The task is to find the sequence of functions $\lbrace f_n \rbrace _{n\geqslant1} \subset \mathbb{K}\mathbb{C}^{1} \left[ 0;1 \right]$ (where $\mathbb{K}\mathbb{C}^{1} \left[ 0;1 \right]$ is a set of piecewise continuously differentiable on $\left[ 0;1 \right]$ functions) that satisfies three conditions:

a)$\underset{t\in [0;1]}{\max} \vert f_n(t) \vert \to 0,\, n \to \infty$;

b)$f_n(0)=f_n(1)=0,\, n \geqslant 1$ ;

c)$\int \limits_{0}^{1} \left( 3(f_n'(t))^2 + (f_n'(t))^3 \right)dt < 0$ for all $n \geqslant 1$.

I tried some simple attempts like $f_n(t)=\frac{t(1-t)}{n}$ or $f_n(t)=-t(t-1)I_{[0;\frac{1}{n}]}(t)$, but none of them where suitable (in particular, indicator function "spoils" $f_n$ in the sense of belonging to the class $\mathbb{K}\mathbb{C}^{1} \left[ 0;1 \right]$).

I encountered this task when I tried to show that for the function $I(x)=\int \limits_{0}^{1} \left(x'(t)\right)^3 dt \to \text{extr},\, x(0)=0,\, x(1)=1$ there are no global extremum on $\left[ 0;1 \right]$ (actually, my lecturer of Calculus of variations mentioned that this is in fact Hilbert's counterexample of something). Presenting wanted sequence will prove that this is right (I know that there are other ways of showing this result but I need to find at least one of the sought-for sequences).

I would really appreciate if you can help me to find the sequence I'm looking for.

  • $\begingroup$ NickM Please let me know if my edit to $(a)$ is what you intended. All I did was write $\max$ ({\max}), instead of $max$ $\endgroup$ – amWhy Jun 9 '18 at 23:26
  • $\begingroup$ amWhy Yes, that's what I meant. Thanks a lot. $\endgroup$ – NickM Jun 9 '18 at 23:30

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