Projective norm for Banach spaces The projective norm of tensors from (algebraic) tensor product of Banach spaces $X,Y$ is defined as
$$\|t\|_\wedge = \inf\left\{ \sum\limits_{j=1}^N \|x_j\|\|y_j\| \, : \, t=\sum\limits_{j=1}^N x_j\otimes y_j\right\}.$$
In one place I found slightly different definition:
$$\|t\|_\wedge '=\inf\left\{\left(\sum\limits_{j=1}^N \|x_j\|^2\right)^{1/2}\left(\sum\limits_{j=1}^N \|y_j\|^2\right)^{1/2} \, : \, t=\sum\limits_{j=1}^N x_j\otimes y_j\right\}.$$
Are they indeed equal? It is easy to see one inequality, but why we have the opposite one?
 A: The two norms are indeed equal. On the one hand, by the Cauchy-Schwarz inequality we have
$$\sum_{j = 1}^{N} \lVert x_j\rVert \lVert y_j\rVert \leqslant \Biggl(\sum_{j = 1}^{N} \lVert x_j\rVert^2\Biggr)^{1/2} \Biggl(\sum_{j = 1}^{N} \lVert y_j\rVert^2\Biggr)^{1/2}$$
which yields $\lVert t\rVert_{\wedge} \leqslant \lVert t\rVert'_{\wedge}$. I suppose that's the inequality you saw.
On the other hand, consider when we have equality in the Cauchy-Schwarz inequality. A particular case where we have equality is when $\lVert x_j\rVert = \lVert y_j\rVert$ for $1 \leqslant j \leqslant N$.
We can always achieve that by moving a scalar factor from $x_j$ to $y_j$, unless one of $x_j$ and $y_j$ is zero. But excluding all representations
$$t = \sum_{j = 1}^{N} x_j \otimes y_j \tag{1}$$
where at least one of the $x_j$ or $y_j$ is zero doesn't change either norm. For the first, including or excluding such terms doesn't change $\sum \lVert x_j\rVert \lVert y_j\rVert$ at all, for the second, the value for the sum without zero factors is smaller (or equal, if $x_j$ and $y_j$ are both zero for such terms), so the terms we exclude can't make the infimum, i.e. $\lVert t\rVert'_{\wedge}$ smaller.
Thus for a representation $(1)$ where no $x_j$ or $y_j$ vanishes, define
$$c_j = \sqrt{\frac{\lVert x_j\rVert}{\lVert y_j\rVert}}$$
and $x_j' = c_j^{-1}\cdot x_j$, $y_j' = c_j \cdot y_j$. Then we have
$$t = \sum_{j = 1}^{N} x_j' \otimes y_j'$$
and
$$\sum_{j = 1}^{N} \lVert x_j\rVert \lVert y_j\rVert = \sum_{j = 1}^{N} \lVert x_j'\rVert \lVert y_j'\rVert = \Biggl(\sum_{j = 1}^{N} \lVert x_j'\rVert^2\Biggr)^{1/2} \Biggl(\sum_{j = 1}^{N} \lVert y_j'\rVert^2\Biggr)^{1/2}$$
which shows
$$\lVert t\rVert'_{\wedge} \leqslant \lVert t\rVert_{\wedge}\,.$$
