Prob. 8, Sec. 3.10 in Kreyszig's functional analysis book: An isometric linear operator has its adjoint as its left inverse

Question

Let $H$ be a Hilbert space, and let $T \colon H \to H$ satisfy $$\langle Tx, Tx \rangle = \langle x, x \rangle \ \mbox{ for all } \ x \in H.$$ Then $T$ is bounded and norm $\Vert T \Vert = 1$ (unless $H = \{0\}$ in which case $T = 0$ also and so $\Vert T \Vert = 0$). So the Hilbert adjoint operator $T^*$ of $T$ exists.

Let $I$ denote the identity operator on $H$.

Then how to show that $T^* T = I$?

My effort:

Since $$\langle Tx, Tx \rangle = \langle x, x \rangle \ \mbox{ for all } \ x \in H,$$ therefore, by using the definition of $T^*$, we obtain $$\langle T^* T x, x \rangle = \langle Tx, Tx \rangle = \langle x, x\rangle \mbox{ for all } \ x \in H;$$ so $$\langle T^* T x - x, x \rangle = 0 \ \mbox{ for all } \ x \in H.$$ If $H$ is complex, then this last equality implies that $$T^* T x = x \ \mbox{ for all } \ x \in H,$$ as required. Am I right?

What if $H$ is real?

Answer 1

We have $$\left< T(x+y) , T(x+y ) \right> = \left< x+y , x+y \right>$$ hence $$\mbox{Re}\left< Tx , Ty \right> = \mbox{Re}\left< x , y \right>$$ and analogously $$\left< T(x+iy) , T(x+iy ) \right> = \left< x+iy , x+iy \right>$$ hence $$\mbox{Im}\left< Tx , Ty \right> = \mbox{Im}\left< x , y \right>$$ therefore $$\left< Tx , Ty \right> = \left< x , y \right>$$ for all $x,y\in H.$

So $$\left< x , y \right>=\left< Tx , Ty \right> =\left< T^*Tx , y \right>$$ thus $$0=\left< T^*Tx -x , y \right>$$ for all $y\in H$

hence $$T^*T x=x $$

Answer 2

If $b$ is a complex sesquilinear form (i.e., linear in the first coordinate and conjugate linear in the second) then $$ b(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}b(x+i^{n}y,x+i^{n}y). $$ That's a nice compact form that you can remember and concisely apply. An immediate consequence: if $b$ and $c$ are complex sesquilinear forms for which $b(x,x)=c(x,x)$, then $b=c$. Hence, $(Tx,Tx)=(x,x)$ for all $x$ implies $(Tx,Ty)=(x,y)$ for all $x,y$, at least for complex spaces. For example, if $(Tx,x)$ is real, then $(Tx,x)=(x,Tx)$ for all $x$, which gives $(Tx,y)=(x,Ty)$ for all $x,y$. Or, $\|Ux\|=\|x\|$ for all $x$ gives $(Ux,Uy)=(x,y)$ for all $x,y$. Or, $\|\hat{f}\|=\|f\|$ for the Fourier transform gives $(\hat{f},\hat{g})=(f,g)$.

A real form $b$ which is linear in both coordinates is different because there are real forms such that $b(x,x)=0$ for all $x$ even though $b \ne 0$. Think of a $2\times 2$ rotation matrix $R$ through $\pi/2$ radius. Clearly $(Rx,x)=0$ for all $x\in\mathbb{R}^{2}$. So if it doesn't hold in $\mathbb{R}^{2}$, then you can't expect better in high dimensions. If $b$ and $c$ are real forms for which $b(x,x)=c(x,x)$ for all $x$, then the difference $b-c$ is a general antisymmetric form which may not be identically $0$.