Any piece of a fuction can obviously be concave up or down.
Of course the function should not be too anomalous, so as not to posses a second derivative, at least piecewise.
Generally a function need not be either.
Yes, it can be convex. Take f(x) = |x|. It is a piecewise convex function, and overall, convex. Similarly, g(x) = -|x| is piecewise convex but not convex overall, when you consider the linear sections together.
f(x)=|x| is a continuous function, although it's not differentiable at x=0. A convex function on an open interval needs to be continuous.
Cenk: the first function is not just piecewise convex - it is convex everywhere!
And: f(x) := -|x| is concave everywhere.
Michael Patriksson , you are right. I was thinking about strict convexity. It is convex but not strictly convex.
Convex (or concave) funtions are continuous over the relative interior of their domain. A discontinuous function could not be convex nor concave on all of its domain - but it can of course be piecewise convex (or concave) over it's continuity regions.
Ralf Gollmer : Your assertion is not true. For example, let us take any discontinuous linear functional whose domain is the whole vector space (e.g., see https://en.wikipedia.org/wiki/Discontinuous_linear_map#A_concrete_example ). Such a function is convex, has the whole space as its domain, and it discontinuous. Additionally, even in the finite dimensions, I would use "effective domain" instead of "domain" to avoid possible confusion.
I'm sorry to have dropped an important assumption: the assertion is true in R^n, which in many cases is the space for finite-dimensional optimization problems.
Wrt. the notion of effective domain I agree with you.
https://math.stackexchange.com/questions/2812060/when-a-piece-wise-convex-function-is-convex
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