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A Banach space that satisfies parallelogram law is a Hilbert space.

I know the definitions of the space and also the law, but have no idea about how to use it to prove this fact. Thanks in advance for help!

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First a comment: this is not as easy as it looks.

Suppose that for all $x,y$ in $X$ it is $\|x-y\|^2+\|x+y\|^2=2\|x\|^2+2\|y\|^2$. Define $\langle\cdot,\cdot\rangle:X\times X\to\mathbb{R}$ by $$\langle x,y\rangle=\frac{\|x+y\|^2-\|x-y\|^2}{4}=\text{(by property)} \frac{\|x+y\|^2-\|x\|^2-\|y\|^2}{2}$$

We will prove that our map is an inner product and that it also induces the already known norm in the classical way:

1) $\langle x,x\rangle=\|x\|^2\geq0$ (this also proves that $\|x\|=\sqrt{\langle x,x\rangle}$.)

2) $\langle x,y\rangle=\langle y,x\rangle$, obviously.

Now we need to prove that $\langle x+y,z\rangle=\langle x,z\rangle+\langle y,z\rangle$ and that $\langle cx,y\rangle=c\langle x,y\rangle$ for all $x,y,z\in X$ and $c\in\mathbb{R}$. I will leave the first property unproven, since it is only a matter of simple but harsh calculations and applications of the supposed property. Let's see how to prove the second property (we will use the first).

We begin by proving it for $c\in\mathbb{N}$. Indeed, for $c=1$ it is obviously true; suppose that it is true for some $n\in\mathbb{N}$. Then $\langle (n+1)x,y\rangle=\langle nx+x,y\rangle=\langle nx,y\rangle+\langle x,y\rangle=n\langle x,y\rangle+\langle x,y\rangle=(n+1)\langle x,y\rangle$, so by induction this is true for any positive integer.

Now let's prove this for $c=-1$. It is $\langle -x,y\rangle=\frac{1}{2}(\|-x+y\|^2-\|-x-y\|^2)=\frac{1}{2}(\|x-y\|^2-\|x+y\|^2)=-\langle x,y\rangle$.

Now let's prove that if the property is true for $c_1,c_2\in\mathbb{R}$, then it is also true for $c_1\cdot c_2$. Indeed, it is $\langle (c_1\cdot c_2)x,y\rangle=\langle c_1\cdot(c_2 x),y\rangle=c_1\langle c_2x,y\rangle=c_1c_2\langle x,y\rangle$.

Applying the last property for $c_1=n\in\mathbb{N}$ and $c_2=-1$, we have that our desired property holds for $c\in\mathbb{Z}$.

Let $n\in\mathbb{Z}-\{0\}$. It is $\langle x,y\rangle=\langle \frac{n}{n}x,y\rangle=n\langle \frac{1}{n}x,y\rangle$, therefore $\langle\frac{1}{n}x,y\rangle=\frac{1}{n}\langle x,y\rangle$.

Using the $c_1,c_2$-property for $c_1=1/n$ for some $n\in\mathbb{Z}$ and $c_2=m\in\mathbb{Z}$ we have that our result is true for every $c\in\mathbb{Q}$.

So we have proved that for all $c\in\mathbb{Q}$ it is $\langle cx,y\rangle=c\langle x,y\rangle$. How are we supposed to move on to $\mathbb{R}$? By using the density of $\mathbb{Q}$ in $\mathbb{R}$. Observe that the $\langle\cdot,\cdot\rangle$ map is continuous in every argument (it is expressed as a linear combination of norms, which are continuous maps). So, if $x_n\to x$, then for all $y\in X$ it is $\langle x_n,y\rangle\to\langle x,y\rangle$.

Let $x,y\in X$. For some $r\in\mathbb{R}$, choose a sequence $(q_n)\subset\mathbb{Q}$ such that $q_n\to r$. Then it is $q_nx\to rx$ in $X$, therefore $\langle q_nx,y\rangle\to\langle rx,y\rangle$. But $\langle q_nx,y\rangle=q_n\langle x,y\rangle\to r\langle x,y\rangle$ and we are done, by the uniqueness of the limit.

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  • $\begingroup$ I think you also need argue for the Complex case. $\endgroup$ – reflexive Feb 11 '19 at 4:45
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One direction is easy. For the other, suppose $X$ satisfies the parallelogram law. If the field is $\mathbb R$, define \begin{align} \langle x,y \rangle := \frac{1}{4} \left(\|x + y\|^2 - \|x-y\|^2 \right) \end{align} The use the parallelogram to show that $\langle x + z,y \rangle \, =\, \langle x,y\rangle + \langle z,y\rangle$. Symmetry is obvious. For the $\langle \lambda x,y \rangle\,=\, \lambda \langle x,y \rangle$ you should start with $\lambda \in \mathbb N$,then $\lambda \in \mathbb Z$, then $\lambda \in \mathbb Q$ and then use continuity. Etc.

If the field is $\mathbb C$, then you can do the same with \begin{align} \langle x,y \rangle := \frac{1}{4} \left(\|x + y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right) \end{align} But it takes much more time.

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