I asked how to solve this optimization here. I found this approach by combining @Royi's idea in his answer with KKT's conditions. Personally, I feel my formulation is clearer and easier to understand.
Could you please verify if my proof is correct or contains logical mistake? Thank you so much!
Let $[\![ p ]\!]:=\{1,\ldots,p\}$ and $(x_1,\ldots,x_p) \in \mathbb R^p$. Solve the constrained optimization problem $$\begin{align*} \text{min} &\quad \frac{1}{2}\sum_{i=1}^p (y_i-x_i)^2 \\ \text{s.t} &\quad \sum_{i=1}^p y_i - 1 &&=0\\ &\quad\forall i \in [\![ p ]\!]: -y_i &&\le 0 \end{align*}$$
$\textbf{My attempt}$ Define $$\begin{aligned} f(y) &= \frac{1}{2}\sum_{i=1}^p (y_i-x_i)^2 \\ h(y) &= \sum_{i=1}^p y_i - 1 \\ \forall i \in [\![ p ]\!]: g_i(y) &= -y_i \end{aligned}$$ We have $f,g_i$ are convex and $h$ is linear. Let $a =(1/p, \ldots, 1/p)$. Then $h(a)=0$ and $g(a) <0$ for all $i \in [\![ p ]\!]$. It follows that Slater's condition is qualified. By Karush-Kuhn-Tucker conditions, we have $$\begin{aligned} \begin{cases} \forall i \in [\![ p ]\!]:\mu_i &\ge 0 \\ \forall i \in [\![ p ]\!]: g_i(y) &\le 0\\ h(y) &=0 \\ \forall i \in [\![ p ]\!]:\mu_i g_i(y)&=0 \\ \nabla f (y)- \lambda\nabla h (y)+ \mu_i \nabla g_i (y) &=0 \end{cases} &\iff \begin{cases} \forall i \in [\![ p ]\!]:\mu_i &\ge 0 \\ \forall i \in [\![ p ]\!]:-y_i &\le 0\\ \sum_{i=1}^p y_i - 1&=0 \\ \forall i \in [\![ p ]\!]: -\mu_i y_i &=0 \\ \forall i \in [\![ p ]\!]: (y_i - x_i) -\lambda - \mu_i &= 0 \end{cases} \end{aligned}$$
If $x_i+\lambda = 0$ then $y_i=\mu_i =0$ and thus $y_i = (x_i+\lambda)_+$. If $x_i+\lambda > 0$ then $y_i>0$ and thus $\mu_i=0$. Then $y_i = (x_i+\lambda)_+$. If $x_i+\lambda < 0$ then $\mu_i>0$ and thus $y_i=0$. Then $y_i = (x_i+\lambda)_+$. As such, we always have $y_i = (x_i+\lambda)_+$.
Then $\sum_{i=1}^p y_i - 1=0 \iff \sum_{i=1}^p (x_i+\lambda)_+ - 1=0$. Notice that $(x_i+\lambda)_+ = \max \{x_i+\lambda,0\}$ is continuous in $\lambda$ for all $i \in [\![ p ]\!]$. Hence $\psi(\lambda) = \sum_{i=1}^p (x_i+\lambda)_+ - 1$ is continuous in $\lambda$. Let $\alpha = -\max_{i \in [\![ p ]\!]}|x_i|$ and $\beta =1+ \max_{i \in [\![ p ]\!]}|x_i|$. It follows that $\psi(\alpha)<0<\psi(\beta)$. By Intermediate Value Theorem, the equation $\psi(\lambda)=0$ has a solution. Notice that $\psi$ is strictly increasing, so such solution is unique. We can also solve that equation by applying Intermediate Value Theorem on the interval $[\alpha , \beta]$.
