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Let $X:= \{ u \in C^{1} [0,1] ~\big|\; u(0)=1, u(1) = 0\}$. Define the functional $F: X \to R$ by $F(u) := \int_{0}^{1} \big(e^{u' (x)} + u(x)^2 \big) \; dx$. Shows that $F$ does not get a minimizer on $X \cap C^{2} [0,1].$

I tried to create a sequence $u_n$ of piecewise functions where the integration is calculatable but I cannot guarantee either $u_n \in X \cap C^{2} [0,1]$ or $F(u_n) \to - \infty$.

Red shoes
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    Here is a proof using the Euler-Lagrange equations (necessary conditions for a minimizer) showing that they cannot be met for this functional and boundary conditions: https://math.stackexchange.com/questions/2200505/show-that-a-functional-has-no-minimum-in-a-given-set – whpowell96 Apr 21 '23 at 03:52

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This answer is wrong. You should use @whpowell96's comment instead.

More precisely, the part where I restrict the integral to $[0,\frac 1 n]$ is false because $e^{0} = 1$ so the integral on $[\frac 1n, 1]$ is worth $1-\frac 1n$. The infimum of $F$ over $X \cap C^2$ is not $0$. It seems to be a positive value which is maybe hard to compute, and it might be hard to construct explicit sequences $u_n$ such that $F(u_n) \to \inf_{X \cap C^2} F$.

First, for every $u \in X \cap C^2$, you have $F(u) \geq 0$ (because both terms in the integrand are positive). And in fact, $F(u) > 0$. Indeed, $F(u) = 0$ would imply that the integrand is zero everywhere so $u(x) = 0$ on $[0,1]$, but you have the boundary condition $u(0) = 1$.

Second, let us show that there exists a sequence $u_n \in X \cap C^2$ such that $F(u_n) \to 0$. For this let us take any function say $v \in C^\infty(\mathbb{R}_+;\mathbb{R})$ with $v(0) = 1$, $v' \leq 0$ on $\mathbb{R}_+$ and $v$ compactly supported in $[0,1)$ (so $v$ and all its derivatives vanish at $1$). In particular $0 \leq v(x) \leq 1$. And we set $u_n(x) := v(nx)$ for $n \in \mathbb{N}^*$. Then you have $u_n(0) = 1$, $u_n(1) = 0$ and $u_n \in C^\infty$. Then $$ \begin{split} F(u_n) & = \int_0^1 e^{-|u_n'(x)|} + u_n^2(x) \mathrm{d}x \\ & = \frac{1}{n} \int_0^{\frac{1}{n}} e^{-n |v'(nx)|} + v^2(nx) \mathrm{d} (nx) \\ & = \frac{1}{n} \int_0^1 e^{-n |v'(y)|} + v^2(y) \mathrm{d} y \\ & \leq \frac{2}{n}. \end{split} $$

To conclude, $F$ does not have a minimizer $u^*$, because the only possible value would be $F(u^*) = 0$, which is not achieved.

cs89
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    $F(u) = 0$ would imply that $u=0$ everywhere and $u' = -\infty$ everywhere. Since your sequence of functions and their derivatives become $0$ identically on the interval $[1/n,1]$, we have that $F(u_n)$ approaches $1$ for your sequence. Also you added a negative sign in your exponential. – whpowell96 Apr 21 '23 at 03:32
  • @whpowell96. Can you construct that sequence in an explicit way? – Red shoes Apr 21 '23 at 04:54
  • @whpowell96 Thank you for spotting the sign in the exponential. The argument is still negative because of my assumption on $v'$. So the computations above prove that $F(u_n) \to 0$. Why do you say $F(u_n) \to 1$? – cs89 Apr 21 '23 at 06:00
  • @red-shoes Sure, you can make it explicit. I use a smooth $v$ but $v \in C^2$ is enough. So for example $v(x) = (1-x)^3$ for $x\in[0,1]$ and $v(x) = 0$ for $x \geq 1$ satisfies $v(0) = 1$, $v' \leq 0$ on $\mathbb{R}+$, $0\leq v \leq 1$ on $\mathbb{R}+$, so the sequence $u_n(x) := v(nx)$ works. – cs89 Apr 21 '23 at 06:01
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    @cs89 for your sequence, the integral of $u_n^2$ goes to $0$. However, since $u'_n$ is zero on $[1/n,1]$, the term $e^{u'_n}$ is one on that interval, and less than one on $[0,1/n)$ from the decreasing assumption. Therefore the integral of $e^{u'_n}$ is a monotone increasing sequence converging to one. – whpowell96 Apr 21 '23 at 11:21
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    Or you could just plug in $u=0$ and see that you get one and not zero – whpowell96 Apr 21 '23 at 11:28
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    I think that the sign in the exponent is very important as it penalizes either increasing or decreasing behavior depending on the sign. I don't think that changing it to an absolute value fixes this. The reason for the nonuniqueness is because the exponential part is only zero when the derivative is unbounded in only a single direction. Try to apply these arguments to the original functional and notice that a sequence of functions with vanishing support will result in the exponential integrating to one – whpowell96 Apr 21 '23 at 11:32
  • @whpowell96 Thank you very much. Using a compactly supported function, I totally overlooked the fact that I would have $e^0 = 1$ outside. The Euler-Lagrange argument indeed answers the question, but it would still be interesting to try to compute $\inf F$ and construct explicit sequences. Thanks again. – cs89 Apr 21 '23 at 12:25