Isoperimetric type inequality in $\mathbb{R}^2$

Question

Fix $L \in (0,\infty)$ and consider $\mathcal{C}_L$ defined as follows: \begin{align*} \mathcal{C}_L := \{ \gamma:[0,1] \rightarrow \mathbb{R}^2 |~ \gamma \text{ is smooth and length($\gamma$)$=L$ }\}. \end{align*}

I am interested in the "blow-up" of $\gamma$, denoted $\gamma_{+r}$, defined as follows: For any set $S \subseteq \mathbb{R}^2$ and $r>0$ \begin{align*} S_{+r} := \cup_{z\in S}(z+r\mathbb{D}), \end{align*} where $\mathbb{D}$ is the unit disc in $\mathbb{R}^2$ which is centred at the origin. So $\gamma_{+r}$ is a bounded open set in $\mathbb{R}^2$. My question is for which $\gamma \in \mathcal{C}_L$ is $m(\gamma_{+r})$ maximised? Here $m(\cdot)$ is the Lebesgue measure in $\mathbb{R}^2$. I feel that it should be maximised by the line segment with length $L$.

If this is a version of some well known result, please do indicate it.

The reason for this title is that sometimes the isoperimetric inequality in $\mathbb{R}^2$ is stated as follows: For any Borel subset $A \subseteq \mathbb{R}^2$ with $m(A) < \infty$ and for every $\epsilon >0$, we have $m(A_{+\epsilon}) \geq m(B_{+\epsilon})$. Here $B$ is a Euclidean ball with $m(A) = m(B)$.

Thanks!

Answer 1

The problem of determining the volume of the tubular neighborhood of a submanifold of Euclidean spaces was studied by Hotelling and Weyl (and then others). See https://www.jstor.org/stable/2371513?seq=1

The amazing result of Weyl is that, as long as the tubular neighborhood has no "overlaps" (in the sense of non-injective normal exponential map) the volume of the tubular neighborhood is independent of (isometric) embedding. (So in the one dimensional case the maximizer is non-unique, since all embeddings are isometric to each other.)

His theorem applies when the submanifold is closed (without boundary); but in your setting of a curve, the end-caps can be explicitly added.