I can't solve this problem from my calculus of variations class. We would like to find a minimizing sequence $u_n$

Question

Let $f \in L^1(\mathbb{R})$ be a continous and positive function, $f(\xi)>0$ for all $\xi \in \mathbb{R}$. Prove that the functional $F:X\to\mathbb{R}^+\cup\{\infty\}$ $$F(u)=\int_0^1 f(u'(x))dx$$ has no minimum on $X=\{u\in Lip([0,1]):u(0)=u(1)=0\}$.

Answer 1

Since $f$ is continuous and integrable, it vanishes at infinity. Now, if we take the hat functions $u_n=n-2n|x-\frac{1}{2}|$, which are Lipschitz and satisfy $u_n(0)=u_n(1)=0$, we can evaluate $$F(u_n)=\int_0^1 f(u_n’(x))\,dx=\int_0^{1/2} f(2n)\,dx+\int_{1/2}^1 f(-2n)\,dx=\frac{1}{2}f(2n)+\frac{1}{2}f(-2n).$$ Use this to show that the infimum of your variational problem is zero, which is not attained since $f$ is strictly positive.