Let $r:[0,1]\to {ℝ}^2$ be a parametrisation of $C$. As the shape is closed, we have $r(0)=r(1)$.

For $\theta \in S^1$, the projection map $P_{\theta }$ is defined by:

This is clearly continuous in $\theta$. Thus $P_{\theta }(C)$ is the image of the continuous function:

The domain of this function is $S^1\times [0,1]$. It therefore lives in the space of continuous functions $C(S^1\times [0,1],ℝ)$. The key property of this space that we want to use is that there is an isomorphism:

Namely, the function:

is continuous as a map from $S^1$ to $C([0,1],ℝ)$.

This is a standard result in functional analysis (with overtones of category theory). For completeness, we shall prove it at the end.

The functions $f\mapsto \max (f)$ and $f\mapsto \min (f)$ are continuous on $C([0,1],ℝ)$ almost by definition:

and similarly for the minimum.

This means that the following compositions are continuous:

whence, finally, the following map is continuous:

This is precisely $\theta \mapsto d_{\theta }(C)$, and so this function is continuous as required.

The map is defined in the obvious manner: given $f:S^1\times [0,1]\to ℝ$ we define $\hat{f}:S^1\to C([0,1],ℝ)$ by ${\hat{f}}_{\theta }(t)=f(\theta ,t)$. The inverse is: given $g:S^1\to C([0,1],ℝ)$, define $\check{g}:S^1\times [0,1]\to ℝ$ by $\check{g}(\theta ,t)=g_{\theta }(t)$.

Assuming that these are well-defined, they are clearly inverse to each other.

We then need to prove the following:

1. For $\theta \in S^1$, $t\mapsto \hat{f}(\theta )(t)$ is continuous.

2. The map $\check{g}$ is continuous.

3. The maps $f\mapsto \hat{f}$ and $g\mapsto \check{g}$ are continuous. Once it is established that they are well-defined, it is enough to establish that, say, $f\mapsto \hat{f}$ is an isometry.

As everything in sight is a metric space, we will use the characterisation of continuity as that of taking convergent sequences to convergent sequences.

1. Fix $\theta \in S^1$ and $f\in C(S^1\times [0,1],ℝ)$. Let $(t_n)\to t$ be a convergent sequence in $[0,1]$.

   Then the sequence ${\tau }_n=(\theta ,t_n)$ is convergent in $S^1\times [0,1]$ with limit $\tau =(\theta ,t)$. As $f$ is continuous, $f({\tau }_n)\to f(\tau )$.

   Then ${\hat{f}}_{\theta }(t_n)=f(\theta ,t_n)=f({\tau }_n)$ and so ${\hat{f}}_{\theta }(t_n)\to f(\tau )=f(\theta ,t)={\hat{f}}_{\theta }(t)$. Hence ${\hat{f}}_{\theta }$ takes convergent sequences to convergent sequences, whence is continuous.

2. Fix $g\in C(S^1,C([0,1],ℝ))$. Let $({\tau }_n)\to \tau$ be a convergent sequence in $S^1\times [0,1]$.

   Write ${\tau }_n=({\theta }_n,t_n)$ and $\tau =(\theta ,t)$. Then $({\theta }_n)\to \theta$ and $(t_n)\to t$.

   Now $\check{g}({\theta }_n,t_n)=g_{{\theta }_n}(t_n)$. Since $({\theta }_n)\to \theta$, and $g$ is continuous, $(g_{{\theta }_n})\to g_{\theta }$. Convergence in $C([0,1],ℝ)$ is uniform convergence, and therefore we have that:

   $$\begin{array}{rl}|g_{{\theta }_n}(t_n)-g_{\theta }(t)|& =|g_{{\theta }_n}(t_n)-g_{\theta }(t_n)+g_{\theta }(t_n)-g_{\theta }(t)|\\ & \le |g_{{\theta }_n}(t_n)-g_{\theta }(t_n)|+|g_{\theta }(t_n)-g_{\theta }(t)|\\ & \le \parallel g_{{\theta }_n}-g_{\theta }\parallel +|g_{\theta }(t_n)-g_{\theta }(t)|\end{array}$$

   Therefore as $(g_{{\theta }_n})\to g_{\theta }$ and $(g_{\theta }(t_n))\to g_{\theta }(t)$ we have that $(g_{{\theta }_n}(t_n))\to g_{\theta }(t)$. Hence $\check{g}$ is continuous.

3. To show that $f\mapsto \hat{f}$ is an isometry, we just need to consider the definitions of the norms on each side.

   $$\begin{array}{rl}\parallel \hat{f}\parallel & =\sup \{\parallel \hat{f}(\theta )\parallel :\theta \in S^1\}\\ & =\sup \left\{\sup \right\{|\hat{f}(\theta )(t)|:t\in [0,1]\}:\theta \in S^1\}\\ & =\sup \left\{\sup \right\{|f(\theta ,t)|:t\in [0,1]\}:\theta \in S^1\}\\ & =\sup \left\{\right|f(\theta ,t)|:t\in [0,1],\theta \in S^1\}\\ & =\parallel f\parallel .\end{array}$$

   Hence $f\mapsto \hat{f}$ is an isometry and so as it has an inverse it is an isomorphism.