Consider a Sobolev function with $p$-integrable first and second weak derivative on some open interval I. Is the following assertion true:
If such function is convex on some sub-interval $(a,b)$ and on sub-interval $(b,c)$, then it is convex on $(a,c)$? This should be true, i think. This means that Sobolev functions in ${\rm W}^{m,p}(I)$, where $m\geq 2$(!), do not allow cusps at point $b$. Thanks.
The derivative of $f$ is increasing in $(a,b)$ and in $(b,c)$, but as you said, since $f$ is $C^1$, the derivative is continuous at $b$, and so the derivative of $f$ is increasing in $(a,c)$. This implies that $f$ is convex in $(a,c)$.