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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 $\epsilon-\delta$.

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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
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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

1 Answer 1

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.

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