… to construct a plane figure for which all internal surfaces are convex?

I was thinking about the Koch Snowflake - a curve of infinite length enclosing a finite space by means of inserting a wiggly bit every time the scale makes the line look smooth.
Is it possible to take, say, a square, bend the sides inwards slightly so that they are convex (from the point of view of someone inside the figure), but this creates more sharply concave corners, so can these be replaced by convex curves, and so on ad infinitum.

I’m not quite sure what you’re asking. It sounds like you want a space filling curve (ie. smooth (non-intersecting) line through every point in an area) such that the inner edges are convex?

I’m not convinced you can have any shape where all the inner edges are convex: pick a point inside, draw a line through it. This must intersect the boundary at at least two points. Then the line between these points goes inside the shape, hence its not “inwardly convex”.

Oh, sorry, I was thinking of the wrong fractal: we’re not talking about space filling, just finite interior, infinite length. My bad. My second point makes sense now - there actually is an interior.

Imagine a slice of toast; you nibble the sides so that they curve inwards, but this makes the corners sharper, so you bite them off - creating an inward curve, but also creating two new corners, so you bite them off… and so on.

It’s easy to create a 2-D figure where “all internal surfaces are concave” without resorting to fractal curves. Some isolated points of discontinuity may not be concave, but IIRC, the radius of curvature for an osculating circle at a point of discontinuity isn’t defined, so it isn’t technically concave or convex. (At least according to the one definition of concavity I semi-clearly recall. IIRC, it isn’t the only one)

A trivial example is a equilateral triangle (or any regular equal-sided convex polygon) whose flat sides have been replaced by concave arcs of a circle. The circle can be of arbitrary diameter, so the figure can be indistinguishable from a true triangle, but it is provably concave except at the vertices, where the radius of curvature is undefined.

No, I’m not just being a wisacre. This actually may lead somewhere. Substituting a smooth concave curve at the verticeswill not create a concave figure. Not “may not”, but WILL NOT. (Nor will substituting a convex curve, though that’s pretty obvious) As far as I can tell, the reason for this (which is beyond my ability to express mathematically in ASCII in less than 10 pages <dammit, Jim, I’m a doctor, not a geometer>) rules out any uniformly and universally concave curve, even (especially?) a fractal one. Only closed curves with points of undefined curvature can be composed entirely of concave surfaces.

But, of course, there may be some other defenition of concavity that permits it.

Each iteration of the fractal results in twice as many acute angles, ad infinitum; every time you try to make all the angles convex, you wind up on the *outside. *

Or maybe something like a spiral, where there’s ambiguity between internal and external.

Of course, we can redefine “internal” and “external,” so that the whole universe is considered “internal” and that little area is the “external.”