Vallée Poussin's Theorem on Uniform Integrablity

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.

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

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 $$k, $$\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.