Proof that the average of a convex curve is inside the curve.

Question

Let $\boldsymbol{\gamma}(t): [a,b] \rightarrow \mathbb{R}^2$ be closed curve in the plane. The region inside the closed curve is $D$, therefore $\boldsymbol{\gamma} = \partial D$ (boundary of $D$).

Let's say that $D$ is convex, that is $\forall \mathbf{x},\mathbf{y} \in D$ implies that $ \mathbf{x}\lambda + (1-\lambda)\mathbf{y} \in D$ for $\lambda \in [0,1]$.

The average of the boundary of $D$ is: $$\boldsymbol{\mu} = \frac{1}{b-a} \oint_{\partial D} \boldsymbol{\gamma}(t) dt $$

I want to show that $\boldsymbol{\mu} \in D$.

Can anyone help me with this proof, maybe give me a hint on how to start?

Answer 1

You also have to assume that $D$ is closed.

Since $\gamma$ is continuous and $[a,b]$ is compact, one has the following equality: $$\mu=\frac{1}{b-a}\int_a^b\gamma(t)\,\mathrm{d}t=\lim_{n\to+\infty}\underbrace{\frac{1}{n}\sum_{k=1}^n\gamma\underbrace{\left(a+k\frac{b-a}{n}\right)}_{\in D}}_{\textrm{barycenter}\in D},$$ so that $\mu$ is a limit of barycenters of points in $D$, whence a limit of points in $D$ (since $D$ is convex) and finally the average $\mu\in D$ (since $D$ is closed).

Reminders.

1. A barycenter of points in a convex set is itself in the convex set.
2. A limit of points in a closed set is itself in the closed set.