I'm taking an introductory course in the calculus of variations and have this question (From Bruce van Brunt "The calculus of variations" 2.2.5.b):
Let J be the functional defined by $$ J(y)=\int_{-1}^{1} x^4(y')^2 dx $$ Without resorting to the Euler-Lagrange equation prove J cannot have a global minimum in the set $$ S=\{y\in C^2[-1,1]:y(-1)=-1, y(1)=1\} $$
First I tried to show that for every $y_1\in S$ there exists a $y_2\in S$ such that $J(y_2)<J(y_1)$. This would prove J doesn't have a global minimum. One $y_2$ I tried was $y_2=y_1+k(x+1)(x-1)$ for some constant $k$ but it didn't work.
$J(y)>0$ since $y'$ cannot be $0$ everywhere so I also tried to prove that $J(y)$ can become arbitrarily close to $0$, i.e. for all $\epsilon>0$ we have $J(y)<\epsilon$ for some $y\in S$. This would also prove that $J$ has no global minimum. I tried functions of the form $y=x^n$ with $n$ odd but it didn't work since $J$ I got that $J$ tend to infinity as n does.
I'm out of ideas, so any help would be appreciated.
If we consider $f_n(x)=\frac{\tanh(nx)}{\tanh(n)}$ (constructed according to the principles outlined in my comment) we have
$$ \int_{-1}^{1}x^4 f_n'(x)^2\,dx = \frac{2n^2}{\tanh(n)^2}\int_{0}^{1}\frac{x^4}{\cosh(nx)^4}\,dx=\frac{2}{n^3\tanh(n)^2}\int_{0}^{n}\frac{z^4}{\cosh(z)^4}\,dz\leq\frac{2C}{n^3\tanh(n)}$$ with $C=\int_{0}^{+\infty}\frac{z^4}{\cosh(z)^4}\,dz=\frac{7\pi^4}{360}-\frac{\pi^2}{6}<\frac{1}{4}$. It follows that the functional may attain arbitrarily small values, but of course not the value $0$, since $J(y)=0$ implies $y'=0$ a.e. in $(-1,1)$, which contradicts $2=y(1)-y(-1)=\int_{-1}^{1}y'(x)\,dx$.
$f_n(x)$ can be regarded as a $C^2$-version of the piecewise-linear function $g_n(x)$ which equals $nx$ on $\left[-\frac{1}{n},\frac{1}{n}\right]$, $1$ on $\left[\frac{1}{n},1\right]$ and $-1$ on $\left[-1,-\frac{1}{n}\right]$. Indeed we may construct suitable $y$s also by considering the convolution between $g_n(x)$ and well-behaved kernels.