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?

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

  • $\begingroup$ 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. $\endgroup$ – Chris Jul 26 '19 at 10:48
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    $\begingroup$ 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)? $\endgroup$ – Chris Jul 26 '19 at 13:37
  • $\begingroup$ @Chris Yes, I had overlooked this. Well spotted. $\endgroup$ – Matthew Daws Jul 27 '19 at 6:41
  • $\begingroup$ @Chris Probably I am just being slow but I have to say that I don't see how Minkowski helps $\endgroup$ – Matthew Daws Jul 27 '19 at 8:17
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    $\begingroup$ 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 $X,Y=\mathbb{N}$. $\endgroup$ – Chris Jul 27 '19 at 15:51

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