This is a hard problem I found and the solution does not seem natural for me can any tell me where the idea came from or at least if the construction of such a sequence is true or an alternative solution here is the problem : Let $f:[0,1]\rightarrow R$ be continuous with $f(0)=0$
show that there is a continuous concave function $g:[0,1]\rightarrow R$ such that $g(0)=0$ and $g(x)\ge f(x)$ for all $x\in[0,1]$
the solution I found for this problem (which I didn't find natural ) will be sent in a picture I hope you can see it clearly
Seginus
The intuition behind the solution is the construction of a continuous function $g$ with straight lines in each sub-interval $[x_{n+1},x_{n})$.
The condition $|f(x)|<M/2^n$ seems to ensure that in each sub-interval $[x_{n+1},x_{n})$ we have that $g(x)\ge f(x)$ for all $x\in [x_{n+1},x_{n})$. To be honest is hard to read this solution (at least to me).
A far more simple solution is just construct an straight line such that $g(x)\ge f(x)$, that is
$$g(x):=nx$$
for suitable $n\in\Bbb N$. We only need to choose some $n\in\Bbb N$ such that
The equation $nx=f(x)$ has no solutions for $x>0$, and
$nx>f(x)$ for some $x\in(0,1]$.
Then the function $g(x)\ge f(x)$ for all $x\in[0,1]$, and $g$ satisfies trivially that
$$\frac{n-nx}{1-x}=n\le\frac{nx-0}{x}=n,\quad\forall x\in[0,1]$$
hence $g$ is concave.
Note: the definition of concavity of the book you took the exercise seems to require that the above inequality must be strict, then for this case we can choose instead of an straight line some strictly concave function as $g(x)=n\sqrt x$, where we choose $n$ following the construction ideas of the two points of above, that is, $g(x)=f(x)$ had no solutions for $x>0$ and $g(x)>f(x)$ for some point in $x\in(0,1]$.