Pre-Hilbert space and polarization identities

Question

These questions were asked in my assignment of a self-study course in Functional Analysis, and I need help in it.

Questions
(a) Let $( H, \|\cdot\|\,)$ be a pre-Hilbert space on $K=\mathbb{C}$, and let $x,y \in H$. Express $\langle x,y\rangle\,$ in terms of $\,\|x+y\|$ , $\|x-y\|$ , $\|x+iy\|$, and $\|x-iy\|\,$.

(b) Let $H$ be a complex Hilbert space. Show that if $\,T\in L(H)\,$ and $\,x,y \in H$, then we have the polarization identity $$ 4\langle Tx,y\rangle \:=\: \langle T(x+y) , x+y\rangle - \langle T(x-y), x-y\rangle + i\,\langle T(x+iy), x+iy\rangle -i\,\langle T(x-iy), x-iy\rangle\,.$$

Attempt: I don't have much to show as attempt (I am sorry), just give me a few but separate hints so that I can proceed towards the solution in both parts.

Answer 1

It would be economic to verify (b) first, then to infer (a) by choosing $T=id_H$ to be the identity map. It is possible to proceed like this because the polarisation identity is an algebraic identity involving only the inner product structure or the associated norm, but it does not depend on completeness which makes the relevant difference between Hilbert spaces and pre-Hilbert spaces.

To check the given identity, you may expand the summands on the RHS, and see what will happen (cancel). Notice that in complex vector spaces the scalar product is a sesquilinear form, i.e., it is linear in one slot, and conjugate-linear in the other slot. Let's assume linearity in the first argument, and expand for example the third summand as $$\begin{align} i\,\langle T(x+iy), x+iy\rangle & \;=\; i\,\langle Tx, x\rangle \:+\:i\cdot i\cdot (-i)\,\langle Ty,y\rangle \:+\: i\cdot (-i)\,\langle Tx,y\rangle \:+\:i\cdot i\,\langle Ty,x\rangle\\[1.5ex] & \;=\; i\,\langle Tx, x\rangle \:+\:i\,\langle Ty,y\rangle \:+\:\langle Tx,y\rangle \:-\,\langle Ty,x\rangle \end{align}$$ Expanding all summands and collecting terms yields the desired equality in (b).

Since (a) asks for a formulation in terms of the norm you may recall how the norm in a (pre-)Hilbert space is given by the inner product: $$\|z\|^2\:=\:\langle z,z\rangle\quad\forall z\in H$$

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The polarisation identity in (a) is a kind of converse as it yields the inner product in terms of the norm.

Answer 2

Let me give hints for both part a) and for part b) that should also give you some way of solving problems like these on your own in the future.

a) Let's recall that $\left|\left|x\right|\right| := \sqrt{\langle x,x\rangle}$. This is the definition of the norm of a vector. Now, what you want is the inner product $\langle x,y \rangle$. If you look at the definition of the norm above, then the only way to "solve" for the inner product is to square both sides. In other words, to get an inner product of a vector with itself, you need to square the norm.

So it stands to reason that you should be looking at $\left|\left|x+y\right|\right|^2$, $\left|\left|x-y\right|\right|^2$, $\left|\left|x+iy\right|\right|^2$ and $\left|\left|x-iy\right|\right|^2$. Now, what you should do, at this stage, is to expand all of these out using the properties of the inner product. Once you've done this, it's a matter of adding and subtracting things so that you can "solve" $4$ equations to get $\langle x,y \rangle$.

b) For this problem, you could take the easy way out and just plug & chug the right-hand side. That is, you could expand all the inner products on the right-hand side and you would obtain the left-hand side if you do it correctly. This is probably the intended solution.

But now, suppose that I gave you the expression $\langle T(x),y\rangle$. How would you "discover" the expression on the right-hand side? If I were to just read this inner product allowed, I would say that "it is the inner product of elements from the image of $T$ and arbitrary elements from $H$". In particular, it might be interesting to see how a linear map changes the "orientation" of a vector $x$. This can be measured by looking at $\langle T(x),x\rangle$.

So now, if I already know how $\langle T(x),x\rangle$ behaves for every $x \in H$, then I could just try to use that to calculate $\langle T(x),y\rangle$. This serves as the motivation to look at the following inner products; $\langle T(x+y),x+y \rangle$, $\langle T(x-y),x-y \rangle$. So, once you look at these, it then becomes natural to make use of the strategy in part (a) to "cancel" out the extra terms so that you get $\langle T(x),y\rangle$.