Convexity of space of integrals of homotopic paths

Question

Let $B$ be a Banach space. Consider a set $Q \subset B$, a path $\gamma_0:[0,1] \to Q$ and the set of paths $\Gamma_0$ of paths $[0,1] \to Q$ homotopic (with ends fixed) to the path $\gamma_0$. Let $l(\Gamma_0)$ be the set of values of integrals $\int_0^1 \gamma(t)dt \in B, \quad \gamma \in \Gamma_0$. Can we show that the set $l(\Gamma_0)$ is convex?

Answer 1

Assuming $Q$ is a subspace, then indeed we can: Fix $t\in(0,1)$, and let $a_1=\int_0^1\gamma_1(s)ds$ and $a_2=\int_0^1\gamma_2(s)$ be elements of $l(\Gamma_0)$. Define the following homotopies: $H_1(s,y)=((t-1)y+1)\gamma_1(s)$, $H_2(s,y)=(-ty+1)\gamma_2(s)$, and $H_3(s,y)=(1-y)t\gamma_1(s)+y(1-t)\gamma_2(s)$.

This shows that $t\gamma_1(s)+(1-t)\gamma_2(s)\in\Gamma_0$. So, we have:

$ta_1+(1-t)a_2=\int_0^1 (t\gamma_1(s)+(1-t)\gamma_2(s))ds$, and therefore $ta_1+(1-t)a_2\in l(\Gamma_0)$.

Answer 2

I will show that the result is false without further assumptions on $Q$. First I'll try to prove the result to get a better understanding of what goes wrong. To this end, let $\theta \in (0,1)$, $\gamma_1$, $\gamma_2 \in \Gamma_0$, and consider $$\theta \int_0^1 \gamma_1(t)\,dt + (1 - \theta) \int_0^1 \gamma_2(t)\,dt = \int_0^1\theta\gamma_1(t) + (1 - \theta)\gamma_2(t)\,dt.$$

To prove that $l(\Gamma_0)$ is convex the reasonable thing to try is proving that $\gamma(t) = \theta\gamma_1(t) + (1 - \theta)\gamma_2(t) \in \Gamma_0$.

• $\gamma$ has the correct fixed ends.
• $\gamma$ is homotopic to $\gamma_0$. Indeed if $F_i\colon [0,1]^2 \to B$ are the continuous maps that give the homotopy to $\gamma_0$ for $\gamma_i$, $i = 1,2$, then $G(t,s) = \theta F_1(t,s) + (1 - \theta)F_2(t,s)$ has all the required properties.
• Here we fail! $\gamma$ does not necessarily take values in $Q$, so we cannot conclude that $\gamma \in \Gamma_0$.

This does not prove that $l(\Gamma_0)$ is not convex, but gives an idea to cook up a counterexample.

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Consider $B = \mathbb{R}^3$ and let $Q$ be upper unit hemisphere together with the spherical arc that joins $(1,0,0)$, the south pole and $(-1,0,0)$. (think of something like a helmet where the arc is the chin strap). If you take the convex combination with $\theta = \frac12$ of this arc and the one with the same ends that passes through the north pole the value you get for the integral is $(0,0,0)$. You won't get the same value considering only curves with values in $Q$, same fixed ends and homotopic to the arc the passes through the north pole. It is worth noticing that this counterexample allows for homotopies to take values in $Q^c$ during the intermediate deformations from a curve into another.