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Let $H$ be a Hilbert space, let $T \colon H \to H$ be a linear operator such that $T$ is isometric but not unitary.

Then how to show that the image $T[H]$ is a proper closed subspace of $H$?

My effort:

Since $T$ is a linear operator, $T[H]$ is a (vector) subspace of $H$.

Since $T$ is isometric, for each $x \in H$, we have $$ \langle Tx, Tx \rangle = \Vert Tx \Vert^2 = \Vert x \Vert^2 = \langle x, x \rangle.$$

If $H$ were complex, then $T$ would be unitary if and only if $T$ were isometric and surjective.

But $T$ is not unitary. So $T$ is not surjective. Thus $T[H]$ is a proper subspace of $H$. How to tackle the case when $H$ is real?

How to show that $T[H]$ is closed in $H$?

After reading @Braindead's comment:

Suppose that $y \in \mathrm{cl}(T[H])$. Let $y_n \colon= T x_n$ be a sequence in $T[H]$ such that $y_n$ converges to the point $y$. Then $y_n$ is a Cauchy sequence in $H$. And since $T$ is isometric, the sequence $x_n$ is also Cauchy in $H$, which is complete. So $x_n$ converges to some point $x \in H$. Again using the isometric property of $T$, we can conclude that $T$ is continuous, so that $T x_n$ converges to $T x$. But $Tx_n = y_n$. So $Tx = y$; that is, $y \in T[H]$, showing that $T[H]$ is closed.

Is the above proof correct?

Using th so-called polarisation identities, we obtain, for all $x, y \in H$, \begin{align*} \langle Tx, Ty \rangle & = \Re \langle Tx, Ty \rangle + \iota \Im \langle Tx, Ty \rangle \\ &= \frac{1}{4} \left( \Vert Tx + Ty \Vert^2 - \Vert Tx - Ty \Vert^2 \right) + \frac{\iota}4 \left( \Vert Tx + \iota Ty \Vert^2 - \Vert Tx - \iota Ty \Vert^2 \right) \\ &= \frac{1}{4} \left( \Vert T (x + y) \Vert^2 - \Vert T ( x - y) \Vert^2 \right) + \frac{\iota}{4} \left( \Vert T (x + \iota y) \Vert^2 - \Vert T ( x - \iota y) \Vert^2 \right) \\ & = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2 \right) + \frac{\iota}4 \left( \Vert x + \iota y \Vert^2 - \Vert x - \iota y \Vert^2 \right) \ \mbox{ [ since $T$ is isometric ]} \\ &= \Re \langle x, y \rangle + \iota \Im \langle x,y \rangle \\ &= \langle x, y \rangle. \end{align*} If $H$ is real, there would be no terms involving the $\iota$. Thus, irrespective of whether $H$ is real or complex, using the definition of the Hilbert adjoint operator $T^*$ of $T$, we have $$\langle T^* Tx, y \rangle = \langle Tx, T y \rangle = \langle x, y \rangle \ \mbox{ for all } \ x, y \in H.$$ This implies that $$T^* T x = x \ \mbox{ for all } \ x \in H,$$ and so $$T^*T = I. \ \tag{equation 1}$$

Now if $T$ were also surjective (i.e. if $T[H]$ were all of $H$), then $T$ would be bijective (and so $T^{-1}$ would also exist) and in that eventuality, we would obtain $$TT^* = TT^*I = TT^* (TT^{-1}) = T(T^*T)T^{-1} = T I T^{-1} = I,$$ which together with equation (1) would imply $T^* = T^{-1}$ and hence $T$ would be unitary, contrary to our hypothesis.

Saaqib Mahmood
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2 Answers2

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Your proof is correct. Here is a slightly shorter way to do the last step (once you have $\langle Tx,Ty\rangle=\langle x,y\rangle$). Suppose $T$ is surjective. Then it is bijective, since any isometry is injective. Now note that for any $x,y\in H$, $$\langle Tx,y\rangle=\langle Tx,T(T^{-1}y)\rangle=\langle x,T^{-1}y\rangle.$$ By definition of adjoints, this means $T^*=T^{-1}$.

Eric Wofsey
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In general, an isometric linear operator satisfies $T^*T = I$.

By polarization identity we have $\langle Tx , Ty \rangle = \langle x ,y \rangle$, then $\langle Tx,Ty \rangle = \langle x,T^*Ty \rangle $, ie., $\langle x,(T^*T-I)y\rangle = 0$ for all $x,y$. Then $T^*T = I$.

In your case, if $T$ is subjective, then $T^{-1}$ exists and $$T^{-1} = I T^{-1} = (T^*T)T^{-1} = T^*$$

DCK
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