At a high level, I am trying to approximate a convex set $C$ with a strictly convex subset that is an arbitrarily "close" approximation of $C$. (Recall that a set $B$ is strictly convex if for all $\boldsymbol{x},\boldsymbol{y} \in B$ and $t \in (0,1)$, the convex combination $(1-t)\boldsymbol{x}+t\boldsymbol{y}$ belongs to the interior of $B$).
More precisely, let $C$ a (not necessarily strictly) convex and compact subset of $\mathbb{R}^d$ (the big rectangle in the figure) and let $B \subset C$ a closed ball with a positive radius centered at the origin (the circle in the figure --assume it exists, i.e., assume that $\boldsymbol{0}$ belongs to the interior of $C$).
I would like to find a family of strictly convex and compact sets, say $\{B_\lambda\}_{\lambda \in [0,1]}$ such that $B \subset B_\lambda \subset C$ for all $\lambda$ and each point in the interior of $C$ belongs to some $B_\lambda$.
I actually think I found such a family of sets, but I was not able to prove that they are strictly convex, even though I am pretty sure that they are. (If it helps, I did prove that they are compact, lie between $B$ and $C$, and each point in the interior of $C$ belongs to some of them.)
At a high level, each one of these sets is built by considering all rays from the origin (the line in red in the figure), taking a convex combination of the two intersections with $B$ and $C$ respectively (blue dots), then add to this set the entire segment from this point (other blue dot) to the origin.
Rigorously, for all $\lambda \in [0,1]$, let $$B_\lambda := \Biggl\{ t \left( \frac{1-\lambda}{\mu_B(\boldsymbol{x})}\boldsymbol{x} + \frac{\lambda}{\mu_C(\boldsymbol{x})} \boldsymbol{x} \right) : \boldsymbol{x} \in \mathbb{R}^d, \lVert \boldsymbol{x} \rVert =1, t \in [0,1] \Biggr\}$$ where we denoted by $\mu_A(\boldsymbol{x})$ the Minkowski functional of a set $A$ at $\boldsymbol{x}$ (which is well-defined for $A=B$ and $A=C$): $\mu_A(\boldsymbol{x}) := \min \{ \tau \ge 0 : \boldsymbol{x} \in \tau A \}$.
Any ideas on how to prove that all $B_\lambda$ defined this way are strictly convex? Alternatively, do you have another family in mind that satisfies the same properties (in particular, strict convexity)?
A different approach:
Let $$c:=\max_{\mathbf x\in B}\mu_C(\mathbf x).$$ Then $0<c<1$. For $\lambda\ge \frac{2}{1-c}$, let $$f_\lambda(\mathbf x)=\lambda \mu_C(\mathbf x)+\mu_B(\mathbf x) -\lambda$$ and $$B_\lambda=\{\,\mathbf x\in\Bbb R^d \mid f_\lambda(\mathbf x)\le -1\,\}.$$
For $\mathbf x\in B$, we have $$f_\lambda(\mathbf x)\le \lambda c+1-\lambda=-\lambda(1-c)+1\le-1 $$ and so $$\tag1B\subseteq B_\lambda.$$
For $\mathbf x\in C^\complement$, we have $f_\lambda(\mathbf x)> 1$ and therefore $$\tag2 B_\lambda\subset C.$$
For $\lambda<\lambda'$ and $\mathbf x\in C$, we have $$f_{\lambda'}(\mathbf x)-f_\lambda(\mathbf x)=(\lambda'-\lambda)(\mu_C(\mathbf x)-1)<0 $$ and therefore $$\tag3 B_\lambda\subseteq B_{\lambda'}.$$
If $\mathbf x\in C\setminus\partial C$, then $\mu_C(\mathbf x) < 1$. Then for $\lambda\ge \frac{\mu_B(\mathbf x)+1}{1-\mu_C(\mathbf x)}$, we have $$f_\lambda(\mathbf x)=\lambda\mu_C(\mathbf x)+\mu_B(\mathbf x)-\lambda = \mu_B(\mathbf x)-\lambda(1-\mu_C(\mathbf x))\le -1$$ so that $$\tag4 C=\overline{\bigcup_\lambda B_\lambda}.$$
Now note that $\mu_C$ is a convex function and $\mu_B$ a strictly convex function. Then $f_\lambda$ is also strictly convex and therefore so is $B_\lambda$.