Relationship between operator norms For a matrix $A$, define the operator $\ell_p$-norm of $A$ to be
$$
\|A\|_p = \sup_{x \neq 0} \frac{\|Ax\|_p}{\|x\|_p}.
$$
Here $\|x\|_p$ denotes the $\ell_p$ norm of the vector $x$.
For $1 \le p \le q \le 2$ and $x \in \mathbb{R}^n$, we know that $\|x\|_q \le \|x\|_p \le n^{1 / p - 1 / q} \|x\|_q$. 
Is there any similar conclusion for the operator $\ell_p$-norm of a given matrix $A \in \mathbb{R}^{n \times m}$?
Or a more concrete problem: if we know $\|A\|_1$ and $\|A\|_2$, what is the best possible upper bound we can achieve for $\|A\|_p$ if $1 \le p \le 2$?
E.g., 
$$\|Ax\|_p \le n^{1 / p - 1 /2} \|Ax\|_2 \le n^{1 / p - 1 / 2} \|A\|_2 \cdot \|x\|_2 \le n^{1 / p - 1 / 2} \|A\|_2 \cdot \|x\|_p,$$
which implies
$$
\|A\|_p \le n^{1 / p - 1 / 2} \|A\|_2.
$$
Similarly, 
$$\|Ax\|_p \le \|Ax\|_1 \le \|A\|_1 \cdot \|x\|_1 \le m^{1 - 1 / p} \|A\|_1 \cdot \|x\|_p,$$
which implies
$$
\|A\|_p \le m^{1 - 1 / p} \|A\|_1.
$$
Combine them we have $\|A\|_p \le \min\{n^{1 / p - 1 / 2} \|A\|_2, m^{1 - 1 / p} \|A\|_1\}$, which is quite naive. 
Are there any tighter upper bounds?
 A: Your estimate is not naive in general. With $p=1$, $q=2$, take 
$$
A=\begin{bmatrix} 
1&0&\cdots&0\\ 
1&0&\cdots&0\\ 
\vdots&\vdots&\ddots&\vdots\\
1&0&\cdots&0\\ 
\end{bmatrix},
$$
It is well-known that 
$$\|A\|_1=\max\{\|A_j\|_1:\ j\},\ \ \ \|A\|_2=\|A^*A\|_2^{1/2}=\max\sigma(A^*A)^{1/2},$$ where $A_j$ denotes the $j^{\rm th}$ column of $A$, and $\sigma(B)$ is the spectrum (i.e., the list of eigenvalues). 
Thus $\|A\|_1=n$, while $$\|A\|_2=\|A^*A\|_2^{1/2}=\left\|\begin{bmatrix} n&0&\cdots&0\\0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{bmatrix}\right\|^{1/2}_2=n^{1/2}. 
$$
So
$$
\|A\|_1=n^{1/1-1/2}\,\|A\|_2.
$$
A: Using H\"older's inequality, we can prove that these norms are equivalent. Let $1\le p\le q\le +\infty$. Then
$$
\|Ax\|_p \le n^{1/p-1/q} \|Ax\|_q \le n^{1/p-1/q} \|A\|_q\|x\|_q\le n^{1/p-1/q} \|A\|_q\|x\|_p, 
$$
hence
$$
\|A\|_p \le n^{1/p-1/q}\|A\|_q.
$$
Similarly,
$$
\|A\|_q \le m^{1/p-1/q}\|A\|_p.
$$
Equality holds in one of the inequalities if $A$ contains a column (row) of all ones, with the remaining entries zero:
$$
A=\pmatrix{1& 0 & \dots & 0 \\ \vdots &\vdots&&\vdots\\1& 0 & \dots & 0}.
$$
Take $x\ne0$, then $\|A\|_p = |x_1| n^{1/p}$. Taking the supremum over $\|x\|_p\le 1$ yields $\|A\|_p = n^{1/p}$. And the first inequality holds with equality.
Equality holds in both inequalities for $p\ne q$ only if $n=m$ for the same matrix $A$ as above.
