Is it true that $\|A\circ \bar{A}\|_2\leq\|AA^{\dagger}\|_2$? Is it true that $\|A\circ \bar{A}\|_2\leq\|AA^{\dagger}\|_2$? Here $\circ$ is the Hardamard product and $\|•\|_2$ is the Frobenius norm.
 A: I'm assuming that $\|\cdot\|_2$ refers to the spectral norm, i.e. $\|A\|_2 = \sigma_1(A)$.  I am also assuming that $A^\dagger$ is the conjugate-transpose of $A$.
Yes, the inequality that you're looking for holds.  One way to prove this is to note that $A \circ \bar A$ is a principal submatrix of the Kronecker product $A \otimes \bar A$.  
That is, we have 
$$
\|A \otimes \bar A\|_2 = \sigma_1(A \otimes \bar A) = \sigma_1(A) \sigma_1(\bar A) = \sigma_1(A)^2 = \|A\|_2^2
$$
Then, we note that there exists a matrix $P$ such that $P(A \otimes \bar A)P^\dagger = A \circ \bar A$, and that $P$ satisfies $PP^\dagger = I$, which implies that $\|P\| = \|P^\dagger\| = 1$.  It follows that
$$
\|A \circ \bar A\|_2 = \|P(A \otimes \bar A)P^\dagger\|_2 \leq 
\|P\|_2 \cdot \|A \otimes \bar A\|_2 \cdot \|P^\dagger\|_2 = \|A \otimes \bar A\|_2 = \|A\|_2^2
$$
Finally, we note that $\|AA^\dagger\|_2 = \|A\|_2^2$ as well.  So, all together, we have
$$
\|A \circ \bar A\|_2 \leq \|A \otimes \bar A\|_2 = \|A\|_2^2 = \|AA^\dagger\|_2
$$
which means that the desired inequality holds.
