If an operator $T$ satisfy a property, then $\|Tx\|=c\|x\|$

Question

Let $(E,\langle\cdot\;,\;\cdot\rangle)$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all operators on $E$.

Assume that $T\in \mathcal{L}(E)$ and satisfy the following property (P):

$$\langle x,y\rangle=0\Longrightarrow \langle Tx,Ty\rangle=0,$$ for all $x,y\in E$.

I want to prove under the property $(P)$ that there exists $c\geq0$ such that $$\|Tx\|=c\|x\|,$$ for all $x\in E$.

Note that, if there exist $x, y\in E\setminus\{0\}$ such that $\|Tx\|=\alpha\|x\|$ and $\|Ty\|=\beta\|y\|$ for some $0\leq\alpha <\beta$, it can be seen that $\langle x,y\rangle\neq0$.

Answer 1

The claim as stated is false, because the operator $T=0$ provides a counterexample.

I assume for the rest that $T\neq 0$.

You wrote that

Note that, if there exist $x, y\in E\setminus\{0\}$ such that $\|Tx\|=\alpha\|x\|$ and $\|Ty\|=\beta\|y\|$ for some $0<\alpha< \beta$, it can be seen that $\langle x,y\rangle\neq0$.

This is already a very good start. However, this statement is also true if you allow $\alpha=0$ (although i do not know your proof of your statement, I would guess it could be modified to allow $\alpha=0$). Thus we can assume that $\ker T=\{0\}$.

Let $x\in E\setminus\{0\}$ be an arbitrary point and let $c>0$ be such that $\|Tx\|=c\|x\|$. For an element $z\in E\setminus\{0\}$ with $\langle x,z\rangle=0$, we can use your observation to obtain that $\|Tz\|=c\|z\|$. (Otherwise we would have have $\|Tz\|=\alpha\|z\|$ for some $\alpha\neq c$. Then your observation would imply $\langle x,z\rangle \neq0$.)

Let $y\in E$ be given. Then we can find a decomposition $y=\alpha x+z$ for $\alpha\in\mathbb C$ and $z\in \{x\}^\perp$. Note that it follows that $\alpha Tx\perp Tz$. We calculate $$ \|Ty\|^2=\|\alpha Tx+Tz\|^2 = |\alpha|^2\|Tx\|^2+\|Tz\|^2+ \langle \alpha Tx,Tz\rangle + \langle Tz,\alpha Tx\rangle \\= |\alpha|^2 c^2 \|x\|^2+c^2\|z\|^2 = c^2\|\alpha x+z\|^2 = c^2\|y\|^2. $$

Thus it holds for all $y\in E$.

Answer 2

Assuming the statement you made at the end we can complete the proof as follows. We first show that $T$ is one-to-one unless it is the zero operator. Suppose $Tx =0, x\neq 0$. Let $y \neq0$. If $\langle x,y\rangle \neq 0$ and $a= \frac {\langle x,y\rangle } {\|y\|^{2}}$then $x-ay$ and $y$ are orthogonal. The hypothesis now tells us that $Ty=0$. On the other hand if $\langle x,y\rangle = 0$ then there exist $a\neq 0, b\neq 0$ such that $x-ay$ and $x-by$ are orthogonal which again gives $Ty=0$. Hence $T$ is the zero operator in which case we can take $c=0$. Thus $T \neq 0$ implies $T$ is one-to -one. Let $\{e_i\}$ be an orthonormal basis (not necessarily countable). It follows that $\|Te_i\|$ is independent of $i$ (unless $T(e_i)=0$). Also $T(e_i)$ are orthogonal. This immediately gives the result with $c=\|Te_i\|$. Just expand any element in terms of the basis and use orthogonality of $T(e_i)$'s. [Proof of the fact that $ \|Te_i \|\neq \|Te_j \|$ cannot occur: suppose $\|Te_i \|< \|Te_j \|$. Take $x =e_i,y=e_j, \alpha =\|Te_i \|, \beta =\|Te_j \|$. We get the contradiction that $\langle e_i,e_j\rangle \neq 0$].

Answer 3

Suppose there exists $x,y\neq 0$ with $\langle x,y\rangle =0$ and $\|T(x)\|=a\|x\|, \|T(y)\|=b\|y\|, a\neq b$. We can assume $\|x\|=\|y\|=1$.

$$\langle x+y,x-y\rangle=0, \langle T(x+y),T(x-y)\rangle$$ $$ =\langle T(x),T(x)\rangle-\langle T(x),T(y)\rangle+\langle T(y),T(x)\rangle-\langle T(y),T(y)\rangle=a^2-b^2\neq 0$$ contradiction.

Let $x\neq 0$, $y\in Vec(x)^{\perp}$, $\|T(x)\|=c\|x\|, \|T(y)\|=c\|y\|$. For every $z\in H$, $z=ux+vy, y\in Vect(x)^{\perp}$, implies $\|T(z)\|^2=\|T(ux+vy)\|^2=\langle T(ux)+T(vy),T(ux)+T(vy)\rangle=c^2\|z\|^2$.