If $f$ is continuous, why is $f$ with the property $$f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y),$$ where $0\le x,y\le 1$ is convex?
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Martin Sleziak
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derivative
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I think there are two different questions here : 1) If $f$ is continuous and "midpoint convex" then $f$ is convex. 2) If $f$ is "midpoint convex", then $f$ is continuous. I think the OP meant the question 1) and not question 2). – user37238 Dec 02 '13 at 13:12
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@dfeuer You messed up this question with your edit. Be more careful when you edit questions. – TMM Dec 02 '13 at 13:15
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2@TMM, the question was and is ambiguous. I'm sorry if I guessed wrong. – dfeuer Dec 02 '13 at 13:42
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Sorry Guys, It was my fault, i tried to post it with the smart phone. – derivative Dec 02 '13 at 13:45
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See also http://math.stackexchange.com/questions/83383/showing-that-f-is-convex-given-that-f-fracxy2-le-fracfxfy2 – Martin Sleziak Nov 12 '15 at 19:49
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Turn it around, prove:
If $f$ is continuous and not convex, then it is not midpoint convex.
By assumption, we have $x<y$ and $0 < t < 1$ with
$$f(tx+(1-t)y) > tf(x) + (1-t)f(y).$$
Let $z = tx + (1-t)y$, and on $[x,y]$, consider the function
$$h(w) = f(w) - \frac{y-w}{y-x}f(x) - \frac{w-x}{y-x}f(y).$$
We have $h(x) = h(y) = 0$, and by assumption, $h(z) > 0$. Let $a = \sup \{ x\leqslant w < z : h(w) = 0\}$ and $b = \inf \{ z < w \leqslant y : h(w) = 0\}$.
Then by continuity we have $h(a) = h(b) = 0$, $a < z < b$, and $h > 0$ on $(a,b)$.
Therefore, $h\left(\frac{a+b}{2}\right) > 0$, and since $h(a) = h(b) = 0$, that implies
$$f\left(\frac{a+b}{2}\right) > \frac12 f(a) + \frac12 f(b).$$
Daniel Fischer
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Is ''not convex'' equivalent to ''concave'' ? So the first inequality always holds for a not convex function ? – derivative Dec 02 '13 at 13:33
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1No. (Strictly) Concave means the first inequality always holds. "Not convex" means there is at least one combination $x,y,t$ such that the inequality holds. In general, a function is neither globally convex nor globally concave. Generally, a nicely behaved function is convex on some subintervals of its domain, and concave on others. – Daniel Fischer Dec 02 '13 at 13:36
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Thanks a lot Daniel for providing a different and much easier approach through contradiction. For long time I have known the traditional answer to this question which uses induction to prove that $f(ax + by) \leq af(x) + bf(y)$ for positive rationals $a, b$ with $a + b = 1$ and then uses continuity to remove the restriction of "rational" $a, b$. This direct traditional proof is much longer compared to what you have done. +1 to you! – Paramanand Singh Dec 02 '13 at 16:57
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By induction we can prove that: if $k,m, l\in\mathbb{N} , k+m=2^l , x,y\in \mbox{domain} f $ then $$f\left( \frac{k}{2^l} \cdot x +\frac{m}{2^l} \cdot y \right)\leq \frac{k}{2^l} \cdot f(x) +\frac{m}{2^l} \cdot f(y). $$ Indeed the asertion is true when $l=1 .$ Suppose that it is true for some $l\geq 1 .$ And let $k+m=2^{l+1} , k=2^l +s , m=2^l-s ,s\in\mathbb{N} . $ We have: $$f\left( \frac{k}{2^{l+1}} \cdot x +\frac{m}{2^{l+1}} \cdot y \right) =f\left( \frac{1}{2} \cdot x +\frac{1}{2}\cdot \left(\frac{s}{2^l} \cdot x+\frac{2^l -s}{2^l} \cdot y\right) \right) \leq \frac{1}{2} \cdot f(x) +\frac{1}{2} \cdot f\left(\frac{s}{2^l} \cdot x+\frac{2^l -s}{2^l} \cdot y\right)\leq \frac{1}{2} \cdot f(x) +\frac{1}{2} \cdot \frac{s}{2^l} \cdot f(x)+\frac{1}{2}\cdot\frac{2^l -s}{2^l} \cdot f(y) = \frac{k}{2^{l+1}} \cdot f(x) +\frac{m}{2^{l+1}} \cdot f(y) .$$ Hence by Induction the asertion holds true for any $l\in\mathbb{N} .$
Now let $1>\alpha >0 , $ and let $\frac{k_l}{2^l} \rightarrow \alpha $ as $l\to \infty .$ Since $f$ continuous we have $$f(\alpha x +(1-\alpha )y ) =\lim_{l\to\infty } f\left(\frac{k_l}{2^l} \cdot x + \left(1-\frac{k_l}{2^l} \right) y\right) \leq \lim_{l\to\infty } \left(\frac{k_l}{2^l} \cdot f(x) + \left( 1-\frac{k_l}{2^l} \right) f(y)\right) =\alpha f(x) +(1-\alpha )f(y) .$$