Proof. We assume that the interior of F is empty, otherwise there is nothing to prove. Let U be the unbounded connected component of $F^c$ . Write $V=F^c\backslash U$ . By the assumption that $F^c$ has at least one bounded connected component, V is a nonempty bounded open subset of $\mathbb {R}^d$ . Set

$$ \begin{align*} W=\{x\in \mathbb{R}^d: \text{there exists } a, b\in E \mbox{ such that }x\in (a+U)\cap (b+V)\}. \end{align*} $$

Clearly W is open. We claim that $W\neq \emptyset $ . To prove this claim, it suffices to show that $(s+V)\cap U\neq \emptyset $ for each nonzero $s\in \mathbb {R}^d$ . To this end, fix a nonzero $s\in \mathbb {R}^d$ and define $P_s:\mathbb {R}^d\to \mathbb {R}$ by $P_s(x)=\langle x,s \rangle $ , where $\langle \cdot , \cdot \rangle $ is the standard inner product in $\mathbb {R}^d$ . Since V is bounded,

$$ \begin{align*}\lambda:=\sup_{x\in V}P_s(x)<\infty.\end{align*} $$

Pick $x_0\in V$ so that $P_s(x_0)>\lambda -\|s\|^2/2$ , and take a small $r\in (0, \|s\|/2)$ so that $B^{o}(x_0, r)\subset V$ , where $B^{o}(x_0, r)$ stands for the open ball centred at $x_0$ of radius r. Then for each $y\in \mathbb {R}^d$ with $\|y-x_0\|<r$ ,

$$ \begin{align*} \langle s+y,s\rangle &= \langle s+x_0,s\rangle +\langle y-x_0,s\rangle\\ &\geq \langle x_0,s\rangle +\|s\|^2 -\|y-x_0\|\cdot \|s\|\\ &\geq \langle x_0,s\rangle +\|s\|^2/2> \lambda, \end{align*} $$

which implies that $s+y\not \in V$ and so $s+B^{o}(x_0, r)\subset V^c$ . Since F has empty interior, it follows that $(s+B^{o}(x_0, r))\cap F^c\neq \emptyset $ , equivalently, $(s+B^{o}(x_0, r))\cap (U\cup V)\neq \emptyset $ . Since $s+B^{o}(x_0, r)\subset V^c$ , we get $(s+B^{o}(x_0, r))\cap U\neq \emptyset $ , so $(s+V)\cap U\neq \emptyset $ , as desired. This proves $W\neq \emptyset $ .

Finally, we prove that $W\subset E+F$ , which immediately implies that $E+F$ has nonempty interior. Suppose this is not true, that is, there exists $x\in W$ so that $x\not \in E+F$ . By the definition of W, there exist $a, b\in E$ so that

(2.1) $$ \begin{align} x\in (a+U)\cap (b+V). \end{align} $$

Since $x\not \in E+F$ , it follows that

(2.2) $$ \begin{align} (x-E)\subset F^c=U\cup V. \end{align} $$

However, according to (2.1), $(x-E)\cap U\supset \{x-a\}$ and $(x-E)\cap V\supset \{x-b\}$ . This together with (2.2) implies that $x-E$ is not connected, leading to a contradiction.

Proof. By taking a suitable rotation and translation if necessary, we may assume that K is contained in the linear subspace $\{(x_1,\ldots ,x_d)\in \mathbb {R}^d: x_d=0\}$ and that at least one bounded connected component of $(F\cup K)^c$ has nonempty intersection with the half-space $H:=\{(x_1,\ldots , x_d)\in \mathbb {R}^d: x_d>0\}$ .

Let U be a bounded connected component of $(F\cup K)^c$ so that $V:=U\cap H$ is nonempty. Since $\partial U\subset F\cup K$ and $K\subset \partial H$ , we easily see that $\partial V\subset F\cup (\overline {U}\cap \partial H)$ . Write $\tilde {K}=\overline {U}\cap \partial H$ . Then $\tilde {K}$ is a compact subset of the linear subspace

$$ \begin{align*}\partial H=\{(x_1,\ldots,x_d)\in\mathbb{R}^d: x_d=0\}\end{align*} $$

and

(2.3) $$ \begin{align} \partial V\subset F\cup \tilde{K}. \end{align} $$

Set

$$ \begin{align*}h:=\sup\Pi(V),\end{align*} $$

where $\Pi : \mathbb {R}^d\to \mathbb {R}$ is the mapping $(x_1, \ldots , x_d)\mapsto x_d$ . Then h is positive and finite.

