5
$\begingroup$

Let $H$ be a Banach space with associated norm $\|-\|.$ Suppose that for any $x,y\in H,$ we have: $$\|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right),$$ which we call the parallelogram law. Then it is a well-known standard fact that $H$ becomes a Hilbert space. This is true both for real and complex coefficients.

I managed to prove this fact for an Hilbert space over $\mathbb{R},$ defining the inner product $$(x,y) \mapsto \langle x,y\rangle=\frac{\|x+y\|^2-\|x-y\|^2}{4}= \frac{\|x+y\|^2-\|x\|^2-\|y\|^2}{2}.$$

Question

How to prove this for the complex case?

The inner product should be in this case $$(x,y)\mapsto\ \alpha(x,y)= \frac{1}{4} \left(\|x + y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right)= \langle x,y \rangle + i \langle x,iy \rangle$$ but I am not able to replicate the proof of the real case.

It would be awesome if there was a slick proof that used the real case to deduce the complex case, but I would still be happy with any kind of direct proof too. Since this is a standard result, if you can provide a reference where a detailed proof is given, that would be excellent too.

$\endgroup$
  • $\begingroup$ @MatheusNunes Mmh, I'm not seeing how $\endgroup$ – user739259 Jan 4 at 2:31
  • $\begingroup$ Sorry, I misunderstood your question. I think that what you want is here math.stackexchange.com/a/506643/523585 $\endgroup$ – Matheus Nunes Jan 4 at 2:40
  • $\begingroup$ @MatheusNunes I don't see how: that question is about deriving the polarization identities once you know you have an inner product; my question is on how to prove that using them you get an inner product. $\endgroup$ – user739259 Jan 4 at 2:44
  • 1
    $\begingroup$ I copied the wrong link. Sorry math.stackexchange.com/a/43848/523585 $\endgroup$ – Matheus Nunes Jan 4 at 2:48
  • 2
    $\begingroup$ To your recent edit: Note if you can show $$c\alpha(x,y) = \alpha(cx,y)$$ for $c \in \mathbb{R}$, then in light of the distributive property and what you've already shown, it suffices to just show it for $c=i$. This should be straightforward to do. $\endgroup$ – Brian Moehring Jan 4 at 4:46
2
$\begingroup$

Let $\|\cdot\|: H\to \mathbb{R}$ be a norm in a complex Banach space $H$ which fulfills for all $x,y\in H$ the parallelogram law \begin{align*} \|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right)\tag{1} \end{align*} We show the map $\alpha:H\times H\to\mathbb{C}$ defined by \begin{align*} \alpha(x,y)= \frac{1}{4} \left\{\|x + y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right\}\tag{2} \end{align*} fulfills for all $x,y,z \in H$ and $c\in\mathbb{C}$ \begin{align*} c\alpha(x,y)=\alpha(cx,y)\tag{3} \end{align*}

Let $x,y,z\in H$. We have \begin{align*} &\color{blue}{\alpha(x,y)+\alpha(z,y)}\\ &\qquad=\frac{1}{4}\left\{\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2\right.\\ &\qquad\qquad\quad\left.+\|z+y\|^2-\|z-y\|^2+i\|z+iy\|^2-i\|z-iy\|^2\right\}\tag{4}\\ &\qquad=\frac{1}{4}\left\{ \left\|\left(\frac{x+z}{2}+y\right)+\frac{x-z}{2}\right\|^2+\left\|\left(\frac{x+z}{2}+y\right)-\frac{x-z}{2}\right\|^2\right.\\ &\qquad\qquad\quad\left.-\left\|\left(\frac{x+z}{2}-y\right)+\frac{x-z}{2}\right\|^2+\left\|\left(\frac{x+z}{2}-y\right)-\frac{x-z}{2}\right\|^2\right.\\ &\qquad\qquad\quad+i\left\|\left(\frac{x+z}{2}+iy\right)+\frac{x-z}{2}\right\|^2+i\left\|\left(\frac{x+z}{2}+iy\right)-\frac{x-z}{2}\right\|^2\\ &\qquad\qquad\quad\left.-i\left\|\left(\frac{x+z}{2}-iy\right)+\frac{x-z}{2}\right\|^2+i\left\|\left(\frac{x+z}{2}-iy\right)-\frac{x-z}{2}\right\|^2\right\}\tag{5}\\ &\qquad=\frac{1}{2}\left\{\left\|\frac{x+z}{2}+y\right\|^2+\left\|\frac{x-z}{2}\right\|^2 -\left\|\frac{x+z}{2}-y\right\|^2-\left\|\frac{x-z}{2}\right\|^2\right.\\ &\qquad\qquad\quad\left.+i\left\|\frac{x+z}{2}+iy\right\|^2+i\left\|\frac{x-z}{2}\right\|^2 -i\left\|\frac{x+z}{2}-iy\right\|^2-i\left\|\frac{x-z}{2}\right\|^2\right\}\tag{6}\\ &\qquad\,\,\color{blue}{=2\alpha\left(\frac{x+z}{2},y\right)}\tag{7} \end{align*}

