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Consider $u:S^1\times (0,T)\to \mathbb R$ a $C^1$ function, and define $$v(t)=\max_{p\in S^1} u(p,t)$$ Prove that $v$ is locally Lipschitz and deduce that $$\frac {dv}{dt}(t_0)=\frac{\partial u}{\partial t}(p_0,t_0)$$ where $p_0$ is any maximum point for $u(\cdot, t_0)$.

For point one my idea was to prove that at least one maximum point at height $t$ (viewing the domain as a vertical cylinder) has a neighbourhood which contains a differentiable curve made of maximum points for $u$. Therefore, in that neighbourhood (i.e. for $t$ near to $t_0$) one could use that $u$ is $C^1$. But is that true? And is this really the best way?

For point two, I know that locally Lipschitz implies absolutely continuous which implies a.e. differentiable. But how could I deduce the (however intuitive) formula for the derivative?

Furthermore, I'm quite sure that the result absolutely continuous $\Longrightarrow$ differentiable a.e. is more advanced than the level at which the problem was given. Isn't there a simpler way for this special case?

Thank you in advance.

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  • $\begingroup$ Thank you: can you just explain me why do you transform $u(p_t,t)$ in $u(p_{t_0},t)$? $\endgroup$ – W. Rether Aug 19 '16 at 21:37
  • $\begingroup$ I think you can show (by contradiction) that as $t\to t_0,\ p_t\to p_{t_{0}}$ so $(p_t,t)\to (p_{t_{0}},t_0)$. Then, look at $\left | \frac{u(p_{t_0+t},t_{0}+t)-u(p_{t_{0}},t_{0})}{t-t_0} \right |$, note that $u$ is $C^{1}$, and apply to this a triangle inequality/MVT argument to prove both differentiabilty and local Lipschitz (by restricting to a compact set containing $t_0$). $\endgroup$ – Matematleta Aug 19 '16 at 22:13
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Let $t_1, t_2 \in (0,1)$. Assume $v(t_1) \ge v(t_2)$ (If not, switch $t_1$ and $t_2$). Let $x_1\in \mathbb S^1$ so that $v(t_1) = u(x_1, t_1)$. Then

$$|v(t_1) - v(t_2)| = v(t_1) - v(t_2) \le u(x_1, t_1) - u(x_1, t_2)\le M |t_1 - t_2|,$$

where $$M=\sup_{(x,t) \in \mathbb S^1 \times [t_1 , t_2]} |\partial _t u|.$$

Thus $v$ is locally Lipschitz.

Let $t_0$ be such that $v$ is differentiable and $x_0 \in \mathbb S^1$ so that $v(t_0) = u(x_0, t_0)$. Then

$$v'(t_0) = \lim_{t\to t_0^+} \frac{v(t) - v(t_0)}{t-t_0} \ge \lim_{t\to t_0} \frac{u(x_0, t)-u(x_0, t_0)}{t-t_0} = \frac{\partial u}{\partial t}(x_0, t_0)$$

similarly, by considering $t\to t_0^-$, we have $v'(t_0) \le \frac{\partial u}{\partial t}(x_0, t_0)$. Thus we have the equality.

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  • $\begingroup$ Thank you very much. The fact that $v$ is differentiable a.e. (which seems to be used in your proof to equate the two limits, right?) comes from the absolute-continuity argument or from a simpler consideration? $\endgroup$ – W. Rether Aug 20 '16 at 11:02
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    $\begingroup$ @W.Rether Yes, it does not come from a simpler observation. $\endgroup$ – user99914 Aug 20 '16 at 16:44

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