Definition of a Convex Body This is the definition as given in my lectures:

A non-empty subset $K$ of a vector space $V$ is convex if for all $x,y\in K$ and all $\lambda\in[0,1]$,
  $$\lambda x + (1-\lambda)y\in K. $$
  That is, $K$ contains all the line segments joining points in $K$.

However, I want to know if it is sufficient to check only the boundary points of $K$. So, suppose that $K$ is a compact and non-empty subset of a vector space $V$. Then, is it true that

For all $x,y\in \partial K$ and all $\lambda\in[0,1]$,
  $\lambda x + (1-\lambda)y\in K \implies K$ is convex?

 A: Let us assume that a subset $K \subseteq V$ (where $V$ is a normed vector space) satisfies $\lambda x + (1 - \lambda) y \in K$ for all $x,y \in \partial K$ and all $\lambda \in [0,1]$ (in particular, $K$ is closed). We want to show that $K$ is convex. For this, let $x,y \in K$ be given and assume that $t_0x + (1 - t_0)y$ is not contained in $K$ for some $t_0 \in [0,1]$. Set $$\lambda_x  = \sup \{t \in [0,1] \:|\: t' x + (1-t') y \in K \text{ for all } 0 \leq t' \leq t\}$$ as well as 
$$\lambda_y  = \inf \{t \in [0,1] \:|\: t' x + (1-t') y \in K \text{ for all } t \leq t' \leq 1\}.$$
Then we have $\lambda_x < t_0 < \lambda_y$ and $\lambda_x x + (1-\lambda_x) y$ and $\lambda_y x + (1 - \lambda_y) y$ both lie in $\partial K$ (otherwise, we could find a small ball around these points contained in $K$ and as balls are convex this would contradict the definition of $\lambda_x$ and $\lambda_y$). But by our assumption $(t \lambda_x + (1 - t) \lambda_y) x + ((1-t) \lambda_x + t \lambda_y) y$ is contained in $K$  for all $t \in [0,1]$ and choosing $t$ such that $(t \lambda_x + (1 - t) \lambda_y) = t_0$ gives the desired contradiction.
A: Lemma: $K$ is convex $\Leftrightarrow$ $\forall$ line $L\colon K\cap L$ is convex.
Proof: $\Rightarrow$ Intersections of convex sets are convex.
$\Leftarrow$ By definition: $\forall z_1,z_2\in K$ take the line $L$ through $z_1,z_2$, so $z_1,z_2\in K\cap L$, hence, the segment $[z_1,z_2]\subset K\cap L\subset K$. $\blacksquare$

Now for a compact set $K$
$$
(\forall x,y\in\partial K\Rightarrow [x,y]\subset K)\qquad\Leftrightarrow\qquad \forall \text{ line } L\colon K\cap L\text{ convex.}
$$ We need only $\Rightarrow$: for any $L$, the set $K\cap L$ is compact $\Rightarrow$ it is empty (convex) or $\exists x,y\in\partial K\colon K\cap L\subset[x,y]$ ($x,y$ can be defined as infimum and supremum of $K\cap L$). Since by assumption $[x,y]\subset K$ we get $[x,y]=K\cap L$ (convex).
