# Convexity of Sobolev functions

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 function $x \mapsto -|\sin x|$ is convex on $[0,\pi]$ and $[\pi,2 \pi]$ but is not convex on $[0,2 \pi]$. – copper.hat Dec 31 '17 at 17:55
• I see. Thanks. How about the following assertion: Consider a Sobolev function $u$ with $p$-integrable first and second weak derivative on some open interval $I$. Then there exists a decomposition of $I$ into countable union of pairwise disjoint open intervals $I=\cup_{n=1}^{+\infty}I_n$ such that for every $n\in\mathbf{N}$ the restriction $u\vert_{I_n}$ is either convex or concave? Thanks. – Andrija Dec 31 '17 at 18:24
• Then again, I don't think that the function $g(x)=-|\sin x|$ belongs to ${\rm H}^2(0,2\pi)$. By the Sobolev embedding, $g$ must be ${\rm C^1}(0,2\pi)$ function, which is not the case. $g'$ has jump discontinuity at point $\pi$. So I still think that, while it is true that $g\in {\rm H}^1(0,2\pi)$, we have a problem with second distributional derivative of $g$, which is not in ${\rm L}^2(0,2\pi)$. I think my original assertion may be true after all. – Andrija Jan 1 '18 at 9:27

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 $(a,c)$.
• Yes, that completes the argument. Is it to much to hope for piecewsie convexity-concavity property of Sobolev functions in ${\rm W}^{m,p}(I)$, where $m\geq 2$, which I mentioned in my initial reply above? I can't find similar assertions in the literature. Thanks. – Andrija Jan 2 '18 at 15:23
• yes, it is too much. If you take any function $g$ in $L^p$ and integrate it twice, you get a function in $W^{2,p}$. There is no reason why the set where $g>0$ to be a countable union of intervals. – Gio67 Jan 3 '18 at 0:43