Comment:

  • In (4) we use the definition of $\alpha$ from (2).

  • In (5) we do some preparatory work in order to apply the parallelogram law.

  • In (6) we apply the parallelogram law (1).

Since we have from (2) \begin{align*} \alpha(x,0)&=\frac{1}{4} \left\{\|x\|^2 - \|x\|^2 + i\|x\|^2 -i\|x\|^2 \right\}=0 \end{align*} we obtain by substituting $z=0$ in (7) \begin{align*} \alpha(x,y)=2\alpha\left(\frac{x}{2},y\right)\tag{8} \end{align*}

We obtain from (7) and (8) by induction that for all $n,m\in\mathbb{N}_0$: \begin{align*} 2^{-n}m\alpha(x,y)=\alpha\left(2^{-n}m x,y\right) \end{align*}

Case $c\geq 0$:

If $c \geq 0$, then we have numbers $c_k=2^{-n(k)}m(k)$ such that $c_k\to c$ as $k\to \infty$. Since the norm $\|\cdot\|$ fulfills for all $x,y\in H$: \begin{align*} \left|\|x\|-\|y\|\right|\leq \|x\pm y\| \end{align*} it follows \begin{align*} &\left|\|c_kx\pm y\|-\|cx\pm y\|\right|\leq \left|c_k-c\right|\|x\|\\ &\left|\|ic_kx\pm y\|-\|icx\pm y\|\right|\leq \left|c_k-c\right|\|x\|\\ \end{align*} and we obtain \begin{align*} \alpha\left(c_kx,y\right)\to\alpha(cx,y)\qquad \text{as }k\to\infty \end{align*} We conclude \begin{align*} \color{blue}{c\alpha(x,y)}=\lim_{k\to\infty}c_k\alpha(x,y)=\lim_{k\to\infty}\alpha\left(c_kx,y\right)\color{blue}{=\alpha(cx,y)} \end{align*}

Case $c\in\mathbb{R}$:

Since \begin{align*} \color{blue}{\alpha(-x,y)}&=\frac{1}{4} \left\{\|-x + y\|^2 - \|-x-y\|^2 + i\|-x + iy\|^2 -i\|-x-iy\|^2 \right\}\\ &=-\frac{1}{4} \left\{-\|x - y\|^2 + \|x+y\|^2 - i\|x - iy\|^2 +i\|x+iy\|^2 \right\}\\ &\,\,\color{blue}{=-\alpha(x,y)} \end{align*} we have $\alpha(cx,y)=c\alpha(x,y)$ for all $c \in \mathbb{R}$.

Case $c\in\mathbb{C}$:

Since \begin{align*} \color{blue}{\alpha(ix,y)}&=\frac{1}{4} \left\{\|ix + y\|^2 - \|ix-y\|^2 + i\|ix + iy\|^2 -i\|ix-iy\|^2 \right\}\\ &=\frac{i}{4} \left\{-i\|x - iy\|^2 + i\|x+iy\|^2 +\|x +y\|^2 -\|x-y\|^2 \right\}\\ &\,\,\color{blue}{=i\alpha(x,y)} \end{align*} we have $\alpha(cx,y)=c\alpha(x,y)$ for all $c \in \mathbb{C}$ and the claim (3) follows.

Note: This post follows closely the proof provided in section 1.2 of Linear Operators in Hilbert Spaces by J. Weidmann.

$\endgroup$
  • $\begingroup$ Thank you so much. If you want, see my edit. $\endgroup$ – user739259 Jan 5 at 0:35
  • $\begingroup$ @marialami: You're welcome. $\endgroup$ – Markus Scheuer Jan 5 at 6:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy