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Let $X=\left\{f\in C^1[0,1]: f(0)=0=f(1)\right\}$.

Define $J:X \to \Bbb R$ by $J(f)=\int\limits^1_0 e^{-f^\prime (x)^2}dx.$

How to show that $J$ does not attain its infimum.

If $F(x,f,f^\prime)=e^{-f^\prime (x)^2}$, using the Euler-Lagrange equation: $F_{f}-\frac{d}{dx}\left(F_{f^\prime}\right)=0$ with $f\in X$, one can obtain $f=0$. But this does not give infimum. Possibly, $f=0$ is the maximum of $J(f)$ as $J(f)\le 1, \forall f\in X$ and $J(0)=1$. Let me know how to approach this?

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user149418
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  • What if you try the same sequence $f_n(x)=n\sin(\pi x)$ as in your other question? Clearly $\inf J(f)\ge 0$, so we need to make $f'^2$ as large as possible to push the exponential to zero. – A.Γ. Jun 23 '18 at 21:33
  • You need an estimate $J(f_n)\le...$ to conclude that it goes to zero. Formally, you may need to exclude a small interval around $x=1/2$ where $\cos(\pi x)$ is small, and on the rest the convergence of the integrand to zero is uniform. The both parts can be made arbitrary small, hence $J(f_n)\to 0$. – A.Γ. Jun 23 '18 at 22:00

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Let $f_n(x)=nx(1-x)$. Then $J(f_n)=\int_0^{n} e^{-n^{2}(1-2x)^{2}} \, dx$ which tends to $0$ as $n \to \infty$ by Dominated Convergence Theorem. Hence the infimum of $J$ is $0$. Obviously, $0$ is not attained.

Kavi Rama Murthy
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