Question is that to prove if $$u'' + e^u = −x \text{ for } 0< x <1$$, then u cannot attain a minimum in $(0,1)$
Step I tried:
First, $u'' + e^u <0.$ I assume exists a $c \in (a,b)$ s.t. $u(c)=m,$ where $m$ is the minimum. Then, we have $u'(c)=0$ and $u''(c)>0$.
I want to show $$u(c)'' + e^{u(c)} >0$$ to get contradiction. Can I just say $u(c)'' + e^{u(c)} >0$ since we get that assume and done the proof? Or some thing need to be explain or prove?
It seems that the point of the exercise is to show that there is no minimum on the open interval $(0,1)$.
As already pointed out in the comments, $u''<0$ on $(0,1)$, therefore the function is concave. If it has a minimum on $[0,1]$, this will have to be on the boundary $x\in\{0, 1\}$, but not on the open interval $x\in (0,1)$, which is a general property of concave functions on $\mathbb R$, see Note 2.
Note 1: due to a lack of initial or boundary conditions, you cannot rule out a minimum on the boundary.
Note 2: A function with $u'' < 0$ can have at most one maximum or minimum on the interior $(0,1)$, because $u'$ is strictly monotonic and can thus only be zero at most once. This point where $u'=0$ will have $u'' < 0$, hence any extremum in $(0,1)$ will be a strict maximum. This leaves the boundary points $\{0, 1\}$ as the only remaining candidates for minima, but they are not in the open interval $(0,1)$.