Injective tensor product

Remarks

The norm is a cross norm: $\varepsilon(x \otimes y) = \norm{x} \, \norm{y}$, by the Hahn–Banach theorem. It is at most the projective norm $\pi$. The bounds on $f$ and $g$ are needed: without them, scaling $f$ makes the supremum infinite for every $u \ne 0$.

Example: let $X = Y = \R^2$ with the Euclidean norm and $u = e_1 \otimes e_1 + e_2 \otimes e_2$. Then $\varepsilon(u) = \sup \abs{\ip{f}{g}} = 1$, over unit vectors $f, g$. The projective norm of $u$ is $2$, so the two norms differ.

For compact Hausdorff spaces $K$ and $L$, the map $f \otimes g \mapsto ((s, t) \mapsto f(s) g(t))$ extends to an isometry of $C(K) \check{\otimes}_\varepsilon C(L)$ onto $C(K \times L)$. More generally, $C(K) \check{\otimes}_\varepsilon X$ is the space of continuous functions from $K$ to $X$ with the sup norm.