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Let G is a locally compact group. Is the following true?

The tensor product of $L^1(G)$ with $L^1(G)$ is $L^1(G \times G)$.

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    $\begingroup$ I believe that you need to specify the type of tensor product to be used. The common ones are the injective and projective tensor products. $\endgroup$ Jul 1, 2015 at 8:33
  • $\begingroup$ The tensor product is certainly dense in $L^1(G\times G)$. This amounts to saying that the algebraic tensor product carries a norm such that the completion of this norm gives $L^1(G\times G)$. The more appropriate object, however, to look at is the $C^*$-algebra of $G$. $\endgroup$
    – user1688
    Jul 1, 2015 at 18:30
  • $\begingroup$ See if this works. Embed the algebraic tensor product $ {L^{1}}(G) \odot {L^{1}}(G) $ into $ {L^{1}}(G \times G) $ by mapping an elementary tensor $ [f] \odot [g] $ to $ [(x,y) \mapsto f(x) g(y)] $ — where $ [\cdot] $ denotes the taking of an equivalence class — and then extending this map by linearity to the entire algebraic tensor product. $\endgroup$ Jul 1, 2015 at 18:34
  • $\begingroup$ Next, show that the image of this map is dense in $ {L^{1}}(G \times G) $ with respect to the $ L^{1} $-norm. The pullback of this norm to $ {L^{1}}(G) \odot {L^{1}}(G) $ is a cross-norm with respect to which the completion of $ {L^{1}}(G) \odot {L^{1}}(G) $ is automatically isomorphic to $ {L^{1}}(G \times G) $ as a Banach space. $\endgroup$ Jul 1, 2015 at 18:34
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    $\begingroup$ Whether this particular completed tensor product is isomorphic to the injective tensor product or the projective one remains to be seen. @Corbennick: I just saw your comment! Hence, what I’ve written above basically repeats what you’ve already said. :) $\endgroup$ Jul 1, 2015 at 18:42

