Converse of pumping lemma for regular expressions
Yes, $L_&gt;=\{0^i1^j: i &gt; j &gt; 0\} $ is a non-regular language.

However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_&gt;$. Consider $w=0^{p+1}1^p$. Let $w=xyz$ with $|y|&gt;1$ and $|xy|&lt;p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.

The most we can say is that all words in $L_&gt;$ can be pumped up.


  If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE? 


No. A simple counterexample is $L=\Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $\{a^{n^2}\mid n\in \Bbb N_0\}$. $LL'$ is still the language of all words, which is regular.

Exercise 1. Find two languages $A$ and $B$ such that 


$A$ is regular
$B$ is not regular
$A\subsetneq B$
$AB$ is regular.


Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.