TITLE: The norm of tensor product operator on Lp spaces
QUESTION [6 upvotes]: Let $X, Y$ be two $\sigma$-finite measure spaces and $p,q\in [1,\infty]$. Let $T_1, T_2:L^p(X)\rightarrow L^q(Y)$ be two bounded linear operators. Then one can define a linear operator $$T_1\otimes T_2: L^p(X)\otimes L^p(X)\rightarrow L^q(Y)\otimes L^q(Y).$$ Here $L^p(X)\otimes L^p(X), L^p(Y)\otimes L^p(Y)$ are subspaces of the Banach spaces $L^p(X\times X), L^p(Y\times Y)$ respectively and equipped with the norms of the latter.
My questions is: would the following equality of operator norms hold? If yes, is there a quick proof? $$\Vert T_1\otimes T_2\Vert=\Vert T_1\Vert \cdot \Vert T_2\Vert$$ 
(When $p=q=2$ and $X, Y$ are finite sets, one can easily show the equality using the property of normal transformations or singular values of matrices).
Edit: When $p\leqslant q$, the above equality is true, as shown in the discussion below. So the question becomes: does the equality also hold when $p>q$, or is there an obvious counter-example?

REPLY [2 votes]: The equality does not need to hold for $p>q$, due to the following
counterexample, which is based on the counterexample from
this article.
We choose $q=1$, $p=\infty$, $X=Y=\{1,2\}$ (equipped with the counting measure) and
$$T=T_1=T_2=
\begin{bmatrix}
    1 & -1 \\ 1 & 2
\end{bmatrix}
: L^p(X)\to L^q(Y).
$$
Then one can show $\|T\|=\|T e_2\|_1 = 3$.
We consider $f:=(e_1+e_2)\otimes (e_1+e_2)-e_1\otimes e_1\in L^p(X)$.
Then one can calculate
$$
(T\otimes T) f = 
9 e_2\otimes e_2
- (e_1+e_2)\otimes (e_1+e_2).
$$
Due to $\|f\|_\infty=1$ we have
$$
\|T\otimes T\| \geq
\|(T\otimes T)f \|_1
= 11 > 9 = \| T\| \cdot \|T\|.
$$
For continuity reasons, this counterexample works also for small $q>1$
and large $p<\infty$.