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

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!

• As long as the curvature of $\gamma$ is less than something of the order $1/r$, and there are no "overlaps", the area of the "sausage" $\gamma_{+r}$ will be equal to $2 r L + \pi r^2$, and it is intuitively clear this is the maximal value. I guess this is standard, I vaguely remember having read that somewhere, but unfortunately I do not have a reference. Apr 13, 2021 at 21:45