# Is the natural map $L^p(X) \otimes L^p(Y) \to L^p(X \times Y)$ injective?

Let $X,Y$ be $\sigma$-finite measure spaces, and let $L^p(X) \otimes L^p(Y)$ be the algebraic tensor product. The product has a natural map into $L^p(X \times Y)$ which takes $\sum a_{ij} f_i \otimes g_j$ to the function $F(x,y) = \sum a_{ij} f_i(x) g_j(y)$. A moment's thought shows that this map is well-defined. Is it also injective?

It seems that this should be true, but I can't see how to prove it. Intuitively, one needs to show that if $\sum a_{ij} f_i(x) g_j(y) = 0$ a.e., then one should be able to cancel all the terms in the sum using bilinearity. It is not quite clear how to do this without knowing anything about the terms.

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OMG! this is as good as it gets, we got LaTeX in question's title!! – Pratik Deoghare Aug 5 '10 at 13:24
Actually, it's not. It lags the loading of the main page, which is why we decided against it on MO. – user126 Aug 6 '10 at 0:20
Could you imagine if we had to wait for our recreational math site to load?! The horror! – user641 Aug 9 '10 at 7:32

The previous version of this answer was incorrect. It has generously been accepted by Nate Eldredge on the ground that "it definitely contains the right idea". This previous version has better remain forgotten forever. The reader should consult Nate's answer, which is correct and complete. I'll just try here to spell out the underlying linear algebra lemma (which is trivial, but which I failed to see).

UNDERLYING LEMMA. Let $U,V,W$ vector spaces over some field $K$, and $\alpha:U\otimes V\to W$ a linear map. Assume that for each nonzero vector $u_0$ in $U$ there are linear maps $\beta:W\to V$ and $\gamma:U\to K$ such that $\gamma(u_0)\not=0$ and $$\beta(\alpha(u\otimes v))=\gamma(u)v\ \forall\ u,v.$$ Then $\alpha$ is injective.

The proof can be very easily extracted from Nate's answer. Here is the most standard application:

APPLICATION. Let $U,V,W$ be respectively the spaces of $K$-valued functions on the sets $X,Y,X\times Y$, and $\beta,\gamma$ are given by $(\beta(w))(y)=w(x_0,y)$, $\gamma(u)=u(x_0)$ where $x_0$ is chosen so that $u_0(x_0)\not=0$.

EDIT. Here is a mild generalization.

Let $X$ be a measure space, let $\mathcal V(X)$ be the vector space of all measurable complex valued functions on $X$, let $\mathcal N$ be the subspace of functions vanishing almost everywhere, and let $V(X)$ be the quotient. [Is there a standard notation for these spaces?]

Let $Y$ be another measure space. The bilinear map from $\mathcal V(X)\times\mathcal V(Y)$ to $\mathcal V(X\times Y)$ sending $(f,g)$ to $(x,y)\mapsto f(x)g(y)$ induces a linear map $\Phi$ from $V(X)\otimes V(Y)$ to $V(X\times Y)$.

THEOREM. Assume $X$ and $Y$ are $\sigma$-finite. Then $\Phi$ is injective.

This slightly more general than the statement proved in Nate's answer because no integrability assumptions are made on the functions.

LEMMA. Let $f_i,\dots,f_n$ be in $\mathcal V(X)$, and $S\subset X^n$ the set of $x\in X^n$ such that $\det(f_i(x_j))\not=0$. Then the $f_i$ are linearly independent in $V(X)$ if and only if the measure of $S$ is positive.

Denote by $|A|$ the measure of $A$, and by $S_g$ the support of the function $g$, that is the set of points where it is nonzero. Put $d(x):=\det(f_i(x_j))$ for $x\in X^n$. The above set $S$ is now $S_d$.

To prove the if part of the lemma, assume the $f_i$ are linearly dependent in $V(X)$ and check $|S_d|=0$ as follows. Let $\lambda$ be a nonzero vector of $\mathbb C^n$ such that $g:=\sum \lambda_i f_i$ satisfies $|S_g|=0$. Denote again by $\lambda$ the 1 by $n$ matrix formed by the $\lambda_i$, and by $A(x)$ the matrix $(f_i(x_j))$. For $x$ in $(S_g^c)^n$, where the superscript c mean "complement", we have $\lambda A(x)=0$. Multiplying on the right by the adjugate of $A(x)$ we get $d(x)=0$. This shows that $S_d$ is contained in the measure zero subset $((S_g^c)^n)^c$ of $X^n$.

