Minimizing the variational problem $ I(y(x)) = \int_{0}^1 e^{-y'(x)^2}dx $ This is a question from a mathematical contest.
Let $X= \{ y \in C^1[0,1]| y(0)=y(1)=0 \}$. Define $I: X \to \mathbb{R}$ by:
$$  I(y(x)) = \int_{0}^1 e^{-y'(x)^2}dx $$.
Then which of the following are true:

*

*$I$ doesn't attain its infimum.

*$I$ attains its infimum at a unique $y \in X$.

*$I$ attains its infimum at exactly two elements $y(x)\in X$.

*$I$ attains its infimum at infinitely many $u \in X$.

What I tried:
Using the Euler's condition the extremal $y(x)$ satisfies:
$$ \frac{\partial F}{\partial y} - \frac{\partial^2 F}{ \partial x \partial y'}- y' \frac{\partial^2 F}{\partial y \partial y'}- y'' \frac{\partial^2 F}{\partial y'^2} =0$$
Now since our $F$ is independent of $x,y$, Euler's equation will give us that $y(x)= Ax+B$.
Now using $y(0)=0=y(1)$, we get the extremal as $y \equiv 0$.
This gives the curve which maximizes the variational problem. I want to know how to proceed when we want to minimize the value? One guess I can make is by taking some function for which $y'$ is large enough. But is there some more formal method?
Thanks!

If you know, that ELG can only have the Solution of $y = 0$, then you know there is no reachable infimum. But as a hint, $\int_0^1 y^\prime(x) \, dx = y(1)-y(0) = 0$ which means you can take like any $y^\prime := g \in C^0([0,1])$ as long the whole area is zero. :)
