# If $L∈\mathfrak L(X⊗_πY,E)$, then $\sup_{\left\|x\right\|_X,\:\left\|y\right\|_Y\:≤\:1}\left\|L(x⊗y)\right\|_E=\sup_{π(u)\:≤\:1}\left\|Lu\right\|_E$

Let $X,Y$ be $\mathbb R$-Banach spaces and $X\otimes_\pi Y$ denote the tensor product $X\otimes Y$ of $X$ and $Y$ with respect to the projective norm $$\pi(u):=\inf\left\{\sum_{i=1}^n\left\|x_i\right\|_X\left\|y_i\right\|_Y:n\in\mathbb N,(x_1,y_1),\ldots,(x_n,y_n)\in X\times Y\text{ with }u=\sum_{i=1}^nx_i\otimes y_i\right\}$$ for $u\in X\otimes Y$.

Now, let $E$ be a $\mathbb R$-Banach space and $L\in\mathfrak L(X\otimes_\pi Y,E)$. I want to show that $$\sup_{\left\|x\right\|_X,\:\left\|y\right\|_Y\:\le\:1}\left\|L(x\otimes y)\right\|_E=\sup_{\pi(u)\:\le\:1}\left\|Lu\right\|_E\tag1\;.$$

Since $\pi(x\otimes y)=\left\|x\right\|_X\left\|y\right\|_Y$ for all $(x,y)\in X\times Y$, we've clearly got "$\le$". How can we show the other inequality?

Let's try the obvious thing. I'll denote by:

$$M := \sup_{\|x\|_X, \|y\|_Y \leq 1} \|L(x \otimes y)\|_E,$$

$$N := \sup_{\pi (u) \leq 1} \|L(u)\|_E,$$

with $M \leq N$. For the other direction, let $u$ be with $\pi (u) \leq 1$ and $\varepsilon >0$. Let $x_i$, $y_i$ nonzero be such that $u = \sum_i x_i \otimes y_i$ and $\sum_i \|x_i\|_X \|y_i\|_Y \leq \pi(u)+\varepsilon$. Then:

$$\|L(u)\|_E = \left\|\sum_i L(x_i \otimes y_i)\right\|_E \leq \sum_i \left\|L(x_i \otimes y_i)\right\|_E$$

$$= \sum_i \|x_i\|_X \|y_i\|_Y \left\|L\left(\frac{x_i}{\|x_i\|_X} \otimes \frac{y_i}{\|y_i\|_Y}\right)\right\|_E \leq M \sum_i \|x_i\|_X \|y_i\|_Y \leq M(\pi(u)+\varepsilon).$$

Since this is true for all $\varepsilon >0$, we get $\|L(u)\|_E \leq M \pi(u)$, so $N \leq M$.