If a bilinear operator $E\times F\to G$ is bounded, is the corresponding linear operator $E\otimes F\to G$ bounded?

Question

Consider a bilinear operator between Hilbert spaces $E\times F\to G$, denoted by $(x,y)\mapsto x\odot y$. There is a corresponding linear operator $\odot':E\otimes F\to G$, given by $\odot'(x\otimes y)=x\odot y$ and extending by linearity. If there is some $M\geq0$ such that, for all $X\neq0$ in $E\otimes F$,

$$\frac{\lVert\odot'(X)\rVert}{\lVert X\rVert}\leq M<\infty,$$

then $\odot$ is also bounded: for $x\in E,\;y\in F$,

$$\frac{\lVert x\odot y\rVert}{\lVert x\rVert\lVert y\rVert}=\frac{\lVert\odot'(x\otimes y)\rVert}{\lVert x\otimes y\rVert}\leq M.$$

Is the converse true? If $\odot$ is bounded, is $\odot'$ bounded?

Using the orthonormal basis $\{e_i\otimes f_j\}$ of $E\otimes F$, we have

$$X=\sum_{i,j}X_{ij}\,e_i\otimes f_j$$

$$\lVert X\rVert^2=\sum_{i,j}|X_{ij}|^2$$

$$\odot'(X)=\sum_{i,j}X_{ij}\,e_i\odot f_j$$

$$\lVert\odot'(X)\rVert\leq\sum_{i,j}|X_{ij}|\,\lVert e_i\odot f_j\rVert$$

$$\leq\sum_{i,j}|X_{ij}|\,M\lVert e_i\rVert\lVert f_j\rVert$$

$$=M\sum_{i,j}|X_{ij}|$$

but we need it to be bounded in the $\ell^2$ norm, not the $\ell^1$ norm.

Answer 1

One counter-example is the inner product itself!

$$x\odot y=x\cdot y=\sum_ix_iy_i$$

$$\odot'(X)=\sum_{i,j}X_{ij}\,e_i\cdot e_j=\sum_iX_{ii}$$

The Cauchy-Schwarz inequality is $|x\odot y|\leq\lVert x\rVert\lVert y\rVert$. But $\odot'$ is clearly unbounded with respect to $\sum_{i,j}|X_{ij}|^2$; for example, take $X_{ii}=1/i$.