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Let $X$ be a vector space, let $M_1 = \left(X, d_1\right)$ be a metric space and let $M_2 = \left(X, d_2\right)$ be another one.

$f : M_1 \to M_2$ is continuous and the origin is a fixed point. $f$ preserves distances in the following way:

$$ d_1(x_1, x_2) = d_2(f(x_1), f(x_2)) $$

$S \in M_1$ is star-like with respect to the origin.

Is $f(S)$ star-like in $M_2$ with respect to the origin?

If it isn't, what restrictions must I place on $f$ to ensure that it preserves star-likeness?

A very lax definition of a star-like region is that given a point of reference, there is a line-segment between that point and another that lies entirely within the region. Convex regions are star-like with respect to all points within the region, for example.

Axoren
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    On the negative side of your question, a continuous function does not preserve star-likeness. Think about taking a disk in the plane and bending it around in to an almost-annulus. – Callus - Reinstate Monica Feb 02 '16 at 22:19
  • That's a good counter-example for the question as written. That's a continuous functions for which the origin is a fixed point and it is indeed non-star-like. My question was missing information. I'll correct it immediately. – Axoren Feb 02 '16 at 22:44
  • @Callus I've corrected question. Is it still possible to bend the disk in such a way that is continuous and preserves the distances in the same way? If so, what constraints can you think of to impose on $f$ that prevents such a bending (the action that breaks star-likeness). – Axoren Feb 02 '16 at 22:49
  • Maybe ( likely ) I'm getting out of my depth, but I don't know what a line segment would be in a general metric space. However, isometries of Euclidean space preserve lines, so also star-shapedness. – Callus - Reinstate Monica Feb 02 '16 at 22:59
  • The definition of convex and starlike does not apply to arbitrary metric spaces. In order for these to be defined you need some concept of "line", which not every metric space has. So you need to put more restrictions on $X$ first. Do you mean $X$ to be a vector space? – Paul Sinclair Feb 03 '16 at 01:27
  • @PaulSinclair Yes. I've corrected the question. – Axoren Feb 03 '16 at 18:12
  • Is the metric compatible with the vector space structure? That is, is $d(ax_1,ax_2)=ad(x_1,x_2)$, and $d(x_1+y,x_2+y)=d(x_1,x_2)$? –  Feb 03 '16 at 18:17
  • @Rahul Yes, it is. – Axoren Feb 12 '16 at 02:21

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