# 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?

• Check sections 7 and 26 in the book Tensor Norms and Operator Ideals of Defant and Floret. – Jochen Wengenroth Jul 26 '19 at 7:06
• Jochen, thanks for recommending the reference book. – Chris Jul 26 '19 at 13:39

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.

• Thank you. It's clever to decompose $T_1\otimes T_2$ as the composition of two simpler tensors. I did not follow the same line and was stuck somewhere similar to the possible unbounded situation of operator $S\cdot x$, which you pointed out. – Chris Jul 26 '19 at 10:48
• Matthew, the argument utilizes an intermediate space $L^p(X)\otimes L^q(Y)$, which is a subspace of $L^p(X,L^q(Y))$ and $L^q(Y,L^p(X))$ at the same time. However, the latter two spaces give the former space two different norms. So there seems to be a hidden map $I_{p,q}$ that maps $L^p(X)\otimes L^q(Y)$ equipped with the first norm to the same space with the second norm. In order for $$T_1\otimes T_2=(T_1\otimes 1)\cdot I_{p,q}\cdot (1\otimes I_2)$$ to be bounded, $I_{p,q}$ needs to be bounded-this seems to require $p\leqslant q$(, due to the condition of Minkowski's integral inequality)? – Chris Jul 26 '19 at 13:37
• @Chris Yes, I had overlooked this. Well spotted. – Matthew Daws Jul 27 '19 at 6:41
• @Chris Probably I am just being slow but I have to say that I don't see how Minkowski helps – Matthew Daws Jul 27 '19 at 8:17
• Matthew, for $h(x,y)\in L^p(X)\otimes L^q(Y)$, let $\Vert h\Vert_{p,q}$ be the norm inherited from $L^p(X,L^q(Y))$ and $\Vert h\Vert_{q,p}$ the norm related to $L^q(Y,L^p(X))$. Set $H(-)=|h(-)|^p$, then $$\Vert h\Vert_{q,p}=(\int_X\Vert h(\cdot,y)\Vert_{L^p(X)}^q dy)^{\frac{1}{q}}=\Vert\int_X H(x,\cdot)dx\Vert_{L^\frac{q}{p}(Y)}^\frac{1}{p}.$$ When $r=\frac{q}{p}\geqslant 1$, $\Vert . \Vert_{L^r(Y)}$ is a norm and the Minkowski yields $$\cdots \leqslant (\int_X \Vert H(x,\cdot)\Vert_{L^r(Y)}dx)^{\frac{1}{p}}=\Vert h\Vert_{p,q}.$$ If $p>q$, $I_{p,q}$ could be unbounded, say $X,Y=\mathbb{N}$. – Chris Jul 27 '19 at 15:51