enter image description here I've started to read Rao's Theory of Orlicz Spaces book. There are two points I could not get well at proof of Vallée Poussin's uniform integrablity theorem. Please find the theorem above.

The two points I referred are in $(ii) \Rightarrow (iii)$. I know they are so easy but I need to see them clearly for understanding. enter image description here

First, I cannot see why $$\varphi(k)\frac{x-k}{x}\leq \frac{\Phi(x)}{x} \ \ \ \text{when} \ \ k<x \ \ \text{and} \ \ k\to \infty$$

Second, I couldn't show that $$\Phi(\alpha x+\beta y) = \int_0^{|\alpha x + \beta y|} \varphi(t)dt\leq \alpha\int_0^{|x|} \varphi(t)dt + \beta\int_0^{|y|} \varphi(t)dt = \alpha\Phi(x)+\beta\Phi(y)$$

Thanks in advance for any explanation.


For the first part, this is two separate statements: first that for $k<x$, $\phi(k)\frac{x-k}{x}\leq\frac{\Phi(x)}{x}$, and second, that as $k\to\infty$ the LHS $\to\infty$ so that the RHS must as well.

For the second statement, we have already that $\phi(n)\to\infty$ with $n$. Then for any $k$, $$\lim_{k\to\infty}\lim_{x\to\infty} \phi(k)\frac{x-k}{x} = \lim_{k\to\infty} \phi(k)=1.$$

The first claim is a little weirder. But note that $\Phi(x)/x$ is exactly the mean value of $\phi(k)$ on the interval $[0,x]$, and also that $\frac{x-k}{x}=\frac{\lvert[k,x]\rvert}{\lvert[0,x]\rvert}$ is the fraction of the interval remaining once you've reached $k$. Recall that $\phi$ is increasing, and think geometrically: $\phi(k)\frac{x-k}{x}$ must be an underestimate of the mean on the whole interval, since $$\phi(k)\frac{x-k}{x}=\frac{\phi(k)}{x}\int_k^x 1\,dt\leq\frac{1}{x}\int_k^x \phi(t)\,dt\leq \frac{1}{x}\int_{0}^x\phi(t)\,dt=\frac{\Phi(x)}{x}.$$

Here, in the first inequality we have used that $\phi$ is increasing, and in the second that it is positive.

OK, now let's deal with the convexity. Note that since $$\Phi'(x)= \operatorname{sgn}(x)\,\phi(\lvert x \rvert),$$ then for $x\neq0$ the second derivative $$\Phi'(x)=0+(\operatorname{sgn}(x))^2\,\phi'(\lvert x \rvert)$$ is always positive, since $\phi'$ is positive.

  • 1
    $\begingroup$ Thanks a lot for your beneficial answer. Is showing convexity with integrals possible? $\endgroup$
    – user519955
    Dec 2 '19 at 17:11
  • $\begingroup$ You know, I tried showing it and didn't come up with much. I've edited to show that the second derivative is positive (which as I am sure you know implies convexity) but I haven't figured out what the author means in the sentence with "change of variables" -- I feel like it should be straightforward, though?? $\endgroup$
    – cwindolf
    Dec 2 '19 at 19:43

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