For any two functions $f_1, f_2 : [0, 1] \to \Bbb R$, define the function $g : [0, 1] \to \Bbb R$ as follows $$g(x) = \max(f_1(x), f_2(x))$$ for all $x \in [0, 1]$. There are three claims to this question,
- when $f_1$ and $f_2$ are linear, then $g$ is linear.
- when $f_1$ and $f_2$ are differentiable then $g$ is differentiable
- when $f_1$ and $f_2$ are convex then $g$ is convex.
I am able to prove the last statement by inequality condition of convex function but i can't think of the counterexample to first and second statement.
Could this be solved using the definition of max function
$$ max{(f(x),g(x))} = [f(x) + g(x)]/2 +\lvert [f(x) - g(x)]/2\rvert $$
Consider the functions $f(x) = x-\frac{1}{2}$ and $g = -x+\frac{1}{2}$ on $[0,1]$.
Then their maximum $h(x) := \max(f,g) = |x-\frac{1}{2}|$ (as you can easily check. Since we get an absolute-value, this function is not differentiable at $x = \frac{1}{2}$. Now assume that h is of the form $h(x) = a \cdot x+b$ for some $a, b \in \mathbb{R}$ (affine-linear). As such, it would have to be differentiable everywhere in $[0,1]$. Which it is not.
Note that I assumed everywhere that you meant "affine-linear" by linear. That means functions of the form $f(x) = a \cdot x+b$. Because for purely-linear functions $f(x) = a \cdot x$ the statement should be true, not false.