# Stationary condition for unit simplex

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 $$\frac{\partial f}{\partial x_{i}}\left(x^{*}\right) \le \frac{\partial f}{\partial x_{j}}\left(x^{*}\right)$$.