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Denote by $\mathcal L$ the set of continuously differentiable real valued functions on $[0, 1]$ with Lipschitz continuous derivative. Does there exist a Borel measurable function $ f: [0, 1] \times \mathbb R \to \mathbb [0, \infty) $ such that

$$\inf_{g \in \mathcal L} \int_{0}^{1} f(t, g(t)) \ dt < \inf_{h \in C^2([0, 1])} \ \int_{0}^{1} f(t, h(t)) \ dt?$$

Note: Here the integrals are allowed to take the value $+\infty$.

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  • $\begingroup$ Don't you want to include $g'$ in the arguments of f? $\endgroup$
    – Alexandre Eremenko
    Commented Jul 23, 2021 at 1:54
  • $\begingroup$ Ah, I intend for $f$ to be independent of $g’$. If we include $g’$ the answer is yes, but the independent case seems much more subtle. $\endgroup$
    – Nate River
    Commented Jul 23, 2021 at 5:24
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    $\begingroup$ Let $u : [0,1] \to \mathbf{R}$ be a function in $C^{1,1} \setminus C^2$. It seems to me that modifying another answer by changing the sentence to 'define $f(x,y,\xi) = 0$ if $y = u(x)$ and $F(y)$ otherwise' is valid. The 'safe path' that this answer refers to would now correspond to the graph of $u$. As any $C^2$-regular function deviates from this, it seems to me that $\int_0^1 f(x,g(x)) \, \mathrm{d} x = + \infty$ for every $g \in C^2$. $\endgroup$
    – Leo Moos
    Commented Jul 24, 2021 at 4:24
  • $\begingroup$ Ah, you are right of course. $\endgroup$
    – Nate River
    Commented Jul 24, 2021 at 4:53

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Edited.

Let $g_0^\prime(x)=|x-1/2|,\; g_0(x)=\int_0^xg_1(t)dt,\; 0\leq x\leq1$. Then $g_0$ has continuus derivative, namely $g_0^\prime$, which is Lipschitz, but the second derivative is discontinuous. Let $L=\{(x,y):y=g_0(x)\}$ be the graph of $g_0$.

Now define $f(x,y)=0$ when $0\leq x\leq 1, y=g_0(x)$ and $$f(x,y)=(\mathrm{dist}(x,y),L)^{-3}+1$$ otherwise. Evidently $f$ is measurable, and $\int_0^1 f(x,g_0(x))dx=0$, while $\int_0^1f(x,g(x))dx>c$ for every $C^2$ function $g$, where $c$ is an absolute constant.

Of course, the last fact requires an accurate proof, but on the other hand, it seems evident. Moreover, one can replace $3$ in the exponent by some larger constant, to make it more evident. The idea is that when the graph of $g$ is too close to $L$, the integral is large.

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  • $\begingroup$ Could you clarify what you intend for $f$ - there seems to be a typo? Also, the last claim seems very strong, could you provide some details? It seems a bit counterintuitive given that a $C^1$ can be approximated by $C^2$ functions. (Personally I thought that the answer given to a related question ought to apply here too.) $\endgroup$
    – Leo Moos
    Commented Jul 23, 2021 at 20:42
  • $\begingroup$ @Leo Moos - this would apply if there was a $g’$ in the argument for f, but the case where it depends only on $t, g$ is much more subtle. $\endgroup$
    – Nate River
    Commented Jul 23, 2021 at 23:14
  • $\begingroup$ @NateRiver I'm not sure I follow - the functional in the linked answer doesn't seem to depend on the derivative. $\endgroup$
    – Leo Moos
    Commented Jul 23, 2021 at 23:31
  • $\begingroup$ Oh what I meant was that if $g’$ was allowed here, we could emulate that construction with $g$ replaced by $g’$ and the proof would hold verbatim. But because we have only $g$ here, and the “regularity gap” is between $C^1$ + Lipschitz and $C^2$, the same proof won’t work. $\endgroup$
    – Nate River
    Commented Jul 23, 2021 at 23:34

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