Does there exist a (onedimensional) integral functional of calculus of variations (with $f$ finite everywhere) $$ F(y)=\int_a^b f(t,y(t),y'(t))\,dt $$ such that $$ \inf_{y\in Lip([a,b])}F(y)<\inf_{y\in C^1([a,b])}F(y) $$ that is, it shows the Lavrentiev phenomenon between $C^1$ and Lipschitz.
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$\begingroup$ Is $f$ allowed to be unbounded discontinuous (like $1/y^2$ corrected to $0$ when $y=0$, say)? $\endgroup$– fedjaCommented Dec 12, 2017 at 14:47
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$\begingroup$ yes, if the Lagrangian $f=f(x,y,\xi)$ is continuous or Caratheodory and bounded on bounded sets the two inf are the same. $\endgroup$– Carlo MantegazzaCommented Dec 12, 2017 at 16:01
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Then trivially yes. Just take $F(y)=1+\sum_{q\in \mathbb Q}\frac{a_q}{|q-y|^2}$ with $a_q>0$ such that the series converges a.e. Then change all $+\infty$ values of $F$ to $1$. Now take $[a,b]=[-1,1]$ and define $$ f(t,y,\xi) = \begin{cases} 0 &\text{if }y=|t|\\ F(y) &\text{otherwise} \end {cases}. $$ A Lipschitz function can just stay on the safe path $y=|t|$ and pay $0$ price for the trip, but a $C^1$ function will have to deviate from this path and either have $y\ne |t|$, $y'\ne 0$ at at least one point, which is enough to blow the integral up to $+\infty$, or stay constant, which gives the cost of at least $2$.
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$\begingroup$ Since I will be possibly put this example in something that could be published, if you send me (in private) your name I could give you proper credit for that. $\endgroup$ Commented Dec 13, 2017 at 11:20
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$\begingroup$ @CarloMantegazza Just quote MO. They'll benefit from the credit more than I ;-) $\endgroup$– fedjaCommented Dec 13, 2017 at 15:23