Mfrhtyj

Let continuous $f:Xtomathbb R$. $X$ is an interval of $mathbb R$.

Are the following two statements equivalent?

1) For almost every $xin X$ $exists$ open interval (neighborhood) $Ini x$ s.t. $f$ is either locally convex or locally concave on $I$. (i.e. only countable number of $xin X$ does not have such neighborhood).

$f$ is locally convex on $I$ if $forall x,yin I$, we have $f(lambda x+(1-lambda)y)leq f(lambda x)+f((1-lambda)y)$

(In another word, epi$f(I)$ is a convex set.)

2) $forall xin X$ $exists$ (non-singleton) closed interval $Ini x$ s.t. $f$ is either locally convex or locally concave over $I$.

If they are equivalent, then, in general, why people tend to use open set rather than closed set to define a neighborhood?

$2Rightarrow 1$. Without loss of generality we can assume that $X$ is an open subset of $Bbb R$. Let $X_{conv}$ (resp. $X_{conc}$) bet a set of all $xin X$ such that there exists an open interval $Ini x$ s.t. $f$ is locally convex (resp. locally concave) over $I$. Assume to the contrary that (1) does not hold, that is a set $X’=Xsetminus(X_{conv}cup X_{conc})$ is uncountable. For a natural $n$ we call a point $xin X’$ an $frac 1n$-corner provided
$(x-frac 1n,x)cap X’=varnothing$ or $(x,x+frac 1n)cap X’=varnothing$. It is easy to check that among any three distinct $frac 1n$-corners we can find two with the distance between them bigger than $frac 1n$. This implies that the set of $frac 1n$-corners is countable, so there exists a point $xin X’$ which is not an $frac 1n$-corner for any natural $n$. By (2), there exists a (non-singleton) closed interval $I’ni x$ s.t. $f$ is either locally convex or locally concave over $I’$. Then $f$ is either locally convex or locally concave over an open interval $I=operatorname{int} I’$. But $X’cap Inevarnothing$, a contradiction.

$1notRightarrow 2$. Define a function $f:Bbb RtoBbb R$ as follows. Put $ f(x)=0$, if $x=0$ or $x=frac 1{2n}$, with $ninBbb Zsetminus{0}$ or $|x|ge 1$. For $kinBbb Ncup{0}$ put $ fleft(pmfrac 1{4k+1}right)= frac 1{4k+1}$ and $ fleft(pmfrac 1{4k+3}right)=-frac 1{4k+3}$. Extend $ f(x)$ piesewise-linearly to the remaining $xinBbb R$. Since $f$ is linear at $Bbb R$ but a countable closed set, it satifies (1). On the other hand, on any open interval $Ini 0$, $f$ has countably infinite many zeros, so $f$ is neither convex, nor concave on $I$.

why people tend to use open set rather than closed set to define a neighborhood?

I don’t see a relation of this with the previous. For me a choice to work with open or closed neighborhoods usually is a matter of convenience. Maybe, to work with open neighborhoods
is often more convenient than with closed these.

By the way, a closed interval containing $x$ is not its neighborhood, because a neighborhood of $x$ must contain $x$ in its interior.