The Tensor product of algebra group

Question

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

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

Answer 1

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 $\lim_{n\rightarrow \infty}f_n\notin L^1(\mathbb{R},\lambda)\hat{\otimes}_{\pi}L^1(\mathbb{R},\lambda)$.