Q:

Context free language false proof

What is wrong with the following proof? 
Show whether $L$ is context-free or not, where $L = \left\{ a^nb^{2n}a^n | n \geq 0\right\}$
We know $\left\{a^nb^n | n \geq 0 \right\} $ and $\left\{b^na^n | n \geq 0 \right\} $ are context-free.  
Notice that, $\left\{a^nb^n \right\} $$\left\{b^na^n \right\} = \left\{ a^nb^{2n}a^n\right\}$
Since CFL's are closed under concatenation we can conclude that $L$ is also context-free.  

A:

$ab$ is in the concatenation language but not in $L$; do you see why? 
For a correct argument I suggest using the pumping lemma for context-free languages.