Suppose we have a concave function $f(x):[0,+\infty)^d\rightarrow [0,+\infty)$ which is nondecreasing. In addition, define another piecewise linear function as $g(x) = \sum_{i\in \{i|x_i \leq \bar{x}_i\}} \lambda_i\cdot x_i + \sum_{i\in \{i|x_i > \bar{x}_i\}} \lambda_i\cdot \bar{x}_i: [0,+\infty)^d\rightarrow [0,+\infty)$, where $\bar{x}$ is a fixed point and $\lambda$ is the parameter I want to design. The only constraint for $\lambda$ is to make $g(\bar{x}) = f(\bar{x})$, i.e. $\sum_{i\in[d]} \lambda_i \cdot \bar{x}_i = f(\bar{x})$
My question is: is it possible to design a function $g(x)$ with appropriate $\lambda$ such that $g(x) \leq f(x)$ for any $x$ while ensuring $g(\bar{x}) = f(\bar{x})$? The answer is very intuitive in 1 dimension, we can simple set $\lambda = \frac{f(\bar{x})}{\bar{x}}$, and $g(x) = \frac{f(\bar{x})}{\bar{x}} \cdot x$ if $x\leq \bar{x}$ and $g(x) = f(\bar{x})$ if $x > \bar{x}$, which is shown in the following figure:
But in high dimension,it seems the task is more intricated, I've tried with $\lambda_i = \frac{f(\bar{x})}{\bar{x}_i \cdot d}$. In this case, although $\sum_{i\in[d]} \lambda_i \cdot \bar{x}_i = f(\bar{x})$ is true, but I wasn't able to prove $g(x') \leq f(x')$ for any $x' \neq \bar{x}$.
Update: $f(x)$ is differentiable everywhere. By non-decreasing, I meant every element in $\nabla f(x)$ is greater than or equal to zero.
If we remove the condition that $f$ is differentiable, then a counterexample can be found (instead of defining nondecreasing via the gradient, we would need to define it by requiring that $f$ is nondecreasing in every variable $x_i$).
I believe even if $f$ is differentiable you will encounter similar problems.
Let $d=2$, $f(x)=\min(x_1,x_2)$, $\bar x=(1,1)$. Suppose that there is a $\lambda$ such that $g$ satisfies the requirements. Then we would have $$ \lambda_1 = g((1,0)) \leq f(x) = 0 \qquad\text{and}\qquad \lambda_2 = g((0,1)) \leq f(x) = 0. $$ This implies $$ g(\bar x)=\lambda_1+\lambda_2 \leq 0 $$ which is a contradiction to $f(\bar x)=g(\bar x)$ and $f(\bar x)=1$.