0
$\begingroup$

Let $\gamma(t) \in W^{1,2}([0,1]; \mathbb{R}^d)$ such that $\gamma(0) \in C$, with $C \subset \mathbb{R}^d$ some closed convex set, and $ \gamma(t) \not \in C, \ \forall t\in (0,1]$. Furthermore we denote $\Pi_C$ the orthogonal projection into this convex set.

How could I prove that that $$ \int_0^1 \lVert \frac{d}{dt} \left( \Pi_C \gamma(t) \right) \rVert^2 < \int_0^1 \lVert \dot \gamma(t) \rVert^2 $$

The importance is in the strictness of this inequality. The inequality can be easily done using the 1-Lipschitzness of the projection into convex sets and taking the a.e. point-wise limit of the quotients $\frac{ \gamma(t +h) - \gamma(t)}{h}$ with the limit existing by dominated convergence, since the quotient is bounded by the 1/2-Holder continuity of any element of $W^{1,2}([0,1]; \mathbb{R}^d)$.

Somehow, for the strict inequality, it is needed to ensure that once projecting, in some neighborhood of the initial time (or exit moment) the projected points will be strictly closer to each other, something that not necessarily happens for any convex set or curve (take a horizontal curve, hovering over a horizontal flat convex set).

$\endgroup$
1
$\begingroup$

Here is an idea. If you would have equality, then you get $$ \| \frac{\mathrm d}{\mathrm d t} \Pi_C\gamma (t) \| = \|\dot\gamma(t)\|$$ for a.a. $t$. Moreover, the projection is firmly non-expansive. Hence, $$ \| \gamma(t_2) - \gamma(t_1) \|^2 \ge \| \Pi_C \gamma(t_2) - \Pi_C \gamma(t_1) \|^2 + \| \Pi_C \gamma(t_2) - \gamma(t_2) - \Pi_C \gamma(t_1) + \gamma(t_1)\|^2 $$ for all $t_1, t_2 \in (0,1)$. This inequality implies $$ \|\dot\gamma(t)\|^2 \ge \|\frac{\mathrm{d}}{\mathrm{d}t} \Pi_C \gamma(t) \|^2 + \|\frac{\mathrm{d}}{\mathrm{d}t} \Pi_C \gamma(t) - \dot\gamma(t)\|^2$$ for a.a. $t$. Together with the first equality, this shows $$\|\frac{\mathrm{d}}{\mathrm{d}t} \Pi_C \gamma(t) - \dot\gamma(t)\| = 0$$ for a.a. $t \in (0,1)$. Hence, $\gamma(t) \in C$ which is a contradiction to your assumptions.

$\endgroup$
  • $\begingroup$ This is quite rough, as I have no clues on how the projection behaves after differentiating it. $\endgroup$ – pancho Jun 11 at 6:40
  • $\begingroup$ One can use the fact that the projection is firmly non-expansive to arrive at the desired conclusion. $\endgroup$ – gerw Jun 11 at 11:01
  • $\begingroup$ Thank you ! I want to point out that I'm unsure how you get the first pointwise equality for all t, as a priori just integrals are equal. Albeit, this is irrevelant because at your 3rd equality, instead of following up with pointwise evaluation, you could integrate and get rid of the same terms, then you would have the integral of a norm equal to zero and then conclude. $\endgroup$ – pancho Jun 11 at 12:16
  • 1
    $\begingroup$ The first equality uses the following fact: if $f \le g$ and $\int f = \int g$, then $f = g$. $\endgroup$ – gerw Jun 11 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.