# Special bound of net and inequality

Suppose that $A$ is a finite dimensional unital commutative Banach algebra and $A\hat{\otimes} A$ is its Projective tensor product by itself. Define $\Delta:A\hat{\otimes} A\to A$ with $$\Delta(\sum_{i=1}^\infty a_i\otimes b_i)=\sum_{i=1}^\infty a_i b_i.$$ There is $m\in A\hat{\otimes} A$ with $$a\Delta(m)=a,\qquad (a\in A)$$ Also there is a net $\{m_\alpha\}\subset A\hat{\otimes} A$ such that $\|m_\alpha-m\|\to 0$ and there is $K>0$ such that $$a\Delta(m_\alpha)\to a,\qquad \|a\Delta(m_\alpha)\|\leq K\|a\|,\qquad \|a\cdot m_\alpha-m_\alpha\cdot a\|\leq K\|a\|,\qquad (a\in A)$$ Could we conclude that $K-\|m\|\geq C$, for some non-negative real number $C$ ?

• Dear @Pietro Majer, I edited my question – Hamid Shafie Asl Dec 12 '16 at 5:10
• I'm a bit confused about the role of the net in the last assumption. From one side, this net may be, in particular, the constant net (which we can't exclude without further information). On the other side, if that assumption is verified by some $\{m_\alpha\}$, then by continuity of $\Delta$ it is also true for the constant net $m_\alpha=m$, with the same $K$. So why this assumption should be any stronger that just asking $\|a\Delta(m)\|\le K\|a\|$? – Pietro Majer Dec 12 '16 at 9:27
• By the way, if a net $(x_i)$ in a Banach space $E$ converges in the norm topology to some $x\in E$, then we always have $\Vert x \Vert \leq \sup_i \Vert x_i\Vert$. This follows from the Hahn-Banach theorem, and is "core functional analysis" that has nothing to do with Banach algebras, or amenability. – Yemon Choi Dec 12 '16 at 15:43
• @YemonChoi: Isn't it really just the triangle inequality? – Nate Eldredge Dec 12 '16 at 16:00
• @NateEldredge Oops, yes you are right :) I tend to use the result with $x_i \to x$ weakly, but since the OP has already used the Mazur trick to improve weak convergence to norm convergence, then there is no need for Hahn-Banach – Yemon Choi Dec 12 '16 at 16:03

Your question appears to be: suppose $A$ is a finite-dimensional unital CBA, and suppose it has an approximate diagonal satisfying certain conditions; does the unique diagonal element $m\in A \otimes A$ satisfy a certain norm bound?
If this is your question then the answer is negative. The reason, as Pietro Majer pointed out, is that if you assume $(m_\alpha)$ satisfies these conditions then the assumption $\Vert m_\alpha -m\Vert\to 0$ implies that $m$ itself satisfies these conditions. So your question becomes
Let $m$ be the diagonal element for $A$ (which in particular forces $a\Delta(m)=a$ and $a\cdot m = m\cdot a$ for all $a\in A$). Assume there is $K>0$ such that $\|a\Delta(m)\|\leq K\|a\|$. Do we have $\Vert m\Vert\leq K$?
This is false because we always have $\Vert a\Delta(m)\Vert=\Vert a\Vert$ but we might have $\Vert m \Vert > 1$. Some explicit calculations of amenability constants can be found in a paper of Ghandehari, Hatami and Spronk: there is a preprint version on the arXiv 0705.4279v2 but be warned that the theorem numbers are different in the final published version.