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

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

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

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

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

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


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


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 $X_{ii}=1/i$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.