To prove that A is totally disconnected, we may assume that $\#A\geq 2$ , since otherwise there is nothing to prove. Let $u,v\in A$ with $u\neq v$ . In the following, we are going to construct an open set $W\subset \mathbb {R}^d$ such that

(2.4) $$ \begin{align} u\in W, \quad v\not\in \overline{W} \quad \text{and} \quad \partial W\cap A=\emptyset. \end{align} $$

Since $u,v\in A$ with $u\neq v$ are arbitrary, it will follow that A is totally disconnected.

To prove (2.4), first notice that the open set $(u+V)\setminus (v+\overline {V})$ is nonempty. Indeed, since $u\neq v$ and $(u+V)\setminus (v+\overline {V})=((u-v+V)\setminus \overline {V})+v$ , it suffices to show that $(a+V)\setminus \overline {V}\neq \emptyset $ for any nonzero $a\in \mathbb {R}^d$ . To this end, fix $a\in \mathbb {R}^d$ with $a\neq 0$ . Define

$$ \begin{align*}\lambda:=\sup_{x\in V}\langle x, a\rangle,\end{align*} $$

where $\langle \cdot , \cdot \rangle $ is the standard inner product in $\mathbb {R}^d$ . Then $\lambda $ is finite as V is bounded. Also, it is clear that $\lambda =\sup _{x\in \overline {V}}\langle x, a\rangle $ . Since $a\neq 0$ , we have $\langle a, a\rangle>0$ . Hence, we can find $x_0\in V$ such that $\langle x_0, a\rangle>\lambda -\langle a,a\rangle $ . Thus, we have

$$ \begin{align*}\langle a+x_0,a\rangle=\langle x_0,a\rangle+\langle a,a\rangle>\lambda,\end{align*} $$

which implies that $a+x_0\in (a+V)\setminus \overline {V}$ . Hence, $(u+V)\setminus (v+\overline {V})$ is a nonempty open set. By the density of D, we can pick a point $z\in D\cap [(u+V)\setminus (v+\overline {V})]$ . Then we have

(2.5) $$ \begin{align} u\in z-V \quad \text{and} \quad v\notin z-\overline{V}. \end{align} $$

Let

$$ \begin{align*}\tilde{D}=\{x\in D: \Pi(x)>\Pi(u)+h\}.\end{align*} $$

Since D is dense in $\mathbb {R}^d$ , $\tilde {D}$ is clearly dense in the open half-space

(2.6) $$ \begin{align} \{x\in \mathbb{R}^d: \Pi(x)>\Pi(u)+h\}. \end{align} $$

Let $x\in \mathbb {R}^d$ with $\Pi (x)>\Pi (u)$ . Since $x+V$ is open and

$$ \begin{align*}\sup\Pi(x+V)=\Pi(x)+h>\Pi(u)+h,\end{align*} $$

we see from (2.6) that $(x+V)\cap \tilde {D}\neq \emptyset $ . Equivalently, $x\in \tilde {D}-V$ . Hence, we have

(2.7) $$ \begin{align} \tilde{D}-V\supset\{x\in \mathbb{R}^d: \Pi(x)>\Pi(u)\}. \end{align} $$

Since $\tilde {K}$ is contained in the linear subspace $\{(x_1,\ldots , x_d)\in \mathbb {R}^d: x_d=0\}$ , V is contained in the open half-space $\{(x_1,\ldots , x_d)\in \mathbb {R}^d: x_d>0\}$ and $z\in u+V$ , we have

(2.8) $$ \begin{align} \Pi(z-\tilde{K})=\{\Pi(z)\} \quad \text{and} \quad \Pi(z)>\Pi(u). \end{align} $$

Then by (2.7), (2.8) and the compactness of $z-\tilde {K}$ , we can find finitely many points $z_1,\ldots , z_k\in \tilde {D}$ so that

(2.9) $$ \begin{align} z-\tilde{K}\subset \bigcup_{i=1}^k(z_i-V). \end{align} $$

Set

(2.10) $$ \begin{align} W:=(z-V)\setminus \bigg(\bigcup_{i=1}^k (z_i-\overline{V})\bigg). \end{align} $$

See Figure 1 for an illustration of the definition of W, where for simplicity we assume that V is an open half disk. Below, we show that the open set W satisfies (2.4).

Figure 1 Definition of W in the proof of Lemma 2.3.

