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 $$\det\begin{pmatrix} 1 & -\lambda \\ a & -\lambda b \end{pmatrix}=\lambda(a-b)=0$$
