Take two Banach spaces $X,Y$ and consider their tensor product $W=X\otimes Y$. Any $w\in W$ can be written, generally not in a unique way, as a linear combination of simple tensors: $$w=\sum_{i=1}^n (x_i\otimes y_i)\ \ \ \ \text{ with }x_i\in X \text{ and } y_i\in Y\ \forall i\quad\quad\quad (1)$$
Under which assumptions is it possible to say that the function $f$ defined on $W$ as: $$f(w)=\big\{\max ||x_i||_X||y_i||_Y : \text{$x_i,y_i$ satisfy (1)}\big \}$$ is such that $|f(w)|<\infty$ for every $w\in W$?
This happens almost never -- in particular, $f(w)$ is finite for all $w \in W$ if and only if $X = 0$ or $Y = 0$.
One direction is clear -- if $X = 0$ or $Y = 0$ then $f(w) = \{0\}$ for all $w \in W$.
In the other direction, suppose $X \neq 0$ and $Y \neq 0$. Then pick nonzero vectors $x \in X$ and $y \in Y$. Now we can write $x \otimes y$ as $(n x) \otimes y - ((n-1)x) \otimes y$ for any positive integer $n$. This gives
$$n\lVert x \rVert \lVert y \rVert = \max\{\lVert nx \rVert \lVert y \rVert, \lVert (n-1)x \rVert \lVert y \rVert\} \in f(x \otimes y)$$
for all positive integers $n$. Since $\lVert x \rVert \lVert y \rVert \neq 0$, we conclude that $f(x \otimes y)$ is an infinite set.