# Polarization identity if operator

Let $$\mathcal{H}$$ be a Hilbert space over $$\mathbb{C}$$. I know (and proved) the following polarization identity: $$\begin{eqnarray} \langle x, y \rangle = \frac{1}{4}\sum_{k=0}^{3}i^{k}\langle x+i^{k}y, x+i^{k}y \rangle \tag{1}\label{1} \end{eqnarray}$$ for $$x,y \in \mathcal{H}$$. Now, I'd like to prove the following result as a consequence of (\ref{1}): $$\begin{eqnarray} \langle Ax, y \rangle = \frac{1}{4}\sum_{k=0}^{3}i^{k}\langle A(x+i^{k}y), x+i^{k}y \rangle \tag{2}\label{2} \end{eqnarray}$$ where $$A$$ is any given bounded linear operator on $$\mathcal{H}$$. If you change $$x$$ to $$Ax$$ in (\ref{1}) you get: $$\begin{eqnarray} \langle Ax, y \rangle = \frac{1}{4}\sum_{k=0}^{3}i^{k}\langle Ax+iy, Ax+iy \rangle \tag{3}\label{3} \end{eqnarray}$$ but not (\ref{2}) and I'm stuck at this point. How does (\ref{1}) imply (\ref{2})?

• Nice format of question
– R.W
Commented Aug 18, 2020 at 17:08
• Hint. Check that the polarisation identity only needs sesquilinearity of $\langle \cdot, \cdot \rangle$ (and not positive definiteness), thus you can replace it by the sesquilinear form $\langle\cdot,\cdot\rangle_A= \langle A\cdot,\cdot\rangle$. Commented Aug 18, 2020 at 17:13
• Is there a typo in (1)? Commented Aug 18, 2020 at 17:21
• @JackyChong yes! Edited it! Thanks! Commented Aug 18, 2020 at 17:48
• There is a typo in 2 also it should be $x + i^k y$ Commented Apr 22, 2023 at 23:39

Thanks to the comments, I believe I got it. First, we must show that the polarization identity also holds in a more general context, namely if $$T:\mathcal{H}\times \mathcal{H}\to \mathbb{C}$$ satisfies the following properties:

(a) $$T(x,\alpha y + \beta z) = \bar{\alpha}T(x,y)+\bar{\beta}T(x,z)$$

(b) $$T(\alpha x + \beta z, y) = \alpha T(x,y) + \beta T(z,y)$$

then, it follows that:

$$\begin{eqnarray} T(x,y) = \frac{1}{4}\sum_{k=0}^{3}i^{k}T(x+i^{k}y,x+i^{k}y) \tag{1'}\label{1.1} \end{eqnarray}$$

Proof of (\ref{1.1}): Write: $$T(x,y) = T\bigg{(}\frac{1}{2}(x+iy+x-iy), \frac{1}{2i}(x+iy - (x-iy))\bigg{)}$$ Using properties (a) and (b), we get: $$T(x,y) = -\frac{1}{4i}[T(x+iy,x+iy)-T(x+iy,x-iy)+T(x-iy,x+iy)-T(x-iy,x-iy)]$$

Now, note that: $$-\frac{1}{4i}[T(x+iy,x+iy)-T(x-iy,x-iy)] = \frac{1}{4}i[T(x+iy,x+iy)-T(x-iy,x-iy)]$$ and, using (a) and (b) again, we also have: $$\frac{1}{4}i [-T(x+iy,x-iy)+T(x-iy,x+iy)] = \frac{1}{4}i[2T(x,iy)-2T(iy,x)] = \frac{1}{4}i[-2iT(x,y)-2iT(y,x)] = \frac{1}{2}[T(x,y)+T(y,x)] = \frac{1}{4}[T(x+y,x+y)-T(x-y,x-y)]$$ and (\ref{1.1}) follows.

Now, the result follows by setting $$T(x,y) := \langle Ax,y\rangle$$ for all $$x,y \in \mathcal{H}$$.