Show that the infimum of a functional is zero, but this infimum is never achieved. Show that the infimum of the integrals $$\int_0^1(y'^2(x)-1)^2dx$$ among all $y(x)\in C^2[0,1]$ such that $y(0)=y(1)=0$, is zero, but is not achieved by any function in this set.  What I've worked on: Zero is obviously a lower bound for these integrals but I'm currently trying to show that any positive number cannot be a lower bound for these integrals.  I thought had it when I used the function $y=x^\alpha-x$ and then let $\alpha\rightarrow0$, but I realized that these functions do not have continuous second derivatives at zero.  Any ideas or advice would be appreciated.  
 A: Start with functions of the form of a Triangle wave, with slopes $\pm 1$, and smoothen the peaks using very short intervals near each peak. The integrand will be zero except at those short intervals - making the value of the functional very small (as small as desired actually).
Now, to show that $0$ isn't achieved:
Suppose it is, then $y'=\pm1$. it can't take both values because of This theorem. But if if it takes only one (i.e. $y'=+1$ or $y'=-1$) then $y$ is monotonic, and can't satisfy both of the boundary conditions.
A: zero can not be achieved because if so, we should have $ y'(x)^2 = 1 $ so $ y'(x)=1 or -1 $ 
which just one of them can happen, $ y'(x) $ is continuous, and you can solve this equation and get a contradiction
you can try to define functions with corresponding value arbitrary close to zero by considering functions which are equal to $x$ in the interval $ [0,1/2 - \epsilon] $ and are equal to $ 1/2 - x$ in the interval $ [1/2 + \epsilon , 1] $ and you should define them appropriately in the remaining interval