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The answer is yes if we consider the projective tensor product: Let $E$ be a Banach space. According to Theorem 3 in http://archive.numdam.org/ARCHIVE/AIF/AIF_1952__4_/AIF_1952__4__73_0/AIF_1952__4__73_0.pdf there is an isometric isomorphism $L^1(G,\mu)\hat{\otimes}_\pi E\cong L^1_E(G,\mu)$, where $L^1_E(G,\mu)$ denotes the $E$-valued integrable functions. Set $E=L^1(G,\mu)$. Then we get $$L^1(G,\mu)\hat{\otimes}_\pi L^1(G,\mu)\cong L^1_{L^1(G,\mu)}(G,\mu).$$ It remains to verify that $L^1_{L^1(G,\mu)}(G,\mu)\cong L^1(G\times G,\mu\otimes \mu)$. This can be done by using the theorem of Fubini if $G$ is $\sigma$-finite. If $G$ is not $\sigma$-finite with respect to the measure $\mu$ then we can use Theorem 5.8 in http://www.ams.org/journals/tran/1966-123-01/S0002-9947-1966-0197669-0/S0002-9947-1966-0197669-0.pdf instead of the ordinary Fubini theorem. However, if we consider a different topology resp. cross norm on $L^1(G,\mu)\otimes L^1(G,\mu)$ then the answer is no (even if $G$ is $\sigma$-finite): For the sake of simplicity set $G=\mathbb{R}$ and let $\mu=\lambda$ be the Lebesgue measure.
Consider $L^1(\mathbb{R},\lambda)\hat{\otimes}_{\varepsilon} L^1(\mathbb{R},\lambda)$. If $L^1(\mathbb{R},\lambda)\hat{\otimes}_{\varepsilon} L^1(\mathbb{R},\lambda)\cong L^1(\mathbb{R}\times \mathbb{R},\lambda\otimes \lambda)$ then $L^1(\mathbb{R},\lambda)\hat{\otimes}_{\varepsilon} L^1(\mathbb{R},\lambda)\cong L^1(\mathbb{R},\lambda)\hat{\otimes}_{\pi} L^1(\mathbb{R},\lambda)$. Thus it comes down to check if $L^1(\mathbb{R},\lambda)\hat{\otimes}_{\varepsilon} L^1(\mathbb{R},\lambda)\cong L^1(\mathbb{R},\lambda)\hat{\otimes}_{\pi} L^1(\mathbb{R},\lambda)$ holds or not. We are going to show that there is no such isometric isomorphism. Note that $$\varepsilon(f)=\sup\left\{\left|(a_1^\ast\otimes a_2^\ast)(f)\right|,a_1^\ast\in L^1\left(\mathbb{R},\lambda\right)^\ast,a_2^\ast\in L^1(\mathbb{R},\lambda)^\ast,\left\|a_1\right\|=1,\left\|a_2\right\|=1\right\}$$ and $$\pi(f)=\inf\left\{\sum_{i=1}^n\left\|f_i\right\|_1\left\|g_i\right\|_1,f=\sum_{i=1}^nf_i\otimes g_i\right\}.$$ Since $L^\infty(\mathbb{R},\lambda)\cong L^1(\mathbb{R},\lambda)^\ast$, we get $$\varepsilon(f)=\sup_{\left\|\phi_1\right\|_\infty=1,\left\|\phi_2\right\|_\infty=1}\left\{\left|\sum_{i=1}^n\int_{\mathbb{R}}f_i\phi_1\mathrm{d}\lambda\int_{\mathbb{R}}g_i\phi_2\mathrm{d}\lambda\right|:\phi_1\in L^\infty(\mathbb{R},\lambda),\phi_2\in L^\infty(\mathbb{R},\lambda)\right\},$$ where $f=\sum_{i=1}^nf_i\otimes g_i$. Set $f_n=\sum_{i=1}^{2n}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\otimes \sqrt{i}\exp(-x^2i)$ (We will take equivalence classes, without actually mentioning it), where $n\geq 1$. Then $\pi(f_n)=\sum_{i=1}^{2n}\left\|(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\right\|_1\left\|\sqrt{i}\exp(-x^2i)\right\|_1=\sum_{i=1}^{2n}\frac {1} {i}\sqrt{\pi}$. Thus $\pi(f_n)$ tends to $\infty$ as $n$ tends to $\infty$. We are going to show that $\lim_{n\rightarrow \infty}\varepsilon(f_n)\in \mathbb{R}$. Without loss of generality we can assume that $\left|\phi_j\right|\neq \mathrm{id}_{\mathbb{R}}$ for $j=1,2$. Then we get $$-\int_{\mathbb{R}}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda<\int_{\mathbb{R}}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\phi_1\mathrm{d}\lambda<\int_{\mathbb{R}}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda$$ if $i$ is even and $$\int_{\mathbb{R}}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda<\int_{\mathbb{R}}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\phi_1\mathrm{d}\lambda<-\int_{\mathbb{R}}(-1)^i\chi_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda$$ if $i$ is odd, where $\phi_1\in L^\infty(\mathbb{R},\lambda)$ with $\left\|\phi_1\right\|_\infty=1$. Futhermore, we have $\int_{\mathbb{R}}\sqrt{i}\exp(-x^2i)\phi_2\mathrm{d}\lambda<\int_{\mathbb{R}}\sqrt{i}\exp(-x^2i)\mathrm{d}\lambda$, where $\phi_2\in L^\infty(\mathbb{R},\lambda)$ with $\left\|\phi_2\right\|_\infty=1$. Hence $$\sum_{i=1}^\infty\left|\frac {1} {i}-\frac {1} {i+1}\right|\sqrt{\pi}\geq \varepsilon(f_n)$$ for all $n\geq 1$. Obviously, $\sum_{i=1}^\infty\left|\frac {1} {i}-\frac {1} {i+1}\right|\sqrt{\pi}<\infty$. Therefore $\lim_{n\rightarrow \infty}f_n\in L^1(\mathbb{R},\lambda)\hat{\otimes}_{\varepsilon}L^1(\mathbb{R},\lambda)$, while $\lim_{n\rightarrow \infty}f_n\notin L^1(\mathbb{R},\lambda)\hat{\otimes}_{\pi}L^1(\mathbb{R},\lambda)$.

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  • $\begingroup$ Are you saying that the canonical image of $L^1(G,\mu)\hat{\otimes}_\pi L^1(G,\mu)$ in $L^1(G\times G,\mu\times\mu)$ may be a proper subspace if $G$ is not $\sigma$-finite? I thought this worked for all locally compact groups, not just the $\sigma$-finite ones. $\endgroup$
    – Yemon Choi
    Jul 24, 2015 at 0:52
  • $\begingroup$ My bad. It holds for all locally compact groups. $\endgroup$
    – Slunak
    Jul 25, 2015 at 15:38

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