have $(e^i)^T s^0 = (e^i)^T D\psi$, $i = 1, \ldots, N$, which implies that the $i$th component of $s^0$ is the same as the $i$th component of $D\psi$ for every $i$, so $s^0 = D\psi$.
More succinctly, observe that if $h^T s^0 = h^T D\psi$ for all $h \in \mathbb R^n$, then for any matrix $A \in \mathbb R^{K \times N}$, we have $As^0 = h^T D\psi$ (by comparing columns of the products on both sides). Take $A = I_n \in \mathbb R^{N \times N}$ (identity matrix).
Just take $h=s_0-D\psi$. You get that $0=h^T(s_0-D\psi)=\|s_0-D\psi\|^2$. Hence $s_0-D\psi=0$.