Equivalent characterisation of hypothesis & conclusion in Grothendieck's inequality

Question

The basic version of Grothendieck's inequality states that for a $m\times n$ matrix $[a_{i, j}]$, if

$$\bigg|\sum_{i, j}a_{i, j}x_i y_j\bigg|\le 1,$$

for all $x_i, y_i\in \{-1, 1\}$, then

$$\bigg|\sum_{i, j}a_{i, j}\langle u_i, v_j\rangle\bigg|\le K,$$

for any unit vectors $u_i, v_j$ in a Hilbert space $H$ (i.e. $\|u_i\| = \|v_j\| =1$), with $K$ an absolute constant.

The following conditions are equivalent to the hypothesis and conclusion of the prior statement respectively:

hypothesis: if $$\bigg|\sum_{i, j}a_{i, j}x_i y_j\bigg|\le \max_i |x_i| \max_j|y_j|,$$

for all $x_i, y_i\in \mathbb{R}$.

conclusion: then $$\bigg|\sum_{i, j}a_{i, j}\langle u_i, v_j\rangle\bigg|\le K\max_i\|u_i\|\max_j\|v_j\|,$$

for any vectors $u_i, v_j\in H$, a Hilbert space, with $K$ an absolute constant.

This equivalence seems like it ought to be very simple to prove, however I seem to be having a mental block which is preventing me from proceeding & presume that I'm just not seeing something pretty obvious here. I'd be grateful for any hints or suggestions on how to approach the proof.

Answer 1

I arrived to same problem of Vershynin's book. This was my guess for that problem:

Grothendieck’s inequality states that $$\bigg|\sum_{i, j}b_{i, j}x_i y_j\bigg| \leq 1 \text{ where $b_{i,j} \in \mathbb{R}$ and $x_i,y_j \in \{-1,1\}$.}$$

The maximum value can be found when the sign of $x_i y_j$ is the same as $b_{i,j}$, so we obtain $\bigg| \sum_{i,j} b_{i,j} x_i y_j \bigg| = 1$.

Renaming the variable $a_{i,j} = b_{i,j} x_i y_j$, we can conclude $$\bigg|\sum_{i, j}a_{i, j}x_i y_j\bigg| \leq \max_{i,j} |a_{i,j}| \cdot \max_i |x_i| \cdot \max_j |y_j| \leq \max_i |x_i| \cdot \max_j |y_j|$$ since $\max_{i,j} |a_{i,j}| \leq 1$.

In the second part, we can use the fact that $\vert \langle u, v\rangle \vert \leq \|u\| \cdot \|v\|$: $$\bigg|\sum_{i, j}a_{i, j}\langle u_i, v_j\rangle\bigg| \leq \bigg|\sum_{i, j}a_{i, j} \|u_i\| \|v_j\| \bigg|$$

And then, we can conclude as previously.

Answer 2

To see the equivalence, consider the following lemmas, the first of which is well-known.

Lemma 1. $|\sum_i\pm a_i|\le 1\iff |\sum_it_ia_i|\le1,\quad \forall t_i\in[-1,1]$.

Proof: $\Leftarrow$ is obvious. $\Rightarrow$: $|\sum_it_ia_i|\le\sum_i|a_i|=|\sum_i\pm a_i|\le1$ where a choice of signs is made so that $|a_i|=\pm a_i$.

Lemma 2. $|\sum_i\langle u_i,w_i\rangle|\le K, \forall u_i$ unit $\iff |\sum_i\langle u_i,w_i\rangle|\le K$, $\forall u_i$, $\|u_i\|\le1$.

Proof: $\Leftarrow$ is obvious. $\Rightarrow$: The inequality is equivalent to $|\sum_i\pm\langle u_i,w_i\rangle|\le K$ by considering $\pm u_i$ instead of $u_i$. So by lemma 1, one can take $t_i=\|u_i\|$ and write $u_i=t_i\hat{u}_i$.

Both lemmas can be extended to multiple indices:

Proposition 1. $|\sum_{ij}a_{ij}x_iy_j|\le 1,\ \forall x_i,y_j\in\{-1,+1\}\iff |\sum_{ij}a_{ij}x_iy_j|\le1\ \forall x_i,y_j\in[-1,1]$.

Proof: $\Rightarrow$: $|\sum_ix_i\sum_ja_{ij}y_j|\le1$ by lemma 1. when $y_j\in\{-1,1\}$.
So $|\sum_j\pm\sum_ia_{ij}x_j|\le1$. Again by lemma 1., $|\sum_jy_j\sum_ia_{ij}x_j|\le1$ when $y_j\in[-1,1]$.

Proposition 2. $|\sum_{ij}\langle u_i,a_{ij}v_j\rangle|\le K,\ \|u_i\|=1=\|v_j\|\iff |\sum_{ij}\langle u_i,a_{ij}v_j\rangle|\le K,\ \|u_i\|,\|v_j\|\le1$.

Proof: $\Rightarrow$: When $v_j$ are unit, but $\|u_i\|\le1$, lemma 2. implies $|\sum_i\langle u_i,\sum_ja_{ij}v_j\rangle|\le K$.
So $|\sum_j\langle\sum_ia_{ij}u_i,v_j\rangle|\le K$. Hence again by lemma 2., this last inequality is valid for $\|v_j\|\le1$.

The equivalence between the two sets of statements should now be clear.