Let $T$ be a normed vector space, $K\subseteq T$ compact and convex and $O\subseteq K$ convex and open in $K$ (i.e. open w.r.t. the subspace topology of $K$ inherited by $T$).


Can we find an open set $O'\subseteq T$ such that $O' \cap K = O$? Can we also find such an $O'$ which is convex?


Edit: The convexity of $K$ is indeed necessary. There exists a counterexample in case of $K$ not being convex.



Existence of an open convex set
A convex $O'$ need not exist: a counterexample is given by setting $K=[-1,1]\times[0,2]\subseteq\mathbb{R}^2$ and $O=\{(x,y)\in K;y>x^3\}$. Indeed, any open $O'$ with $O'\cap K=O$ would contain some nhood of $(-\frac{1}{2},0)$, so it would contain some point $q=(-\frac{1}{2},-\varepsilon)$ with $\varepsilon>0$. Note that the tangent line from $q$ to the function $f:[0,\infty)\to\mathbb{R};f(x)=x^3$ passes below the point $(0,0)$. This easily implies that if $O'$ was convex, it would contain $(0,0)$, contradicting $O'\cap K=O$.


However, the conjecture is true for lots of convex sets $K$ (for example [strictly convex](https://en.wikipedia.org/wiki/Convex_set#strictly_convex) sets, which satisfy the lemma below). I will assume that the normed space is $\mathbb{R}^n$, with $n\geq2$ with norm $||\cdot,\cdot||\_2$, since the norm is irrelevant for the problem. First I prove the result for $\overline{\mathbb{B}^n}$, where it's easier to visualize, and then I generalize below.


**Proposition:** For each convex set $O\subseteq\overline{\mathbb{B}^n}$ open in $\overline{\mathbb{B}^n}$ we can find a convex open set $O'$ such that $O=O'\cap\overline{\mathbb{B}^n}$.


Proof: For each $v\in\mathbb{S}^{n-1}$ let $X\_v:=\{x\in\mathbb{R}^n;\langle x,v\rangle<1\}$. Let $B=\mathbb{S}^{n-1}\setminus O$, and let $A$ be the interior of $\bigcap\_{v\in B}X\_v$. Clearly $A$ is open and convex, and $O\subseteq A$. We will define $O'=O\cup(A\setminus\overline{\mathbb{B}^n})$.


By definition, $O=O'\cap\overline{\mathbb{B}^n}$, and $O'$ is open because $A$ is open, $O\setminus\mathbb{S}^{n-1}$ is open and any $p\in O\cap\mathbb{S}^{n-1}$ is in the interior of $A$, so it is in the interior of $O'$.


So we just need to deduce that $O'$ is convex: let $p,q\in O'$.


* If $p,q\in O$ then the segment $[p,q]$ is contained in $O\subseteq O'$.
* If $p\in A\setminus\overline{\mathbb{B}^n},q\in O$, then $[p,q]\subseteq A$. If $[p,q)$ doesn't intersect $\mathbb{S}^{n-1}$, then $[p,q)\subseteq A\setminus\overline{\mathbb{B}^n}\subseteq O'$ so we are done. If not, let $r$ be que unique intersection of $[p,q)$ with $\mathbb{S}^{n-1}$. As $r\in A\cap\mathbb{S}^{n-1}$, $r$ cannot be in $B$, so $r\in O$. Thus $[r,q]\subseteq O\subseteq O'$ and $[p,r)\subseteq A\setminus\overline{\mathbb{B}^n}\subseteq O'$, so as we wanted $[p,q]\subseteq O'$.
* A similar argument in the case $p,q\in A\setminus\overline{\mathbb{B}^n}$ (this time considering the up to $2$ points of intersection of $[p,q]$ with $\mathbb{S}^{n-1}$, which again will be in $O$) allows us to conclude that $O'$ is convex. $\square$


Now let $K$ be a general convex set with nonempty interior. We say an open half space $X$ is tangent to $K$ at $p\in\partial K$ if $K\subseteq\overline{X}$ and $p\in\partial X$. Let $\mathfrak{X}$ be the set of all half spaces tangent to $K$ at points of $\partial K\setminus O$ and let $A$ be the interior of $\bigcap\_{X\in\mathfrak{X}}X$. Then again, $A$ is open and convex, but it need not be true that $O\subseteq A$. If indeed $O\subseteq A$, we can still define $O'=O\cup(A\setminus K)$ and the rest of the proof works the same way (when proving convexity, we can define $r$ as the point of $[p,q)\cap\partial K$ closest to $p$).


A sufficient condition for $O\subseteq A$ is the following:


**Lemma:** Suppose that every hyperplane tangent to $K$ intersects $K$ at just one point. Then $O\subseteq A$.


Proof: Suppose some $p\in O$ is not in $A$. The statement of the lemma implies that $p\in X$ for every $X\in\mathfrak{X}$, so $p\in\bigcap\_{X\in\mathfrak{X}}X$. Thus $p$ is in the boundary of $A$, and there is a sequence of half spaces $X\_n\in\mathfrak{X}$ tangent to $K$ at some points $p\_n\in\partial K\setminus O$, such that $d(p,\partial X\_n)\to0$. Taking a subsequence if necessary, we can suppose that the vectors $v\_n$ normal to the hyperplanes $\partial X\_n$ converge to $v\_\infty$, and that the points $p\_n$ converge to $p\_\infty$ (note that $p\_\infty\not\in O$, so $p\_\infty\neq p$). So the half space $X$ based at $p\_\infty$ with normal vector $v\_\infty$ is tangent to $K$, and $p\in\partial X$. Which contradicts the statement of the lemma.$\square$


