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Suppose $(x_n)$ is a positive sequence of reals converging to $x$. Furthermore we have a measure space $(E,\mathcal{E},\mu)$ given, with finite measure $\mu$. There are a measurables nonnegative and real valued maps $(f_{x_n})$ and a $f_x$ (depending on $x_n$ and $x$) such that for some $\epsilon >0$ given

$$\lim\sup_n \mu\{y\in E:|f_{x_n}(y)-f_x(y)|>\epsilon\}>\epsilon\tag{1}$$

Moreover

$$\lim\sup_n \mu\{y\in E:|f_{x_n}(y)-f_x(y)|>\epsilon,f_{x_n}(y)+f_x(y)\le \frac{1}{\epsilon}\}>\epsilon\tag{2}$$

Furthermore there is a strictly convex map $H$ given, hence

$$H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))<\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))\tag{3}$$

Now we should deduce from $(2)$ and $(3)$ the existence of $\delta>0$ such that

$$\lim\sup_n\mu\{H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))\le\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta\}>\delta$$

It is clear that I can find for very $n$ a $\delta(n)>0$ s.t.

$$H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))\le\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta(n)$$

However I'm struggling with the uniformity in $n$ and the measure of this event.

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have you solved the problem? I'm faced with a similar one and I'm interested in a solution. –  math Jun 17 '13 at 7:13
    
@math no sorry, so far I can not provide a complete solution. –  user20869 Jun 17 '13 at 7:14
    
I'm a bit confused by your notation. Why are you using $f_{x_n}$ and $f_x$? Would it make any difference to your question if you used $f_n$ and $f$? –  Tim Jun 17 '13 at 8:30
    
No I think not, the $x_n$ indicates, that $f_{x_n}$ depends on an element $x_n$. –  user20869 Jun 17 '13 at 11:17

1 Answer 1

up vote 1 down vote accepted

If $H$ is both strictly midpoint-convex and measurable then there exists a strictly increasing weak derivative $H'$.

Therefore for fixed $y$ the function $$ g_y(z) = \frac{H(z) + H(f(y))}2 - H\left(\tfrac12 (f(y)+z)\right) $$

Is strictly decreasing and continuous with $g_y(f(y)) = 0$ therefore $g_y$ has a continuous inverse and we may find some $\delta_y>0$ such that $g_y(z)\le\delta_y \Rightarrow \left|z-f(y)\right|\le\epsilon$.

Furthermore as $\mu$ is finite we may choose some $\delta <\frac 12\epsilon$ such that $\mu\{y\in E:\delta_y < \delta\}\leq\frac 12 \epsilon$.

So if $$H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))>\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta$$ we must have $g_y(f_{x_n}(y)) <\delta$ hence either $\left|f_{x_n}(y)-f(y)\right|\leq\varepsilon$ or $\delta_y<\delta$.

That is $$\begin{align} \{y\in E:g_y(f_{x_n}(y)) <\delta\} &\subset \{y\in E:\left|f_{x_n}(y)-f(y)\right|\leq\varepsilon\} \cup \{y\in E:\delta_y<\delta\}, \\ \{y\in E:g_y(f_{x_n}(y)) \geq \delta\} &\supset \{y\in E:\left|f_{x_n}(y)-f(y)\right|>\varepsilon\} \setminus \{y\in E:\delta_y<\delta\}. \end{align}$$

Therefore for there exist infinitely many $n$ such that $$\begin{align} \mu\{y\in E:g_y(f_{x_n}(y)) \geq \delta\} &\geq \mu\{y\in E:\left|f_{x_n}(y)-f(y)\right|>\varepsilon\} - \mu\{y\in E:\delta_y<\delta\}. \\ &\geq \epsilon - \frac 12\epsilon \\ &>\delta \end{align}$$ and $$\lim\sup_n\mu\{H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))\le\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta\}>\delta$$as required.

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