Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:(a,b)\rightarrow \mathbb{R}$ be a continuous function. Suppose $\exists k\in (0,1)$ such that $\forall x,y\in (a,b), f(kx+(1-k)y)≦kf(x)+(1-k)f(y)$.

Let $A=\{\lambda\in [0,1]|\forall x,y\in (a,b) , f(\lambda x + (1-\lambda)y)≦\lambda f(x) + (1-\lambda)f(y)\}$.

Then $A$ is dense in $[0,1]$.

I have proved that $f$ is convex when $k$ is a rational, but what if $k$ is irrational? (in ZF)

I constructed a sequence in $A$ which is convergent to some fixed $p$ in $(0,1)$ when $k\in \mathbb{Q}$, but there must be a better proof using the definition of $\epsilon-\delta$.

share|cite|improve this question
I don't see how you did it when $k\in\mathbb Q$, but I also fail to see why there would be any distinction from the general case. For example, $\mathbb Q+k\cap(0,1)$ would be a countable dense subset which contains $k$. – Asaf Karagila Nov 12 '12 at 18:20
@Asaf Since $[i,j\in A \Rightarrow ki+(1-k)j\in A]$, $A$ contains a countable dense subset (only when $k$ is a rational), and this is how i proved it for $k\in \mathbb{Q}$. – Katlus Nov 12 '12 at 18:28
And how do i conclude that $\mathbb{Q} + k\cap (0,1)$ is contained $in$ $A$? (I mean i cannot find a countable dense subset of $A$) – Katlus Nov 12 '12 at 18:31
I don't see how this is a problem when $k\notin\mathbb Q$. – Asaf Karagila Nov 12 '12 at 18:48
@Asaf Since I only knew the relation above in my comment, i had no information whether any rational is in $A$, so I couldn't choose an element for each $B(p,1/n)\cap A$ and construct a sequence – Katlus Nov 12 '12 at 19:03

I was so foolish that i couldn't think of this argument.

Fix $p\in (0,1)$ and $x,y \in (a,b)$.

Suppose $f(px+(1-p)y)>p f(x) + (1-p) f(y)$.

Let $\alpha = f(px+(1-p)y) - [p f(x) + (1-p) f(y)]$.

Then, there exists $\delta_1$ such thay $d(p,z)<\delta_1 \Rightarrow d(pf(x) + (1-p)f(y), zf(x)+(1-z)f(y))<\frac{\alpha}{2}$.

Also, there exists $\delta_2$ such that $d(p,z)<\delta_2 \Rightarrow d(f(px+(1-p)y),f(zx+(1-z)y))<\frac{\alpha}{2}$.

Let $\delta = \min \{\delta_1,\delta_2\}$.

Since $A$ is dense, $B(p,\delta)\cap A ≠ \emptyset$.

So, there exists $z\in A$ such that $f(zx+(1-z)y>zf(x) + (1-z)f(y)$. This leads a contradiction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.