Let $X\subset \Bbb R^2$. We define $V(x)$ to be the set of points in $X$ that $x$ 'can see', i.e. $V(x)=\{y\in X\mid [x,y]\subset X\}$.
The kernel is then $\text{ker}(X)=\{x\in X\mid V(x)=X\}$.
I want to show that the kernel is a convex set.
This is an exercise in a section of my textbook that covers only Helly's and Radon's theorems, so I would imagine they are relevant, but cannot work out how to use them. So I tried to just attack it directly:
If $\text{ker}(X)=\emptyset$ (the space is not star shaped), then this is convex vacuuously. If $|\text{ker}(X)|=1$ then this holds trivially. So say $\text{ker}(X)$ has at least two points $x,y$. We need to show that all the points $tx+(1-t)y$ are in $\text{ker}(X)$ for $t\in [0,1]$. For any other point $z\in X$, we have $[x,z]$ and $[y,z]$ contained in $X$. Then either $x,y,z$ are affine dependent, or $X$ contains the convex hull of $\{x,y,z\}$ in which case $\text{ker}(\text{convex}(\{x,y,z\}))=\text{convex}(\{x,y,z\})$ in which case $tx+(1-t)y$ can see $z$, for each $t\in[0,1]$. For any pair $x,y\in\text{ker}(X)$, we can do this for all $z\in X$, and so for each $t\in[0,1]$ we have that $tx+(1-t)y\in \text{ker}(X)$ so we are done.
I am not really sure if my argument is sound honestly, but maybe the approach is almost there?
Please tell me if my proof is OK, or otherwise give a nicer proof.
If $a,\ b\in {\rm Ker}\ X$ and $z\in X$, then $$ [az],\ [bz]\subset X $$
If $c=ta+(1-t)b,\ 0<t<1$, then assume that $[cz]$ is not in $X$.
That is, there is $x\in (cz)$ not in $X$.
Since $[az]\subset X$, so $[sa+(1-s)z\ b]\subset X$ by assumption on $b$.
For suitable $s$, $[sa+(1-s)z\ b]$ contains $x$.