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

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.

$$x\odot y=x\cdot y=\sum_ix_iy_i$$
$$\odot'(X)=\sum_{i,j}X_{ij}\,e_i\cdot e_j=\sum_iX_{ii}$$
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 $$X_{ii}=1/i$$.