[proofplan]
We prove the first-order condition by testing the minimizer against every sufficiently small affine perturbation inside the open feasible set. This reduces the constrained problem to an unconstrained one-variable minimization in the perturbation parameter. The resulting directional derivative vanishes in every direction, forcing the gradient vector to be zero. For uniqueness, positive definiteness of the Hessian makes the objective strictly convex along every nonconstant line segment, so two distinct stationary points cannot both solve the optimality equation.
[/proofplan]
[step:Reduce the minimization problem to one-variable perturbations through the interior point]
Define the objective
\begin{align*}
F_t:C^\circ \to \mathbb{R}
\end{align*}
by
\begin{align*}
F_t(x)=t\,c\cdot x+\Phi(x).
\end{align*}
Let $h \in \mathbb{R}^n$ be arbitrary. Since $C^\circ$ is open and $x(t) \in C^\circ$, there exists $\delta_h>0$ such that
\begin{align*}
x(t)+\varepsilon h \in C^\circ
\end{align*}
whenever $|\varepsilon|<\delta_h$. Define
\begin{align*}
\varphi_h:(-\delta_h,\delta_h)\to \mathbb{R}
\end{align*}
by
\begin{align*}
\varphi_h(\varepsilon)=F_t(x(t)+\varepsilon h).
\end{align*}
Because $x(t)$ minimizes $F_t$ on $C^\circ$, the point $\varepsilon=0$ minimizes $\varphi_h$ on $(-\delta_h,\delta_h)$.
[guided]
We want to use the fact that $x(t)$ is an interior minimizer. The word “interior” is what allows us to move in every direction $h \in \mathbb{R}^n$ while remaining feasible for small positive and negative values of the perturbation parameter.
Define
\begin{align*}
F_t:C^\circ \to \mathbb{R}
\end{align*}
by
\begin{align*}
F_t(x)=t\,c\cdot x+\Phi(x).
\end{align*}
Fix an arbitrary vector $h \in \mathbb{R}^n$. Since $C^\circ$ is open and $x(t) \in C^\circ$, there is a radius $\delta_h>0$ such that the whole perturbed segment
\begin{align*}
x(t)+\varepsilon h \in C^\circ
\end{align*}
for every real number $\varepsilon$ with $|\varepsilon|<\delta_h$. This is the exact point where openness of $C^\circ$ is used.
Now define the one-variable function
\begin{align*}
\varphi_h:(-\delta_h,\delta_h)\to \mathbb{R}
\end{align*}
by
\begin{align*}
\varphi_h(\varepsilon)=F_t(x(t)+\varepsilon h).
\end{align*}
Because $x(t)$ minimizes $F_t$ over all of $C^\circ$, every feasible perturbation has value at least $F_t(x(t))$. Therefore
\begin{align*}
\varphi_h(\varepsilon)\geq \varphi_h(0)
\end{align*}
for every $|\varepsilon|<\delta_h$. Thus $\varepsilon=0$ is a one-variable minimizer of $\varphi_h$.
[/guided]
[/step]
[step:Differentiate the perturbation function and force every directional derivative to vanish]
The function $\varphi_h$ is differentiable at $0$ because $\Phi$ is differentiable at $x(t)$ and the affine map $\varepsilon \mapsto x(t)+\varepsilon h$ is differentiable. Since $0$ is an interior minimizer of $\varphi_h$, its derivative at $0$ is zero:
\begin{align*}
0=\varphi_h'(0).
\end{align*}
By the chain rule,
\begin{align*}
\varphi_h'(0)=t\,c\cdot h+\nabla\Phi(x(t))\cdot h.
\end{align*}
Hence
\begin{align*}
(t c+\nabla\Phi(x(t)))\cdot h=0
\end{align*}
for every $h \in \mathbb{R}^n$. Taking $h=t c+\nabla\Phi(x(t))$ gives
\begin{align*}
|t c+\nabla\Phi(x(t))|^2=0.
\end{align*}
Therefore
\begin{align*}
t c+\nabla\Phi(x(t))=0.
\end{align*}
[/step]
[step:Use positive definiteness of the Hessian to rule out two solutions]
Assume now that $\Phi$ is twice differentiable and that the Hessian matrix $D^2\Phi(x)$ is positive definite for every $x \in C^\circ$. Suppose $x_0,x_1 \in C^\circ$ both solve
\begin{align*}
t c+\nabla\Phi(x)=0.
\end{align*}
Since $C^\circ$ is convex, the line segment from $x_0$ to $x_1$ lies in $C^\circ$. Define
\begin{align*}
\psi:[0,1]\to \mathbb{R}
\end{align*}
by
\begin{align*}
\psi(s)=F_t((1-s)x_0+s x_1).
\end{align*}
Let $v=x_1-x_0$. Then $\psi$ is twice differentiable and
\begin{align*}
\psi'(s)=(t c+\nabla\Phi((1-s)x_0+s x_1))\cdot v.
\end{align*}
Also,
\begin{align*}
\psi''(s)=v^\top D^2\Phi((1-s)x_0+s x_1)v.
\end{align*}
If $x_0\neq x_1$, then $v\neq 0$, so positive definiteness gives
\begin{align*}
\psi''(s)>0
\end{align*}
for every $s \in [0,1]$. Hence $\psi'$ is strictly increasing on $[0,1]$. But the optimality equation at $x_0$ and $x_1$ gives
\begin{align*}
\psi'(0)=0
\end{align*}
and
\begin{align*}
\psi'(1)=0.
\end{align*}
This contradicts strict increase of $\psi'$. Therefore $x_0=x_1$, and the equation has at most one solution in $C^\circ$.
[/step]
