What Does the Polarization Identity Imply for Operators?

The polarization identity says that the inner product can be recovered from norm values. Its most common operator-theory use is to show that a linear map preserving norms also preserves inner products. For a bounded operator $T$ on a Hilbert space, that conclusion is equivalent to $T^*T=I$.

This does not, by itself, show that $T$ is onto or that $TT^*=I$. The distinction between an isometry and a unitary operator matters in infinite-dimensional Hilbert spaces.

The polarization identity

Use the convention that the inner product is linear in its first argument. For a complex Hilbert space,

$$\langle x,y\rangle=\frac{1}{4}\sum_{k=0}^{3} i^k\|x+i^k y\|^2.$$

Although the right-hand side uses only norms, it recovers the full inner product. In particular, knowing the norm of every vector determines the inner product.

Norm preservation implies inner-product preservation

Suppose $T:H\to K$ is linear between complex Hilbert spaces and

$$\|Tx\|=\|x\|\qquad\text{for every }x\in H.$$

Apply the polarization identity to $Tx$ and $Ty$. By linearity, $T(x+i^k y)=Tx+i^k Ty$, and by the norm-preserving assumption, $\|Tx+i^k Ty\|=\|x+i^k y\|$. Therefore

$$\begin{aligned} \langle Tx,Ty\rangle &=\frac{1}{4}\sum_{k=0}^{3}i^k\|Tx+i^k Ty\|^2\\ &=\frac{1}{4}\sum_{k=0}^{3}i^k\|x+i^k y\|^2\\ &=\langle x,y\rangle. \end{aligned}$$

So $T$ is an isometry: it preserves both norms and inner products. In particular, it is injective, since $Tx=0$ implies $\|x\|=\|Tx\|=0$.

Why this gives $T^*T=I$

Now suppose $T:H\to H$ is a bounded linear operator. For every $x,y\in H$, the adjoint satisfies

$$\langle Tx,Ty\rangle=\langle x,T^*Ty\rangle.$$

If $T$ preserves inner products, then $\langle x,T^*Ty\rangle=\langle x,y\rangle$ for all $x,y$. Nondegeneracy of the inner product implies $T^*Ty=y$ for every $y$, hence

$$T^*T=I.$$

Conversely, if $T^*T=I$, then

$$\|Tx\|^2=\langle Tx,Tx\rangle=\langle x,T^*Tx\rangle=\|x\|^2,$$

so $T$ preserves norms. Thus for a bounded operator on a Hilbert space, norm preservation, inner-product preservation, and $T^*T=I$ are equivalent.

Worked example: the unilateral shift

On $\ell^2(\mathbb{N}_0)$, define the right shift by

$$S(x_0,x_1,x_2,\ldots)=(0,x_0,x_1,x_2,\ldots).$$

Its norm is unchanged:

$$\|Sx\|^2=0+|x_0|^2+|x_1|^2+\cdots=\|x\|^2.$$

The polarization identity therefore gives $\langle Sx,Sy\rangle=\langle x,y\rangle$, and $S^*S=I$. But $S$ is not onto: every vector in its range has first coordinate zero. In fact, $SS^*$ is the projection that sets the first coordinate to zero, not the identity. This example shows why an isometry need not be unitary.

A related consequence for quadratic forms

The complex polarization identity can also recover an operator from its diagonal values. If $A$ and $B$ are bounded operators on a complex Hilbert space and $\langle Ax,x\rangle=\langle Bx,x\rangle$ for every $x$, then applying polarization to the sesquilinear forms $\langle Ax,y\rangle$ and $\langle Bx,y\rangle$ gives $\langle Ax,y\rangle=\langle Bx,y\rangle$ for all $x,y$. Hence $A=B$. This conclusion uses complex scalars; over a real Hilbert space, diagonal values of a general operator determine only its symmetric part. For self-adjoint real operators, the quadratic form does determine the operator.

Common mistakes

• Assuming norm preservation directly proves surjectivity. It proves injectivity, not surjectivity. The unilateral shift is a counterexample.
• Confusing $T^*T=I$ with $TT^*=I$. The first says $T$ is an isometry; both equations together say it is unitary.
• Using the complex formula unchanged over the reals. The four-term formula uses $i$ and does not apply to real Hilbert spaces. In real spaces, the polarization identity is $\langle x,y\rangle=(\|x+y\|^2-\|x-y\|^2)/4$.