Showing the polarization of (complex) quadratic form is sesquilinear. Let $q: V\times V \to \mathbb{C}$ on a complex vector space $V$ be a quadratic form. Define $\tilde q$ by the polarization identity:
$$
\begin{equation}
\tilde q(\phi,\psi)  = \frac{1}{4} [q(\phi + \psi) -q(\phi - \psi) + iq(\phi + i\psi) - iq(\phi - i\psi)]  
\end{equation}
$$
My question is how can I show that $\tilde q$ is sesquilinear?
My plan is to show that (1) $\tilde q$ is skew-symmetric, and (2) linear in the second argument. However, to show (2), I'm not exactly sure how to do so given that $q$ is defined in complex space. I found a hint that tells me to show that

a. $\tilde q(\phi, 2\psi) = 2\tilde q(\phi, \psi)$ 
b. $\tilde q(\phi, \psi + \psi') = \tilde q(\phi, \psi) + \tilde q(\phi, \psi')$
c. $\tilde q(\phi, \pm i\psi ) = \pm i\tilde q(\phi, \psi)$
d. $\tilde q(\phi, \alpha\psi ) = \alpha\tilde q(\phi, \psi )$ for all dyadic rationals in $\mathbb C$.

It makes sense to me that we need to take the complex part into account for $c$ and $d$. However, I have no idea what $d$ means. Also for $a$, is the number 2 just some random real integer?

PS: Here's the definition I want to use for (complex) quadratic form:


*

*$q(\lambda x) = |\lambda|^2q(x)\quad \forall \lambda\in\mathbb C, x\in V$

*$q(\phi+\psi) + q(\phi-\psi) = 2q(\psi)+2q(\phi)$

From 2, we could rewrite the polarization identity as
$$
\tilde q(\phi,\psi) = \frac{1}{2}[q(\phi+i\psi)-q(\phi)-q(\psi)] - \frac{i}{2}[q(\phi+i\psi)-q(\phi)-q(i\psi)]
$$
But I'm unsure if that's helpful for showing $\tilde q$ is sesquilinear.
Thanks for the help!
 A: I think the given hints are quite helpful. Nevertheless a complete proof is somewhat lengthy. Here we show part (a).
We consider $\tilde q: V\times V\to \mathbb{C}$ which fulfills the polarisation identity:
\begin{align*}
\tilde q(\phi,\psi)  = \frac{1}{4} [q(\phi + \psi) -q(\phi - \psi) + iq(\phi + i\psi) - iq(\phi - i\psi)]  \tag{1}
\end{align*}
We obtain for $\phi,\psi,\xi\in V$:
\begin{align*}
\color{blue}{\tilde{q}}&\color{blue}{(\phi,\psi)+\tilde{q}(\phi,\xi)}\tag{2.1}\\
&=\frac{1}{4}\left(q(\phi+\psi)-q(\phi-\psi)+iq(\phi+i\psi)-iq(\phi-\psi)\right.\\
&\quad+\left.q(\phi+\xi)-q(\phi-\xi)+iq(\phi+i\xi)-iq(\phi-\xi)\right)\tag{2.2}\\
&=\frac{1}{4}\left(q\left(\left(\phi+\frac{\psi+\xi}{2}\right)+\frac{\psi-\xi}{2}\right)
+q\left(\left(\phi+\frac{\psi+\xi}{2}\right)-\frac{\psi-\xi}{2}\right)\right.\\
&\quad-q\left(\left(\phi-\frac{\psi+\xi}{2}\right)-\frac{\psi-\xi}{2}\right)
-q\left(\left(\phi-\frac{\psi+\xi}{2}\right)+\frac{\psi-\xi}{2}\right)\\
&\quad+iq\left(\left(\phi+i\frac{\psi+\xi}{2}\right)+i\frac{\psi-\xi}{2}\right)
+iq\left(\left(\phi+i\frac{\psi+\xi}{2}\right)-i\frac{\psi-\xi}{2}\right)\\
&\quad\left.-iq\left(\left(\phi-i\frac{\psi+\xi}{2}\right)-i\frac{\psi-\xi}{2}\right)
-iq\left(\left(\phi-i\frac{\psi+\xi}{2}\right)+i\frac{\psi-\xi}{2}\right)\right)\tag{2.3}\\
&=\frac{1}{2}\left(q\left(\phi+\frac{\psi+\xi}{2}\right)+q\left(\frac{\psi-\xi}{2}\right)
-q\left(\phi-\frac{\psi+\xi}{2}\right)-q\left(\frac{\psi-\xi}{2}\right)
\right.\\
&\quad\left.+iq\left(\phi-i\frac{\psi+\xi}{2}\right)+iq\left(\frac{\psi-\xi}{2}\right)
-iq\left(\phi+i\frac{\psi+\xi}{2}\right)-iq\left(\frac{\psi-\xi}{2}\right)\right)\tag{2.4}\\
&\,\,\color{blue}{=2\tilde{q}\left(\phi,\frac{\psi+\xi}{2}\right)}\tag{2.5}
\end{align*}

