Integration of a continuous function whose values lie on the boundary of convex set. In the proof of a specific holonomic approximation, I encountered the following claim:

Let $\sigma\colon[0,1]\rightarrow\mathbb{S}^1$ be non-constant and continuous, then $\displaystyle\int_0^1\sigma(u)\,\mathrm{d}u$ lies in the interior of the unit disk.

While this statement is intuitive, I struggle to establish it formally ; here is what I have done:
Proof. Since $\sigma$ is continuous and $[0,1]$ is compact, one has:
$$I:=\int_0^1\sigma(u)\,\mathrm{d}u=\lim_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n-1}\sigma\left(\frac{k}{n}\right)\tag{1}.$$
Besides, since $\sigma$ is non-constant continuous and since $\displaystyle\left\{\frac{k}{n};n\in\mathbb{N};k\in\{0,\cdots,n-1\}\right\}$ is dense in $[0,1]$, there exists $n\in\mathbb{N}\setminus\{0\}$ and $k,k'\in\{0,\cdots,n-1\}$ such that: $$\sigma\left(\frac{k}{n}\right)\neq\sigma\left(\frac{k'}{n}\right)\tag{2}.$$
Which motivates me to proove the following lemma:

Lemma. Let $\{z_1,\cdots,z_n\}$ be a collection of points in $\mathbb{S}^1$ with at least two distincts elements, then any barycenter of $z_1,\cdots,z_n$ lie in the interior of the unit disk.

Since for all $N\geqslant n$, $\displaystyle\frac{k}{n},\frac{k'}{n}\in\left\{\frac{\ell}{N};\ell\in\{0,\cdots,N-1\}\right\}$, according to $(1),(2)$ and the lemma, $I$ is a limit of points being in the interior of the unit disk. $\Box$
However, I do not get why the elements involved in $(1)$ cannot accumulate on the boundary of the unit disk.
I was also wondering whether or not the following generalization is true:

Proposition. Let $C\subseteq\mathbb{R}^n$ be convex set and let $\sigma\colon[0,1]\rightarrow\partial C$ be non-constant and continuous, then $\displaystyle\int_0^1\sigma(u)\,\mathrm{d}u$ lies in the interior of $C$.

I am preferably looking for a proof that can be adapted to the last proposition, but any enlightenment would be greatly appreciated!
 A: What you need for the conclusion is strict convexity of $C$, the boundary must not contain any convex subset with more than one point.
Then take a supporting hyperplane $H = \{ x : \lambda(x) = c\}$ at $x_0 \in \partial C$, for a linear functional $\lambda \colon \mathbb{R}^n \to \mathbb{R}$ and some $c \in \mathbb{R}$. Assume that $C$ lies in the half-space $\{ x : \lambda(x) \leqslant c\}$. If $C$ is strictly convex, then $H \cap C = H\cap \partial C = \{x_0\}$, and hence $\lambda\circ\sigma \leqslant c$, and since $\sigma$ is not constant, $\lambda(\sigma(t_0)) < c$ for some $t_0 \in [0,1]$. By continuity, it follows that $\lambda \circ \sigma < c$ on some non-degenerate subinterval of $[0,1]$, and therefore
$$\lambda\Biggl(\int_0^1 \sigma(t)\,dt\Biggr) = \int_0^1 \lambda(\sigma(t))\,dt < c,$$
which shows
$$\int_0^1 \sigma(t)\,dt \neq x_0.$$
Since $x_0\in \partial C$ was arbitrary, it follows that the integral lies in the interior of $C$.
If $\partial C$ contains a convex set $K$ with more than one point, take a non-constant $\sigma$ with image contained in $K$, then also $\int_0^1 \sigma(t)\,dt\in K \subset \partial C$.
A: This can be read as Schwarz' inequality $\lvert \langle \sigma, \mathbb{1} \rangle \rvert \leq \lVert \sigma \rVert \cdot \lVert \mathbb{1} \rVert = 1$ with equality if and only if $\sigma$ is a multiple of $\mathbb{1}$, i.e. constant. Your proposition is false if $\partial C$ contains a line segment.
