Convex set, differentiable in one point, inner direction I have a convex set $\mathcal{C}$ with non-empty interior and I have shown that its boundary is differentiable in a point $c^*$, in the sense that there is a vector $n^*$(the normal vector), such that $n^*\frac{c_n-c^*}{\|c_n-c^*\|}\to 0$ for every sequence $c_n\to c^*$ on the boundary $\partial\mathcal{C}$. 
My first question is, if this notion of differentiability makes sense; i.e., does it imply that $\partial\mathcal{C}$ is locally the graph of a function that is differentiable in one point?
My second question is, if my notion of differentiability implies that $c^*-\epsilon d$ is in the convex set for small $\epsilon$ and ${n^*}^\intercal d<0$?  Would this follow from the graph property above?
 A: True on both counts, with pretty straightforward, if somewhat tedious, proofs.
For any convex set with nonempty interior, the boundary is locally the graph of a continuous function. Indeed, let $x$ be an interior point and $y$  a boundary point. We have $B(x,r)\subset\mathcal C$ for some $r>0$. Let $L$ be the half-line starting at $x$ and passing through $y$. For every $w$ with $|w|<r$, the translate $L+w$ is a half-line with vertex in $\mathbb C$; therefore it meets $\partial \mathcal C$ (by convexity, at exactly one point). This gives a representation of $\partial \mathcal C$ as a graph $Z=F(X)$, with $Z$-axis being parallel to $xy$. The convexity of $\mathcal C$ implies that $F$ is concave, hence continuous.
Now let $c^*$ and $n^*$ be as in the question.  Let $P$ be the plane passing through $c^*$ and orthogonal to $c^*$. We can choose $x$ above so that $x\notin P$ (in fact, $P$ does not meet the interior of $\mathcal C$ at all, but I don't want to argue this point). Write  $P$ as a graph $Z=L(X)$, with $Z$ axis being parallel to  $xc^*$ direction being the dependent variable axis. The tangency property implies that $F-L$ is appropriately small to make $F$ differentiable. 
For the second question: once we have the set locally represented as $Z<F(X)$ with differentiable $F$,  you can rewrite this equation as $Z-F(X)<0$ and then observe that the condition ${n^*}^Td<0$ says precisely that the directional derivative of $Z-F(X)$ in the direction $d$ is negative.