First notice that $v\notin \overline {W}$ since $v\notin z-\overline {V}$ (see (2.5)). Moreover, since $z_1,\ldots ,z_k\in \tilde {D}$ ,

$$ \begin{align*}\inf\Pi\bigg(\bigcup_{i=1}^k(z_i-\overline{V})\bigg)=\min_{1\leq i\leq k}(\Pi(z_i)-h)>\Pi(u),\end{align*} $$

which implies that

$$ \begin{align*}u\notin \bigcup_{i=1}^k(z_i-\overline{V}).\end{align*} $$

Since $u\in z-V$ , it follows that $u\in W$ . In the following, we show that $\partial W\cap A=\emptyset $ .

To see this, observe that by (2.3) and (2.10),

(2.11) $$ \begin{align} \partial W\subset (z-F)\cup (z-\tilde{K})\cup\bigg(\bigcup_{i=1}^k(z_i-F)\bigg)\cup\bigg(\bigcup_{i=1}^k(z_i-\tilde{K})\bigg). \end{align} $$

By (2.9), (2.10) and the compactness of $z-\tilde {K}$ , we see that W is disjoint from a neighbourhood of $z-\tilde {K}$ . In particular, this implies that

$$ \begin{align*} \partial W\cap (z-\tilde{K})=\emptyset. \end{align*} $$

Next we show that

(2.12) $$ \begin{align} \partial W\cap \bigg(\bigcup_{i=1}^k (z_i-\tilde{K})\bigg)=\emptyset. \end{align} $$

To see this, since $\tilde {K}\subset \{(x_1,\ldots ,x_d)\in \mathbb {R}^d: x_d=0\}$ , we have for $i\in \{1,\ldots ,k\}$ ,

(2.13) $$ \begin{align} \Pi(z_i-\tilde{K})=\{\Pi(z_i)\}. \end{align} $$

Moreover, since $z_1,\ldots , z_k\in \tilde {D}$ , we see that

$$ \begin{align*} \Pi(z_i)>h+\Pi(u), \quad i=1,\ldots, k. \end{align*} $$

However, since $V\subset \{(x_1,\ldots ,x_d)\in \mathbb {R}^d: x_d>0\}$ and $z\in u+V$ ,

(2.14) $$ \begin{align} h+\Pi(u)\geq \Pi(z)\geq \sup\Pi(z-\overline{V})\geq \sup\Pi(\overline{W}), \end{align} $$

where the last inequality is by (2.10). Now (2.13)–(2.14) imply that for $i\in \{1,\ldots , k\}$ ,

$$ \begin{align*}\inf\Pi(z_i-\tilde{K})=\Pi(z_i)>\sup\Pi(\partial W)\end{align*} $$

and from this, (2.12) follows.

By (2.11)–(2.12), we see that

(2.15) $$ \begin{align} \partial W\subset (z-F)\cup\bigg(\bigcup_{i=1}^k(z_i-F)\bigg). \end{align} $$

Since $z, z_1,\ldots , z_k\in D$ , the definition of A implies that A has no intersection with the right-hand side of (2.15). As a consequence, $A\cap \partial W=\emptyset $ . Hence, (2.4) is proved and we finish the proof of the lemma. For an illustration of the proof, see Figure 1.

Proof of Lemma 2.2

We prove the lemma by contradiction. Suppose that $(E+F)^{\circ }=\emptyset $ . Then we have, equivalently, that the set

$$ \begin{align*}D:=(E+F)^c\end{align*} $$

is dense in $\mathbb {R}^d$ . Hence, by Lemma 2.3, $\bigcap _{z\in D}(z-F)^c$ is totally disconnected. However, from the definition of D, we see that

$$ \begin{align*} E\subset \bigcap_{z\in D}(z-F)^c. \end{align*} $$

This contradicts the assumption that E is a connected set with cardinality greater than $1$ . Hence, we have $(E+F)^{\circ }\neq \emptyset $ , completing the proof of the lemma.

Proof. The result might be well known. However, we are not able to find a reference, so we include a proof. By Carathéodory’s theorem, x can be represented as a convex combination of three elements of S. Equivalently, x lies in a triangle with vertices in S. Since x is on the boundary of $\mathrm { conv}(S)$ , it follows that x lies on one edge of the triangle. Let $u,v\in S$ be the endpoints of this edge. Since $x\not \in S$ , we have $u\neq v$ .

