Well it is, by drawing one and looking at it. But how about starting from the definition: the sum of distances to two points is constant. A more general question is to start with an $n$-ellipse (some examples here: What are curves (generalized ellipses) with more than two focal points called and how do they look like?).

Suppose $n$ points $\mathbf{p}_i$ in a two dimensional plane, define the general ellipse $E$ by $$ \sum_i^n d_i = D_0 \quad \mathrm{with} \quad d_i = |\mathbf{r}-\mathbf{p}_i|, \quad \mathbf{r} = (x,y) $$ Questions

  1. is $E$ convex?
  2. does $E$ enclose all points?
  3. if "no" to question 2, what's the value $D_0$ such that $E$ cross a point? And which point?
  4. To higher dimenions?

I know that's lots of questions, and I doubt there is existing knowledge on those questions. Could someone explain or point out some references?

Update: clarification on question 2. For the standard ellipse, when $D_0$ is very small, it does not exist. It come to existence firstly as a line segment connecting the two points---I still count this as "enclose". A point is not enclosed only if it totally falls outside of the shape.

  • $\begingroup$ Your definition of ellipse is a 1D curve, which cannot be convex in 2D. $\endgroup$
    – Kenny Lau
    May 6, 2017 at 2:06
  • $\begingroup$ @KennyLau No, because $r = (x,y)$ is 2D. $\endgroup$
    – Taozi
    May 6, 2017 at 2:12
  • 1
    $\begingroup$ $r=(x,y)$ is 2D, but the points satisfying your equation are 1D (the edge of the ellipse). $\endgroup$
    – Kenny Lau
    May 6, 2017 at 2:14
  • 1
    $\begingroup$ @KennyLau Although curves like the circle cannot be convex in $\Bbb R^2$ nonetheless they are often described as "convex curves". For the purposes of this question I would propose we agree that a Jordan curve whose interior is convex be regarded as a "convex curve". $\endgroup$ May 6, 2017 at 2:25
  • $\begingroup$ @LordSharktheUnknown Thank you for clarifying, only now do I understand the question. Your point it correct and let's follow that definition. $\endgroup$
    – Taozi
    May 6, 2017 at 2:45

2 Answers 2


The ellipse $E = \{\mathbf r : \sum_i d_i(\mathbf r) = D_0\}$ is generally not a convex set in $\mathbb R^n$, but one may ask whether the filled ellipse $F = \{\mathbf r : \sum_i d_i(\mathbf r) \le D_0\}$ is convex. In fact, it always is, because

  1. $d_i(\mathbf r)=\|\mathbf r - \mathbf p_i\|$ is a convex function of $\mathbf r$,
  2. the sum of convex functions is convex, and
  3. the sublevel set of a convex function is a convex set.

However, $F$ includes the point $\mathbf p_i$ if and only if $D_0 \ge \sum_j \|\mathbf p_i - \mathbf p_j\|$. Therefore, $F$ includes all points if and only if $D_0 \ge \max_i \sum_j \|\mathbf p_i - \mathbf p_j\|$. This property does not depend on the dimensionality of the space $\mathbb R^n$.

  • $\begingroup$ $F$ (and $E$) are non empty as long as $D_0$ is greater than the minimum of $\sum_i d_i$, and the point realising the minimum need not be one of the $\mathbb p_i$. So $F$ may very well contain none of these points. $\endgroup$
    – N.Bach
    May 6, 2017 at 9:07
  • $\begingroup$ @Rahul Thanks. In point 2, by "sum of convex function" do you also mean "sum of convex set"? Is this sum similar to the idea of Minkowski addition? $\endgroup$
    – Taozi
    May 6, 2017 at 17:23
  • $\begingroup$ @Rahul Think two filled circles, to me the union of them is impossible to be convex, so why the sum of them is convex? Any references? $\endgroup$
    – Taozi
    May 6, 2017 at 17:25
  • $\begingroup$ @Taozi: No, I mean the sum of convex functions. As I said, the function $d_i(\mathbf r)=\|\mathbf r - \mathbf p_i\|$ is a convex function of $\mathbf r$. Therefore, the sum of such functions $f(\mathbf r)=\sum_i d_i(\mathbf r)$ is also a convex function, and its sublevel set $\{\mathbf r : f(\mathbf r) \le D_0\}$ is a convex set. $\endgroup$
    – user856
    May 6, 2017 at 19:39

This is an attempt at proving that the set of all points on the interior and the boundary of the general ellipse $E$ is indeed a convex set.

Define the set $S$ as follows: $$S= \left\{\mathbf{r} :\sum_i^n d_i(\mathbf{r}) \leq D\right \},$$ where $d_i(\mathbf{r})=\mid \mathbf r - \mathbf{p_i} \ \mid$, $D$ is some constant, and $\{\mathbf{p_1}, \mathbf{p_2} , ..., \mathbf{p_n}\}$ are the $n$ foci of the general ellipse $E$.

Now, we wish to prove that the set $S$ is convex. Using the definition of convexity, for any pair of elements $\mathbf{r_1}, \mathbf{r_2} \in S$, we have to show that $\lambda \mathbf{r_1}+(1-\lambda)\mathbf{r_2}$ is also in $S, 0 \leq \lambda \leq 1$. This can be done by a simple application of the Triangle-Inequality, as follows:

$$\sum_i^n d_i\left(\lambda \mathbf{r_1}+(1-\lambda)\mathbf{r_2}\right)$$ $$=\sum_i^n \mid \lambda \mathbf{r_1}+(1-\lambda)\mathbf{r_2} - \mathbf{p_i} \ \mid$$ $$=\sum_i^n \mid \lambda \mathbf{r_1} - \lambda \mathbf{p_i} + (1-\lambda) \mathbf{r_2} \ -(1-\lambda) \mathbf{p_i} \ \mid$$ $$\leq \sum_i^n \lambda \mid \mathbf{r_1} - \mathbf{p_i} \ \mid + \ (1-\lambda) \mid \mathbf{r_2} - \mathbf{p_i} \ \mid$$ $$= \lambda \sum_i^n \mid \mathbf{r_1} - \mathbf{p_i} \ \mid + \ (1-\lambda) \sum_i^n \mid \mathbf{r_2} - \mathbf{p_i} \ \mid$$ $$\leq \lambda D + (1-\lambda) D$$ $$= D.$$

Where the second last inequality follows since $\mathbf{r_1}, \mathbf{r_2}$ are elements of $S$.


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