Infinite matrix defining bounded linear operator on $\ell^p$ Am I right that if I have an infinite complex matrix that defines a bounded linear operator on $\ell^p$, then its conjugate transpose does not necessarily define a bounded linear operator on the same $\ell^p$? In fact, I am only considering infinite matrices in which the nonzero entries are all in a band of finite width about the diagonal.
I feel like the matrix needs to satisfy some conditions on the sequence of norms of its rows/columns (regarded as vectors in $\ell^p$ or $\ell^q$ where $\frac{1}{p}+\frac{1}{q}=1$) in order for the conjugate transpose to also be a bounded linear operator on the same space. But I don't have a definitive answer either way.
 A: An infinite matrix $A$ satisfies a bound 
$$\|Ax\|_p \le C\|x\|_p \tag1$$
if and only if it satisfies the bilinear bound
$$|y^*Ax| \le C\|x\|_p\|y\|_q \tag2$$
(with same $C$ in both equations). Here   $1/p+1/q=1$ and ${}^*$ stands for conjugate transpose. Rewriting $(2)$ as
$$|x^*A^*y| \le C\|x\|_p\|y\|_q \tag3$$
we see that it is also equivalent to 
$$\|A^*y\|_q \le C \|y\|_q\tag4$$
Simply put, the adjoint of a bounded operator is a bounded operator on the dual space. As David C. Ullrich said, there is no reason for it to be bounded on the original space.
For a concrete example with $A^*$ not bounded on the same $\ell^p$ space, let $p=4/3$ and 
$$
A = \begin{pmatrix} 1 & 1/\sqrt{2} & 1/\sqrt{3} & 1/\sqrt{4}  & 1/\sqrt{5} & \cdots \\
0 & 0 & 0 & 0 & 0 & \cdots \\
0 & 0 & 0 & 0 & 0 & \cdots 
\end{pmatrix}
$$
where all rows except the first are zero. This is  a bounded rank-1 operator on $\ell^{4/3}$, because the first row entries are a vector from the dual space $\ell^{4}$. However, trying to apply the transpose $A^*$ on $\ell^{4/3}$, we see that it maps the first basis vector to $(1,  1/\sqrt{2},   1/\sqrt{3},  1/\sqrt{4}, 1/\sqrt{5},\dots)$ which is not an element of $\ell^{4/3}$.
