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Let $f:I \rightarrow \mathbb{R}$, where $I\subset \mathbb{R}$ is an interval, be midconvex, that is $$f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$$ for all $x,y \in I$. Assume that for some $x_0, y_0 \in \mathbb{R}$ such that $x_0 < y_0$ holds equality $$f\left(\frac{x_0+y_0}{2}\right)= \frac{f(x_0)+f(y_0)}{2}.$$ What can we say about $f$ restricted to $[x_0, y_0]$ ? Maybe $f|_{[x_0,y_0]}$ is a sum of some additive function and some constant?

Thanks.

Added.

Maybe then $f(qx_0+(1-q)y_0)=qf(x_0)+(1-q)f(y_0)$ for $q=\frac{k}{2^n}$, where $n \in \mathbb{N}$, $k=0,1,...,2^n$ ?

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Note that $c_b(qx+ry)=qc_b(x)+rc_b(y)$ for any $q,r\in\mathbb Q$, hence Robert Israel's example fulfills the condition, that you've added in your recent edit, too. –  Martin Sleziak Dec 7 '11 at 14:43
    
Thanks for comment. Yes $f$ satisfies this condition. But I look for a proof in general case or for counterexample. Continuous midconvex (that is convex) functions have such property for all $q \in [0,1]$. I want to know what is for midconvex function. It was a reason for my question. –  Richard Dec 7 '11 at 21:46
    
So you are asking (in the edited question) whether $f\left(\frac{x_0+y_0}{2}\right)= \frac{f(x_0)+f(y_0)}{2}$ implies $f(qx_0+(1-q)y_0)=qf(x_0)+(1-q)f(y_0)$ for $q=\frac{k}{2^n}$? ...Maybe I misunderstood something. –  Martin Sleziak Dec 7 '11 at 21:53
    
Yes about this condition. –  Richard Dec 7 '11 at 22:01

2 Answers 2

up vote 2 down vote accepted

If I understand your new question correctly (after the addition in your question and some clarification in the comments) you want to know, whether the following is true:

Let $f$ be a midpoint convex (a.k.a. Jensen convex) function on $[x_0,y_0]$, i.e. for any $x,y\in[x_0,y_0]$ $$ f\left(\frac{x+y}2\right)\le \frac{f(x)+f(y)}2 \qquad (1)$$ holds. Suppose that, moreover, $$ f\left(\frac{x_0+y_0}2\right)=\frac{f(x_0)+f(y_0)}2 \qquad (2).$$ Then $f(qx_0+(1-q)y_0)=qf(x_0)+(1-q)f(y_0)$ for $q=\frac{k}{2^n}$, where $n \in \mathbb{N}$, $k=0,1,...,2^n$.


This can be shown by induction on $n$.

For $n=1$ the claim is equivalent to (1).

For $n=2$ we want to show that $f\left(\frac{x_0+3y_0}4\right)=\frac{f(x_0)+3f(y_0)}4$ and $f\left(\frac{3x_0+y_0}4\right)=\frac{3f(x_0)+f(y_0)}4$. Using (1) for the points $\frac{x_0+3y_0}4$ and $\frac{3x_0+y_0}4$ we get $$2f\left(\frac{x_0+y_0}2\right) \le f\left(\frac{x_0+3y_0}4\right) + f\left(\frac{3x_0+y_0}4\right). \qquad (3)$$ We also get from (1) that $$2f\left(\frac{x_0+3y_0}4\right) \le f\left(\frac{x_0+y_0}2\right) + f(y_0) \qquad (4)$$ and $$2f\left(\frac{3x_0+y_0}4\right) \le f(x_0)+f\left(\frac{x_0+y_0}2\right) \qquad (5) $$ Combining (3), (4), (5) we get that $$4f\left(\frac{x_0+y_0}2\right) \le 2f\left(\frac{x_0+3y_0}4\right) + 2f\left(\frac{3x_0+y_0}4\right) \le f(x_0)+2f\left(\frac{x_0+y_0}2\right)+f(y_0).$$ But since from (2) we know that in fact the equality holds for the leftmost and rightmost expression, we get that equality holds in (4) and (5), too. This implies our claim for $n=2$.

Suppose that the claim is true for $n$, we will show it for $n+1$: Let $q=\frac{k}{2^{n+1}}$. If $k$ is even, the claim follows from induction hypothesis. If $k$ is odd, we can use what we proved in the case $n=2$, using either $\frac{k-1}{2^{n+1}}$ and $\frac{k+3}{2^{n+1}}$ or $\frac{k-3}{2^{n+1}}$ and $\frac{k+1}{2^{n+1}}$ instead of $x_0$ and $y_0$. (We can choose whichever pair belongs to the original interval. For these point the claim is true, since $k\pm1$, $k\pm 3$ are even.)


The proof is basically a modification of the proof that every midpoint convex function is rationally convex, see this question: Showing that $f$ is convex (You can try whether other proofs mentioned there can be modified to you situation too.)

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Very thanks for answer. –  Richard Dec 7 '11 at 23:19

Consider the following example. Let $B$ be a Hamel basis of $\mathbb R$ over the rationals $\mathbb Q$. Thus each $x \in \mathbb R$ can be written uniquely as $x = \sum_{b \in B} c_b(x) b$, where all but finitely many of the $c_b(x)$ are $0$ for any particular $x$. Let $f(x) = c_{b_0}(x)^2$ for some given $b_0 \in B$. Since $c_b\left(\frac{x+y}{2}\right) = \frac{c_b(x) + c_b(y)}{2}$, $f$ is midpoint-convex. For any $x_0$ and $y_0$ such that $c_b(x_0) = c_b(y_0) = 0$ we have $f\left(\frac{x_0+y_0}{2}\right) = \frac{f(x_0)+f(y_0)}{2} = 0$. But there is a dense set of points $ y_1$ where $f\left(\frac{x_0+y_1}{2}\right) < \frac{f(x_0)+f(y_1)}{2} $

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Thanks for answer. –  Richard Dec 7 '11 at 23:24

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