The operator $A(L)= \{w \mid ww \in L\}$
As an amplification of sdcvvc's hint:  You can find a context-free language $L$ such that $A(L) = \{a^m b^m c^m \mid m\geq 0\}$, which as you probably know is not context-free.

So we want to find $L$ such that if $ww\in L$, then $w$ has the form $a^m b^m c^m$.  We might as well have $L$ be a sublanguage of $a^*b^*c^*a^*b^*c^*$.  Now we just need to put enough restrictions on the words in $L$ so that words of the form $ww$ in $L$ will have the form we require, while ensuring $L$ is still context-free. 

Partial spoiler:


  A word $a^{m_1} b^{m_2} c^{m_3} a^{m_4} b^{m_5} c^{m_6}$ has the form $ww$ if and only if $m_1 = m_4$, $m_2 = m_5$ and $m_3 = m_6$.  Now we can add some restrictions on the $m_i$ for words in $L$ to force them to all be equal in words of the form $ww$.  (Hint: we can require $m_1 = m_2$, but then we can't require $m_1 = m_3$ or $m_2 = m_3$, because then $L$ wouldn't be context-free anymore.  This isn't a problem though, because we can make use of some of the equalities we already have between the $m_i$.)