# 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.

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.

• I have to admit, "smoothing" a function is a pretty foreign concept for me, this is my first sequence in real analysis. Could you expand upon this? Mar 23, 2013 at 22:26
• Define a new function which agrees with the triangle wave except near the peaks. And make sure it is smooth (in fact $C^2$ is enough) Mar 23, 2013 at 22:28

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

• Yes, zero cannot be achieved but what is to say that it is truly the infimum? That is the problem I'm dealing with. Mar 23, 2013 at 22:06
• infimum may not be taken by definition, if it is taken we call it minimum. Mar 23, 2013 at 22:10