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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.

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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.

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  • $\begingroup$ 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? $\endgroup$
    – KarlG
    Mar 23, 2013 at 22:26
  • $\begingroup$ 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) $\endgroup$
    – user1337
    Mar 23, 2013 at 22:28
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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

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

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