Let $L_{u,v}$ denote the straight line passing through the points $u, v$ . We show that S lies completely on one side of $L_{u, v}$ . Suppose, in contrast, that S does not lie on one side of $L_{u,v}$ . Then we can pick $w_1, w_2\in S$ such that $w_1, w_2$ lie on different sides of $L_{u,v}$ . Clearly, the point x lies in the interior of the quadrilateral with vertices $u, v, w_1, w_2$ (see Figure 2). However, this quadrilateral is a subset of $\mathrm {conv}(S)$ , contradicting the assumption that $x\in \partial (\mathrm {conv}(S))$ .

Proof. We may assume that $F^c$ has no bounded connected components, otherwise we simply take $K=\emptyset $ .

Let $\mathrm {conv}(F)$ denote the convex hull of F. Since F is not contained in a straight line, $\mathrm {conv}(F)$ has nonempty interior and there exists a homeomorphism $h:\mathbb {R}^2\to \mathbb {R}^2$ so that $h(\mathrm {conv}(F))$ is the unit closed ball centred at the origin (see, for example, [Reference Borwein and Lewis4, Exercise 8.11]).

We claim $\partial (\mathrm {conv}(F))\not \subset F$ . Suppose, in contrast, that $\partial (\mathrm { conv}(F))\subset F$ . As $\partial (\mathrm {conv}(F))$ is homeomorphic to the unit circle, $\mathbb {R}^2\backslash \partial (\mathrm {conv}(F))$ has exactly two connected components $V_1, V_2$ , where $V_1$ is unbounded and $V_2$ bounded. Since $\partial (\mathrm {conv}(F))\subset ~F$ , it follows that $F^c\subset V_1\cup V_2$ . Since F is compact, $F^c\cap V_1\neq \emptyset $ and so $F^c\subset V_1$ by the connectedness of $F^c$ . This implies that $V_2\subset F$ , contradicting the assumption that F has empty interior.

Pick $z\in \partial (\mathrm {conv}(F))\backslash F$ . By Lemma 3.2, there exist $u,v\in F$ such that the straight line $L_{u,v}$ passes through z and F lies completely on one side of $L_{u,v}$ . By taking a suitable rotation and translation to F, we may assume that $z=(0,0)$ , $L_{u,v}$ is the x-axis, u is on the negative part of the x-axis and v is on the positive part of the x-axis, and F lies entirely on the upper half plane.

Choose a large $R>0$ such that F is contained in the closed half disc

$$ \begin{align*} S:=\{(x, y)\in \mathbb{R}^2: y\geq 0,\; x^2+y^2\leq R^2\}. \end{align*} $$

Let K be the line segment $[-R, R]\times \{0\}$ , which is the bottom edge of S. Below, we show that $(F\cup K)^c$ has at least one bounded connected component.

Since the origin is not contained in F, there exists a small $r>0$ such that the open half disc

$$ \begin{align*} T:=\{(x, y)\in \mathbb{R}^2: y> 0,\; x^2+y^2< r^2\} \end{align*} $$

is contained in $(F\cup K)^c$ . Let V be the connected component of $(F\cup K)^c$ that contains T. Notice that $S^c$ is contained in the unbounded connected component U of $(F\cup K)^c$ . To show that V is bounded, it is enough to show that $V\neq U$ (keep in mind that $(F\cup K)^c$ has a unique unbounded connected component, due to the compactness of $F\cup K$ ).

Suppose, on the contrary, that $V=U$ . Then $U\supset T\cup S^c$ . Pick $a\in T$ and $b\in S^c$ . Since U is open and connected, there exists a simple curve $\gamma \subset U$ such that $\gamma $ consists of finitely many line segments and $\gamma $ joins the points $a, b$ (see, for example, [Reference Ahlfors1, page 56] for a proof). Clearly, $\gamma $ must intersect the open half circle

$$ \begin{align*}\Gamma:=\{(x,y): x^2+y^2=R^2,\; y>0\}\end{align*} $$

at one or more than one points. As $\gamma $ is a polygon, we may choose a sub-polygon $\gamma _1$ which joins a and a point $c \in \Gamma $ such that c is the unique intersection point of $\gamma _1$ and $\Gamma $ . Connect the point a and the origin by a simple polygon $\gamma _2\subset \overline {T}$ such that $\gamma _2$ intersects $\gamma _1$ only at the point a, and $\gamma _2$ intersects K only at the origin.

Let $\eta =\gamma _1\cup \gamma _2$ . Then $\eta $ is a simple polygon, joining the origin and the point c. Except for the endpoints, points of $\eta $ are contained in $U\cap S^o$ . Hence, $\eta \cap F=\emptyset $ .

