Let $X\subset\mathbb{R}$ is convex and $f:X\rightarrow\mathbb{R}$ is a twice continuously differentiable function. Is $f(X)$ convex?
Asked
Active
Viewed 37 times
0
-
For the compactness. Because $f$ is differentiable, then is continuous. How $X$ is compact, then, $f(X)$ is compact. – Carlos Jiménez Nov 23 '16 at 05:20
-
Compact convex subsets of $\Bbb R$ are not a very rich class of objects... – Nov 23 '16 at 05:40
-
@G.Sassatelli;I was wondering whether continuous image of a convex set is convex or not?Do you know any such result – Learnmore Nov 23 '16 at 05:51
-
1If the image is a subset of $\Bbb R$, you have the luck that convexity and connectedness coincide. In higher dimension there is no such hope, if the function is not an affine map: consider $f:[0,1]\to \Bbb R^2$ given by $f(x)=(x,x^2)$ (or any plane curve, for that matter). – Nov 23 '16 at 07:12
-
The convex subsets of the real lines are intervals, so by the IVT it should hold. It should be asked what happens for convex subsets $X \subset \Bbb R^n$ with $n>1$. – Watson Jul 23 '18 at 19:16