Putting $\xi=0$ in (1) we obtain
\begin{align*}
\color{blue}{\tilde{q}(\phi,0)}=\frac{1}{4} [q(\phi) -q(\phi) + iq(\phi) - iq(\phi)] \color{blue}{=0}\tag{2.6}
\end{align*}
Combining (2.1), (2.5) and (3) we finally obtain
\begin{align*}
\color{blue}{\tilde{q}(\phi,\psi)=2\tilde{q}\left(\phi,\frac{\psi}{2}\right)}\tag{2.7}
\end{align*}
and the hint (a) follows.

Comment:

*

*In (2.2) we use the polarisation identity (1) for both terms $\tilde{q}(\phi,\psi)$ and $\tilde{q}(\phi,\xi)$.


*In (2.3) we split the terms conveniently as preparation for the next step.


*In (2.4) we apply the parallelogram identity
\begin{align*}
q\left(\phi+\psi\right)+q\left(\phi-\psi\right)=2\left(q\left(\phi\right)+q\left(\psi\right)\right)
\end{align*}


*In (2.5) we cancel terms and apply (1) again.
Note: This derivation follows closely section 1.2 of Linear Operator in Hilbert Spaces by Joachim Weidmann. The other parts can also be derived from this section.


Add-on [2022-04-17]:
With respect to comments to this post here is some additional information. In order to show sesquilinearity of $\tilde{q}:V\times V\to \mathbb{C}$ we have to show for all $\phi,\psi,\xi\in V$ and for all $a,b\in\mathbb{C}$:
\begin{align*}
\color{blue}{\tilde{q}\left(a\phi+b\psi\xi\right)}&\color{blue}{=a\tilde{q}\left(\phi,\xi\right)+b\tilde{q}\left(\psi,\xi\right)}\tag{*}\\
\color{blue}{\tilde{q}\left(\phi,a\psi+b\xi\right)}&\color{blue}{=a^{*}\tilde{q}\left(\phi,\psi\right)+b^{*}\tilde{q}\left(\psi,\xi\right)}\tag{**}\\
\end{align*}
which is linearity in the first component and antilinearity in the second component. We now list the steps which can be used to show (*) and (**) together with some details.

additivity second component
In order to show antilinearity in the second component we show at first additivity
\begin{align*}
\color{blue}{\tilde{q}\left(\phi,\psi+\xi\right)=\tilde{q}\left(\phi,\psi\right)+\tilde{q}\left(\phi,\xi\right)}\tag{3}
\end{align*}

We obtain from (2.1) and (2.5)
\begin{align*}
\color{blue}{\tilde{q}(\phi,\psi)+\tilde{q}(\phi,\xi)}&=2\tilde{q}\left(\phi,\frac{\psi+\xi}{2}\right)\tag{$\to 2.5$}\\
&\,\,\color{blue}{=\tilde{q}(\phi,\psi+\xi)}\tag{$\to 2.7$}
\end{align*}
and additivity (3) follows.

additivity first component
We want to show
\begin{align*}
\color{blue}{\tilde{q}\left(\phi+\psi,\xi\right)=\tilde{q}\left(\psi,\xi\right)+\tilde{q}\left(\psi,\xi\right)}\tag{4.1}
\end{align*}
and do so by establishing at first
\begin{align*}
\color{blue}{\tilde{q}\left(\phi,\psi\right)^{*}=\tilde{q}\left(\psi,\phi\right)}\tag{4.2}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\tilde{q}(\phi,\psi)^{*}}
&=\frac{1}{4}\left(q\left(\phi+\psi\right)-q\left(\phi-\psi\right)
+i q\left(\phi+i\psi\right)-i q\left(\phi-i \psi\right)\right)^{*}\tag{$\to (1)$}\\
&=\frac{1}{4}\left(q\left(\phi+\psi\right)-q\left(\phi-\psi\right)\right)\\
&\qquad+\frac{1}{4}\left(-i q\left((i)\left(\psi-i\phi\right)\right)+i q\left((-i)\left(\psi+i \phi\right)\right)\right)\tag{$\to\rm{def}$}\\
&=\frac{1}{4}\left(q\left(\psi+\phi\right)-q\left(\psi-\phi\right)
+i q\left(\psi+i \phi\right)-i q\left(\psi-i\phi\right)\right)\\
&\,\,=\color{blue}{\tilde{q}(\psi,\phi)}
\end{align*}
and (4.2) follows. We now obtain
\begin{align*}
\color{blue}{\tilde{q}\left(\phi+\psi,\xi\right)}&=\tilde{q}\left(\xi,\phi+\psi\right)^{*}\tag{$\to (4.2)$}\\
&=\left(\tilde{q}\left(\xi,\phi\right)+\tilde{q}\left(\xi,\psi\right)\right)^{*}\tag{$\to (3)$}\\
&=\tilde{q}\left(\xi,\phi\right)^{*}+\tilde{q}\left(\xi,\psi\right)^{*}\tag{$\to \rm{def}$}\\
&\,\,\color{blue}{=\tilde{q}\left(\phi,\xi\right)+\tilde{q}\left(\psi,\xi\right)}\tag{$\to (4.2)$}\\
\end{align*}
and the claim (4.1), additivity in the second component follows.

