convex curve as boundary of a convex set

Question

In short, my question is to know if the following statement is true, and how to show it :

Theorem Let $\gamma$ be a closed simple $C^2$ convex curve in $\mathbb{R}^2$. We denote $\Gamma$ its image. Then $\Gamma$ is the boundary of a convex set.

Question Is it true ? How to show it ?

Here is how I try to show this theorem. For every $p\in\Gamma$ we define $H_p$ as being the the (closed) half plane delimited by the tangent to $\gamma$ at $p$ and containing $\Gamma$. Now let $D$ be the intersection of all these half-planes. As intersection of convex sets, $D$ is convex.

I try to show that $\partial D=\Gamma$. Using compactness I am able to show that $\Gamma\subset \partial D$, but I am unable to show the converse.

question Is $D$ the convex hull of $\Gamma$ ? I guess yes, but I failed to prove.

question If $p\in \Gamma$, is it true that the tangent to $\Gamma$ at $p$ intersects $\Gamma$ only at $p$ itself ? I guess not because a convex curve is allowed to have some affine parts.

EDIT

I guess I found a solution. It that correct ?

Let $p\in \partial D$. Then there exists $ q\in\Gamma$ such that $ p\in l_q$. Here $l_q$ is the tangent to $\Gamma$ at $ q$. If $p=q$ we are done.

If $p\neq q$, notice that $ D$ being close, we have $ p,q\in D$ and thus the segment $[p,q]$ is a subset of $D$. The intersection $[q,p]\cap \Gamma$ being closed, one can pick $r$, the last point of $[q,p]$ being on $ \Gamma$.

We parametrize $\gamma$ starting at $r$, so that $\gamma(0)=r$ and for $t$ in a small interval $I=\mathopen] 0 , \epsilon \mathclose]$ we have $\gamma(t)\notin[q,p]$.

Suppose that for every $s\in\gamma(I)$ we have $l_s=l_q$. In that case the curve $\Gamma$ remains on $[p,q]$ after $r$ which is a contradiction. Thus there exists $s\in \Gamma$ that is arbitrarily close to $r$, but with $ l_s\neq l_q$.

Warning This point has to be much clarified (and as it, it is wrong)

By continuity of $ \gamma$ and $\gamma'$, the lines $l_s$ and $l_q$ can be made arbitrarily close (in the sense that the coefficients of their equation $ax+by+x=0$ can be made arbitrarily close). So the intersection point $ l_s\cap l_q$ is as close as $r$ as I want, and in particular, there exists $s\in \Gamma$ such that $l_s\cap l_q$ lies between $r$ and $p$. This contradicts the fact that $r$ and $p$ are both in $D$, because $D$ is contained in the half-plane $H_s$ delimited by the line $l_s$.

EDIT For who can be interested, I wrote a full solution here, in the French part under the title "Enveloppe convexe".

Answer 1

(I) $c$ is convex curve if curve lies in one half plane wrt tangent line. If $c$ is a simple closed convex curve then $c$ bounds a convex set $K$ : For $p,\ q\in K$, if $l$ is a line segment from $p$ to $q$ and if $l\subset K$, then we are done\

If not, there exists $r\in l$ s.t. $r$ is not in $K$. Then WLOG we have $$ r',\ r''\in l\cap \partial K,\ pr'<pr''$$ and $ pr',\ qr''\subset K$ and $r'r'' -\{ r',r''\}$ is not in $K$

(1) If $l$ is not in $T_{r'}\partial K$, since $c$ is in one half plane $P$ wrt $T_{r'} \partial K$ then $$ c\subset P,\ r''\in P $$

Hence $K\subset P$ so that it is contradiction to the fact that $p$ is not in $P$ (Here $l$ meets $T_{r' } \partial K$ transversally).

(2) If $l$ is in $T_{r'}\partial K$, note that arc $r'r''$ in $c$ is not in $K$ Hence if $r$ is closed to $r'$ then $T_r\partial K$ meets $T_{r'}\partial K$ transversally That is $T_r\partial K$ meets $l$ transversally

(II) Convex hull $C$ of $\Gamma$ is a smallest convex set containing $\Gamma$ Note that since $K$ is convex so $C\subset K$ If $x\in K-C$, then $x$ is interior point of $K$. Note that if $l_p$ is a line segment from $x$ to $p\in \partial K$, then $l_p$ is in $K$ And there exists $-p\in \partial K$ s.t. $l_p\cap l_{-p}$ is line segment. Hence $x\in p(-p)$ So $x\in C$.

Proof of $D=K $ : Since $K$ is convex hull and $D$ is convex containing $\Gamma$, so $K\subset D$. If $x\in D-K$, there exists $h\in K$ s.t. $$ d(h,x)=d(K,x) $$

Since $D$ is in a side of tangent line of $\gamma$ at $h$, so it is contradiction.