The norm of tensor product operator on Lp spaces 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? 
 A: For a Banach space $E$, let $L^p(X,E)$ be the completion of the simple functions $X\rightarrow E$ with the norm $\|f\| = \big(\int_X \|f(x)\|^p\big)^{1/p}$ (which reduces to a sum, as $f$ is simple).  If $g\in L^p(X)$ is a simple function, and $x\in E$, then let $f=g\cdot x$ be the simple function $X\rightarrow E; t\mapsto g(t)x$.  Then $\|f\| = \|g\| \|x\|$ and so by continuity, we can extend this definition to $g\cdot x$ for any $g\in L^p(X)$.  Thus $L^p(X)\otimes E$, the algebraic tensor product, can be identified as a dense (we get all simple functions) subspace of $L^p(X,E)$.
Given $T:E\rightarrow F$ a bounded linear map, it is easy to see that $f\cdot x\mapsto f\cdot T(x)$ is bounded, of norm $\|T\|$, and so extends to $L^p(X,E)$.  Denote this by $1\otimes T$.
In general, for $S:L^p(X)\rightarrow L^p(Y)$ a bounded linear map, the map $f\cdot x\mapsto S(f)\cdot x$ need not be bounded, and it is an interesting question to determine when this is bounded.
Edit: This next argument requires $p=q$ while the original question wants to consider the general case, which at the moment I'm not sure I can say much about.
In our case, though, things become easier.  If we set $E=L^p(X)$ then $L^p(X, E)$ is isometrically isomorphic to $L^p(X\times X)$.  Given your $T_1,T_2$ we first form $1\otimes T_2$.  We then swap around the roles of $L^p(X)$ and $E$, and form $T_1\otimes 1$ (with the obvious notation) which is also bounded.  The composition of these maps is exactly $T_1\otimes T_2$.  So, yes, this is bounded, with norm at most $\|T_1\| \|T_2\|$ (with then obvious equality).
As Jochen Wengenroth says, Defant and Floret is a great resource for more on this, and in particular, for details about my comment about $S$ above.
A: 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 $p<\infty$.
