Consider the minimization problem $$\min_{x \in \Delta_n} f(x)$$ where $f$ is $C^1$ function over the unit simplex $\Delta_n$. Prove that $x^*\in\Delta_n$ is a stationary point of the problem iff there exists $\mu\in\mathbb{R}$ such that $$\frac{\partial f}{\partial x_i}(x^*)=\left\{\begin{array}{rcl} = \mu&x_i^*>0 \\ \geq\mu&x_i^*=0 \end{array}\right.$$

I know the solution for the stationary points of $\mathbb{R}_{+}^{n}$ and for

$$C = \{ x : \sum_{i=1}^{n}x_i = 1 \}$$

and I need to solve this without KKT.
I've tried taking $y\in\Delta_n$ which defined by $y_k=\left\{\begin{array}{rcl} x_k^*&k\notin\{i,j\}\\ x_j^*&k=i\\ x_i^*&k=j\\ \end{array}\right.$
And then looking at the definition of stationary point for $x^*$ and y :
$\nabla f(x^*)(y-x^*)=\sum_{k=1}^{n}\frac{\partial f}{\partial x_k}(x^*)(y_k-x_k^*)=\frac{\partial f}{\partial x_i}(x_j-x_i)+\frac{\partial f}{\partial x_j}(x_i-x_j)=(x_j-x_i)(\frac{\partial f}{\partial x_i}-\frac{\partial f}{\partial x_j})\geq0$ and I tried from this condition to get the desired result but got stuck here.

Definition of Stationarity Let $f$ be $C^1$ function over a closed and convex set $C$ . then $x^*$ is called a stationary point of (P) if $\nabla f(x^*)(x-x^*)\geq0$ for any $x\in C$


Let $e^{(k)}$ denote the vector with $k$-th component 1 and all other components 0. Assume first that $x^*$ is a stationary point. Pick two indices $i\neq j$ for which $x_i^*$ and $x_j^*$ are both positive. Then for small positive $\delta$, $x=x^* + \delta e^{(i)} - \delta e^{(j)}$ and $x^-=x^* - \delta e^{(i)} + \delta e^{(j)}$ are in the simplex. Use the definition of stationarity to show that $\frac{\partial f}{\partial x_{i}}\left(x^{*}\right)=\frac{\partial f}{\partial x_{j}}\left(x^{*}\right)$. Next, assume that $x^*_i \gt 0$ and $x^*_j = 0$, in which case only $x^-$ is in the simplex. Use the definition of stationarity to show that $\frac{\partial f}{\partial x_{i}}\left(x^{*}\right) \le \frac{\partial f}{\partial x_{j}}\left(x^{*}\right)$.

Now assume that the gradient condition is satisfied, and use similar logic to show that the stationarity condition is met.


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