Let $E$ and $F$ be Banach spaces. We define the projective norm of the elementary tensor product of $E\otimes F$ as follows-
For $t\in E\otimes F$, define $\lVert t\rVert_{\wedge} =\text{inf}\{\sum\limits_{i=1}^n \lVert a_i\rVert\lVert b_i\rVert:\ t=\sum\limits_{i=1}^n a_i\otimes b_i, a_i\in E, b_i\in F\}$.Now take the completion with respect to this norm and call it $E\hat\otimes F$.
I have to prove that $\lVert a\otimes b\rVert_{\wedge}=\lVert a\rVert\lVert b\rVert$.
It's obvious from the definition that $\lVert a\otimes b\rVert_{\wedge}\le\lVert a\rVert\lVert b\rVert$
Now I have to show the converse inequality i.e. if $a\otimes b=\sum\limits_{i=1}^n a_i\otimes b_i$, we must have $\lVert a\rVert\lVert b\rVert\le \sum\limits_{i=1}^n \lVert a_i\rVert\lVert b_i\rVert$
Can anyone suggest me a hint to prove this? Thanks for your assistance in advance.
Here I'm writing the solution of the above problem based on the hint provided by @jd27:
For $\phi\in E^\ast$ and $\psi\in F^\ast$, we can define $\phi\otimes\psi:E\hat\otimes F\to\Bbb{C}$ by $\phi\otimes\psi(a\otimes b)=\phi(a)\psi(b)$ and extend linearly. For $t=\sum a_i\otimes b_i\in E\otimes F$, we have $|(\phi\otimes\psi)(t)|=\left|\sum\phi(a_i)\psi(b_i)\right|\le\lVert\phi\rVert\lVert\psi\rVert\sum \lVert a_i\rVert\lVert b_i\rVert\le\lVert\phi\rVert\lVert\psi\lVert t\rVert_\wedge$. This implies $\lVert \phi\otimes\psi\rVert\le\lVert\phi\rVert\lVert\psi\rVert$.
Now for $a\in E,b\in F$, we have $j(a\otimes b)\in (E\hat\otimes E)^*$, then
$\lVert a\otimes b\rVert_\wedge=\lVert j(a\otimes b)\rVert\ge\text{sup}\{|j(a\otimes b)(\phi\otimes\psi)|:\ \lVert\phi\otimes\psi\rVert\le1\}\ge \text{sup}\{|\phi(a)||\psi(b)|:\ \lVert\phi\rVert\lVert\psi\rVert\le1\}\ge\lVert a\rVert\lVert b\rVert$
This completes the proof.