# A question about operators on Hilbert spaces

Let $$H$$ be a Hilbert space and $$T,S \in B(H).$$ Suppose that $$T,S$$ satisfy the following condition: $$\exists c\geq 0:~~|\langle Tu,v\rangle | \leq c |\langle Su,v\rangle |, \text{ for all }u,v\in H.$$ Can we say that one of $$T$$ or $$S$$ is a multiple of the other; i.e., $$T=aS$$ or $$S=aT$$ for some scalar $$a?$$

Thanks

This is correct. First $$Tu, Su$$ must be always linearly dependent, otherwise there exists a vector $$v$$ that is perpendicular to $$Su$$ but not $$Tu$$, which would contradict the inequality. To show the eixstence of such $$v$$, note that $$((Su)^{\perp})^{\perp}=\overline{\text{Span}(Su)}=\text{Span}(Su)\not\ni Tu$$ thus $$Tu$$ is not orthogonal to all elements in $$(Su)^{\perp}$$. Or more intuitively, simply take the orthogonal decomposition $$Tu = a (Su) + v \text{ such that }0\not=v\in \text{Span}(Su)^{\perp}$$
Now we show $$T=aS$$ for some $$a$$. If $$Su=0$$, pick $$v=Tu$$, we conclude from the given inequality $$Tu=0$$. If $$Su\not=0$$, then by above, $$Tu=a_uSu$$ for a unique $$a_u$$, as $$Tu, Su$$ are dependent. Now it's sufficient to show $$a_u$$ doesn't depend on $$u$$. (Note that we have essentially exahusted the power of the inequality, as the inequality would hold for $$c:=\sup_{Su\not=0}\|a_u\|$$ assuming $$c<\infty$$).
Assume $$Tu=aSu$$ and $$Tv=bSv$$ with $$Su\not=0, Sv\not=0$$. If $$Su$$ and $$Sv$$ are linearly independent, then $$T(u+v)=aSu+bSv=\lambda S(u+v)=\lambda Su + \lambda Sv$$ for some $$\lambda\in\mathbb C$$, hence $$a=b=\lambda$$. If $$Su = \lambda Sv$$ where $$\lambda\not=0$$ as $$Su\not=0$$, then $$S(u-\lambda v)=0\Rightarrow T(u-\lambda v)=0\Rightarrow aSu-\lambda bSv=0$$
Hence $$[1:\lambda]=[a: \lambda b]\Rightarrow \frac{a}{1}=\frac{\lambda b}{\lambda}=b$$. Here the homogeneous coordinates (of points in dual space) are used. We can also do e.g. $$\begin{cases} x - \lambda y = 0 \\ a x - \lambda by=0\end{cases}$$ has nontrivial solutions for $$x,y\in H$$, hence $$\det\begin{pmatrix} 1 & -\lambda \\ a & -\lambda b \end{pmatrix}=\lambda(a-b)=0$$