# Strict inequality of the H1 seminorm of a curve emerging from a convex set and its convex projection

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).

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.
• The first equality uses the following fact: if $f \le g$ and $\int f = \int g$, then $f = g$. – gerw Jun 11 at 13:34