Consider the sequence space $\ell_r$ defined by $$\ell_r=\left\{x=(x_n)_{n=1}^{\infty}:x_n\in\mathbb{R}\text{ and }\sum_{n=1}^{\infty}|x_n|^r<\infty\right\}.$$ Let $2\leq p,q<\infty$ such that $\frac{1}{p}+\frac{1}{q}\leq\frac{1}{2}$. I want to find the norm of the bilinear operator $T:\ell_p\times\ell_q\rightarrow\mathbb{R}$ given by $$T(x,y)=x_1y_1+x_1y_2+x_2y_1-x_2y_2.$$
Observe that $\frac{e_1}{2^{1/r}}+\frac{e_2}{2^{1/r}}\in B_{\ell_{r}}=\{x\in\ell_r:\|x\|_r\leq1\}$ and $$T\left(\frac{e_1}{2^{1/p}}+\frac{e_2}{2^{1/p}},\frac{e_1}{2^{1/q}}+\frac{e_2}{2^{1/q}}\right)=2^{1-\left(\frac{1}{p}+\frac{1}{q}\right)}.$$ This fact shows that $\|T\|\geq 2^{1-\left(\frac{1}{p}+\frac{1}{q}\right)}$ and, because of this, I have thought that $\|T\|=2^{1-\left(\frac{1}{p}+\frac{1}{q}\right)}$. However, making some calculations on the computer, I guess I'm wrong.
What is the exact value of the norm of $T$?
Note that $$ \Vert T\Vert=\sup\{\Vert T_x\Vert: x\in S_{\ell_p}\} $$ where $T_x:\ell_q\to\mathbb{R}:y\mapsto T(x,y)$. Thus we have the functional $$ T_x(y)=(x_1+x_2)y_1+(x_1-x_2)y_2 $$ Since $\ell_q^*=\ell_{q'}$, then $\Vert T_x\Vert=\Vert (x_1+x_2, x_1-x_2,0,0,\ldots)\Vert_{q'}$. Therefore $$ \begin{align} \Vert T\Vert &=\sup\{ \Vert (x_1+x_2, x_1-x_2,0,0,\ldots)\Vert_{q'}:x\in S_{l_p}\}\\ &=\sup\left\{\left(|x_1+x_2|^{q'}+|x_1-x_2|^{q'}\right)^{1/q'}:\left(|x_1|^p+|x_2|^p\right)^{1/p}\leq 1\right\}\\ &=\left(\max\limits_{|x_1|^p+|x_2|^p= 1} \left(|x_1+x_2|^{q'}+|x_1-x_2|^{q'}\right)\right)^{1/q'}\\ \end{align} $$ Evaluation of the last maximum is the standard constrained optimization problem. The function to be maximaize and domain doesn't change under transformations $x_1\to -x_1$ and $x_2\to -x_2$, so it is enough to maximize our function on the domain $x_1^p+x_2^p=1$, $x_1,x_2\geq 0$.