Assume $f(x)$ is defined and bounded on $[a, b]$ and $g(x) = \sup_{z \in [x, b]} f(z)$. Then if $g(x)$ is a strictly concave on $[a, b]$ function then $f(x)$ is also a strictly concave on $[a, b]$ function.
As I know $\forall \ x_1, x_2 \in [a, b], \forall \ t \in [0, 1] \to \ g(tx_1 + (1-t)x_2) > tg(x_1) + (1-t)g(x_2)$ and I need to prove that $\forall \ x_1, x_2 \in [a, b], \forall \ t \in [0, 1] \to \ f(tx_1 + (1-t)x_2) > tf(x_1) + (1-t)f(x_2)$
I've managed to conclude that $tf(x_1) + (1-t)f(x_2) \leq tg(x_1) + (1-t)g(x_2) < g(tx_1 + (1-t)x_2) \ge f(tx_1 + (1-t)x_2)$
Could you please give me any hints what to do next? Thanks in advance!
$\mathbf{Update}$ The foregoing statement is false and the counter example is $f(x)=(1−x^2) \mathbb{1}_Q(x)$ not a strictly concave function despite of the strictly concave $g(x) = (1-x^2)$
If $f$ is not continious, then the statement is false. (See my counterexample in the comments.) On the contrary, assuming that $f$ is continious, the statement is true.
Proof: Define $M:= \sup_{h \in [a,b]} f(x)$. By continuity at least one point exists $x \in [a,b]$ with $f(x) =M$. If another $y \in [a,b]$, say $x < y$, exists with $f(x) = f(y)=M$, then $$M= t f(x) +(1-t) = t g(x) +(1-t) g(y) < g(tx+(1-t)y) =f(v)$$ for some $v \in [tx+(1-t)y,b]$. A contradiction, because $M$ is the supremum! So there is only one global maximum.
We have already $x=a$. If the smallest $x \in [a,b]$ with $f(x) = M$ is bigger than $a$, we get $f(y) < f(x)$ for all $y \in [a,x)$, since there is only one global maximum. Thus $g(y) = M$ for every $y \leq x$. That's not a strictly concave function!
We can use the first agument on $[h,b]$ instead of $[a,b]$, since $f$ restricted to $[h,b]$ satisfies the initial condtions, in order to show that already $f(h) = g(h)$.
Note that $g$ is already strictly montone decreasing. For $v < w$ exists $t \in (0,1)$ with $v= ta +(1-t)w$, and by strictly concavity we have $$g(w) < t g(a) + (1-t) g(w) \leq g) < g(v).$$