conjugate homogeneity second component
We want to show for all $\phi,\psi \in V$ and for all $a\in \mathbb{C}$:
\begin{align*}
\color{blue}{\tilde{q}\left(\phi,a\psi\right)=a^{*}\tilde{q}\left(\phi,\psi\right)}\tag{5.1}
\end{align*}
In order to show (5.1) for arbitray $a\in\mathbb{C}$ we proceed in some steps starting with dyadic rationals. We can show
\begin{align*}
\tilde{q}\left(\phi,\frac{m}{2^n}\psi\right)&=\frac{m}{2^n}\tilde{q}\left(\phi,\psi\right)\qquad\qquad m,n\in\mathbb{N_{0}}\tag{5.2}\\
\tilde{q}\left(\phi,a\psi\right)&=a\tilde{q}\left(\phi,\psi\right)\qquad\qquad\quad\  a\in\mathbb{R_{0}^{+}}\tag{5.3}\\
\tilde{q}\left(\phi,a\psi\right)&=a\tilde{q}\left(\phi,\psi\right)\qquad\qquad\quad\  a\in\mathbb{R}\tag{5.4}\\
\tilde{q}\left(\phi,a\psi\right)&=a^{*}\tilde{q}\left(\phi,\psi\right)\qquad\qquad\quad a\in\mathbb{C}\tag{5.5}\\
\end{align*}


*

*(5.2): Based upon (2.7) we can show the validity for dyadic rationals by induction.


*(5.3): Since the set of dyadic rational numbers is dense in $\mathbb{R}$, we can use a continuity argument to show the validity of (5.1) for $a\geq 0$. In order to do so we recall that for a norm $p:V\to\mathbb{R}_0^{+}$ we have due to the triangle inequality
\begin{align*}
 |p(\phi)-p(\psi)|\leq p\left(\phi\pm\psi\right)
 \end{align*}
It follows for all $\phi,\psi\in V$ and for all $a, a_k\geq 0$
\begin{align*}
|p\left(\phi\pm a_k\psi\right)-p\left(\phi\pm a\psi\right)|
&\leq p\left(a_k\psi-a\psi\right)\leq \left|a_k-a\right|p(\psi)\\
|p\left(\phi\pm i a_k\psi\right)-p\left(\phi\pm i a\psi\right)|
&\leq p\left(i a_k\psi-i a\psi\right)\leq \left|a_k-a\right|p(\psi)\\
\end{align*}
which can be used in conjunction with the polarisation identity (1) to show the claim. We use thereby a sequence $(a_k)_{k\geq 0}$ of non-negative dyadic rationals converging to $a\geq 0$.


*(5.4): Here we can use (1) to show $\tilde{q}\left(\phi,-\psi\right)=-\tilde{q}\left(\phi,\psi\right)$ from which (5.4) follows.


*(5.5): Here we can use (1) to show $\tilde{q}\left(\phi,i\psi\right)=-i\tilde{q}\left(\phi,\psi\right)$ from which (5.5) follows.
and the claim (5.1), conjugate homogeneity in the second component follows.

homogeneity first component
We want to show for all $\phi,\psi \in V$ and for all $a\in \mathbb{C}$:
\begin{align*}
\color{blue}{\tilde{q}\left(a\phi,\psi\right)=a\tilde{q}\left(\phi,\psi\right)}\tag{6}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\tilde{q}\left(a\phi,\psi\right)}&=\tilde{q}\left(\psi,a\phi\right)^{*}\tag{$\to(4.2)$}\\
&=\left(a^{*}\tilde{q}\left(\psi,\phi\right)\right)^{*}\tag{$\to(5.1)$}\\
&=a\tilde{q}\left(\psi,\phi\right)^{*}\tag{$\to \rm{def}$}\\
&\,\,\color{blue}{=a\tilde{q}\left(\phi,\psi\right)}\tag{$\to (4.2)$}\\
\end{align*}
and the claim (6), homogeneity in the first component follows.
Conclusion: We have shown $\tilde{q}:V\times V \to \mathbb{C}$ fulfills additivity in both components, homogeneity in the first component and conjugate homogeneity in the second component, so that (*) and (**) follows. This shows that mappings $\tilde{q}$ for which the polarisation identity is valid are sesquilinear forms.