To prove the only if part of the lemma, assume $|S_d|=0$ and check that the $f_i$ are linearly dependent in $V(X)$ as follows. Expand $d(x)$ as $$d(x)=\sum\ d_i(x')\ f_i(x_n)$$ with $x':=(x_1,\dots,x_{n-1})$. Arguing by induction, we can assume $|S_{d_i}|>0$ for all $i$. Let $U$ be the union of the $S_{d_i}$. Denoting by $d(x',\bullet)$ the function $x_n\mapsto d(x)$, we have $$S_d=\bigcup_{x'\in X^{n-1}}\ {x'}\times S_{d(x',\bullet)}.$$ Fubini yields $$0=|S_d|=\int_{X^{n-1}}\ |S_{d(x',\bullet)}|\ dx'.$$ If the $f_i$ were linearly independent in $V(X)$, we would have $|S_{d(x',\bullet)}|>0$ for all $x'$ in $U$, a contradiction. The lemma is proved.

Let's prove the theorem. In the next lines, $i,j,k$ will lie between 1 and $m$, whereas $p,q,r$ will lie between 1 and $n$.

Let $f_1,\dots,f_m$ be in $\mathcal V(X)$; let $g_1,\dots,g_n$ be in $\mathcal V(Y)$; and put $F_{ip}(x,y):=f_i(x)g_p(y)$.

The theorem is equivalent to the statement that the linear independence of the $f_i$ and $g_p$ in $V(X)$ and $V(Y)$ implies that of the $F_{ip}$ in $V(X\times Y)$.

Let $S\subset X^m$ and $T\subset Y^n$ be defined by the conditions $\det(f_i(x_j))\not=0$ and $\det(g_p(y_q))\not=0$. Embedd $U:=S^n\times T^m$ into $(X\times Y)^{mn}$ in the obvious way, and denote the $(i,p)$-th coordinate of $z\in(X\times Y)^{mn}$ by $z_{ip}=(x_{pi},y_{ip})$. Then $z$ is in $U$ iff $x_p\in S$ and $y_i\in T$ for all $(i,p)$, that is iff $$\det\Big(f_j(x_{pk})\Big)_{jk}\not=0\not =\det\Big(g_q(y_{ir})\Big)_{qr}$$ for all $(i,p)$. We claim that the validity of this double nonequality for all $(i,p)$ implies the nonequality $$\det\Big(f_i(x_{qj})\ g_p(y_{jq})\Big)_{(i,p)(j,q)}\not=0.$$ In view of the lemma, the theorem is equivalent to the claim. But the claim doesn't depend on the measures defined on $X$ and $Y$. As the theorem holds when $X$ and $Y$ are equipped with the counting measure (see the application of the "underlying lemma" at the beginning of the post), we are done.

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EDIT: Here is a cleaned-up and corrected version of this answer, based on Pierre-Yves' suggestion (thanks!). His answer above contains a much more complete version.

If $\sum_{i=1}^n a_{i} f_i \otimes g_i$ is not the zero element of $L^p(X) \otimes L^p(Y)$, we may assume without loss of generality that the $f_i$ are linearly independent. We can also assume that $a_1 \ne 0$ and $g_1 \ne 0$.

Suppose that the corresponding function $F(x,y) = \sum_{i=1}^n a_{i} f_i(x) g_i(y) = 0$ a.e. Since $g_1 \ne 0$, there is a measurable $B \subset Y$ of positive finite measure such that $\int_B g_1 \ne 0$ (the integral is finite by Hölder). Then by Fubini's theorem, for a.e. $x$ we have $$0 = \int_{B} F(x,y)dy = \sum_{i=1}^n a_i \left(\int_{B} g_i\right) f_i(x).$$ This contradicts the assumed linear independence of the $f_i$.

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Wonderful!!! You can even assume $a_i=1$ for all $i$. By definition a tensor is a finite sum $\sum f_i\otimes g_i$. – Pierre-Yves Gaillard Aug 11 '10 at 9:26