Figure 3 Illustration for the proof of Proposition 3.3.

Write $c=(c_1, c_2)$ . Let $L^+, L^{-}$ be the vertical half lines $L^+:=\{(c_1, y): y\geq c_2\}$ and $L^-:=\{(0, y): y\leq 0\}$ . Then the union $\eta \cup L^+\cup L^-$ has no intersection with F. Moreover, its complement has two connected components, with $u, v$ being contained in different components. This implies that F is disconnected, leading to a contradiction. See Figure 3 for an illustration of the proof.

Proof of Theorem 1.2

We can assume that $F^{\circ }=\emptyset $ , since otherwise there is nothing to prove. Since F is compact, connected, $F^{\circ }=\emptyset $ and F is not a line segment, by Proposition 3.3, there exists a compact set K contained in a line segment such that $(F\cup K)^c$ has a bounded connected component. Then it follows from Lemma 2.2 that $E+F$ has nonempty interior.

Proof. In one direction, assume that $\inf _{x\in \mathbb {R}}\omega _f(x)>0$ . Let L be a vertical line segment with length $0<\ell <\inf _{x\in \mathbb {R}}\omega _f(x)$ . Below, we show that $(G_f+L)^{\circ }=\emptyset $ .

By applying a suitable translation, we can assume that $L=\{0\}\times [0,\ell ]$ . Suppose, in contrast, that $(G_f+L)^{\circ }\neq \emptyset $ . Then, in particular, $G_f+L$ contains a horizontal line segment, say, $[a,b]\times \{c\}$ for some $a,b,c\in \mathbb {R}$ with $a<b$ . Notice that

$$ \begin{align*}G_f+L=\bigcup_{x\in \mathbb{R}}((x,f(x))+L)=\bigcup_{x\in \mathbb{R}}(\{x\}\times[f(x), f(x)+\ell])\end{align*} $$

is a disjoint union of vertical line segments. By this and our assumption that $G_f+L\supset [a,b]\times \{c\}$ , we easily see that

$$ \begin{align*}(x,c)\in \{x\}\times[f(x), f(x)+\ell] \quad\mbox{for all } x\in [a,b],\end{align*} $$

and thus

(4.1) $$ \begin{align} c-\ell\leq f(x)\leq c \quad\mbox{for all } x\in [a,b]. \end{align} $$

Let $x_0=(a+b)/2$ . Then (4.1) implies that $\omega _f(x_0)\leq \ell $ , contradicting $\inf _{x\in \mathbb {R}}\omega _f(x)>\ell $ . The contradiction yields $(G_f+L)^{\circ }=\emptyset $ .

In the other direction, we will prove that if $\inf _{x\in \mathbb {R}}\omega _f(x)=0$ , then $(G_f+L)^{\circ }\neq \emptyset $ for any vertical line segment L. Assume that $\inf _{x\in \mathbb {R}}\omega _f(x)=0$ . Let L be a vertical line segment with length $\ell>0$ . Again, we can assume that $L=\{0\}\times [0,\ell ]$ .

Since $\inf _{x\in \mathbb {R}}\omega _f(x)=0$ , then by definition, we can find $x_0\in \mathbb {R}$ and $\delta>0$ such that

$$ \begin{align*}\sup_{x\in [x_0-\delta, x_0+\delta]}f(x)-\inf_{x\in [x_0-\delta, x_0+\delta]}f(x)<\ell/4.\end{align*} $$

This clearly implies that

(4.2) $$ \begin{align} f(x_0)-\ell/4<f(x)<f(x_0)+\ell/4 \end{align} $$

for all $x\in [x_0-\delta , x_0+\delta ]$ . We claim that $G_f+L$ contains the rectangle

$$ \begin{align*}R:=[x_0-\delta, x_0+\delta]\times[f(x_0)+\ell/4, f(x_0)+(3\ell)/4].\end{align*} $$

To see this, let $(x,y)\in R$ . Then, $f(x_0)+\ell /4\leq y\leq f(x_0)+(3\ell )/4$ . By this and (4.2), we see that $0\leq y-f(x)\leq \ell $ . Hence, $(0, y-f(x))\in L$ . Therefore, we have

$$ \begin{align*}(x,y)=(x, f(x))+(0,y-f(x))\in G_f+L.\end{align*} $$

Since $(x,y)\in R$ is arbitrary, it follows that $R\subset G_f+L$ . This proves the above claim, and in particular, that $(G_f+L)^{\circ }\neq \emptyset $ .