# Function cannot attain a minimum inside the domain

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?

• $u'' < 0$ for all $x$ in the interval. If $u$ is continuous and differentiable, and it obtains its minimum in an open interval, we require $u'=0$ and $u'' > 0$ Sep 28 at 10:43
• That can't be determined. Nothing says that this function achieves a maximum either. It could be monotonic. Sep 28 at 11:01
• Try a simple contradiction: If $u$ has a minimum $x_0\in (0,1)$, then $u''(x_0) \ge 0$. But as mentioned in the first comment, $u'' < 0$ everywhere, so there cannot be any minimum in $(0,1)$. Sep 28 at 11:18

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 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 $$(0,1)$$.