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I have the following dilemma!

Say, $f_1=\sqrt{1-x^2}$, and $f_2=-\sqrt{1-x^2}$ are two continuous functions on $[-1,1]$

Lets define another function by $F = tf_1 + (1-t)f_2$ where $t=[0,1]$

Clearly $F$ is a homotopy between $f_1$ and $f_2$. But notice that during deformation I have changed the sign of the curvature, which means I have inverted the curve in the process.

Is there any way to put an extra condition on maintaining the sign of curvature (no invertion locally as well as globally), and then check the feasibility of homotopy between them.

Here sign of the curvature I mean a convex curve should stay convex, a concave curve should stay concave. Let's assume that concave means positive sign.

Please guide!

share|cite|improve this question
What is the sign of curvature of say, $x^2$? – Ben Millwood Mar 19 '14 at 13:22
I have rephrased the question for clarity! In $x^2$ case, we can assume it to be positive to start with and then see if it can be negative (means convex)! Thanks for your point. – user135317 Mar 19 '14 at 13:30

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