Consider the following piecewise monotonically increasing, concave, and smooth function defined over $x\in[0,\infty)$: $$ f(x) = \begin{cases} \displaystyle 2\ln\left(\frac{1}{6}x+\frac{5}{3}\right), & \text{if } x\le8,\\ \ln(1+x), & \text{if } x>8. \end{cases} $$
Note that $f(x)$ is continuous and differentiable over any point $[0, \infty)$.
The following figure represents the above function $f(x)$:
In this regard, can it be possible to express the above function in one exact concave function? I guess that it can be done using $\max$ and $\min$ functions:
- $\max$ function is convex,
- $\min$ function is concave.
Actually, I am using the software, called CVX. In addition, my real goal is to find $x\in\mathbb{R}^N$ that maximizes $\sum_{i=1}^{N} f_i(x_i)$ subject to $x_i\ge0, \forall i$, and $\sum_{i=1}^N x_i \le L$, where $L$ is given positive value. For each $i$, $f_i(x)$ is piecewise monotonically increasing, concave, smooth function, like the above $f(x)$.
Ok, let's translate everything to be at the origin. Let $f(x)=2\ln(x/6 + 5/3)$ and $g(x) = \log(x+1)$. Define the following functions,
$$\tilde{f}(x) = f(x+8)$$
and
$$\tilde{g}(x) = g(x+8).$$
Notice that $\tilde{f}(0)=\tilde{g}(0)$ since $f(8)=g(8)$.
Furthermore define
$$\hat{f}(x) = \tilde{f}(x)-\tilde{f}(0)$$
and
$$\hat{g}(x) = \tilde{g}(x) - \tilde{f}(0).$$
Then we can almost write the function we want as,
$$\hat{h}(x) = \max(\hat{g}(x),0) + \min(\hat{f}(x),0),$$
except that this has been translated so we must translate back,
$$h(x) = \hat{h}(x-8) + \tilde{f}(0).$$
So our final function is simply $h(x)$.
I leave it to you to understand why it's still concave and to compute explicitly all the functions.
Edit: made notation more consistent by switching $\hat{h}(x)$ and $h(x)$.