Norms Induced by Inner Products and the Parallelogram Law
Solution 1:
Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove.
So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law $$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x - y \Vert^2$$ for all $x,y \in V$ and put $$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$ We're dealing with real vector spaces and defer the treatment of the complex case to Step 4 below.
Step 0. $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$.
Obvious.
Step 1. The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$.
Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous.
Remark. This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps.
Step 2. We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$.
By the parallelogram law we have $$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y + z \Vert^2 + \Vert x - y + z\Vert^2 .$$
This gives $$\begin{align*} \Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 - \Vert x - y + z \Vert^2 \\ & = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 - \Vert y - x + z \Vert^2 \end{align*}$$ where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get
$$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 - \frac{1}{2}\Vert x - y + z \Vert^2 - \frac{1}{2}\Vert y - x + z \Vert^2.$$
Replacing $z$ by $-z$ in the last equation gives $$\Vert x + y - z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x - z \Vert^2 + \Vert y - z \Vert^2 - \frac{1}{2}\Vert x - y - z \Vert^2 - \frac{1}{2}\Vert y - x - z \Vert^2.$$
Applying $\Vert w \Vert = \Vert - w\Vert$ to the two negative terms in the last equation we get $$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 - \Vert x + y - z \Vert^2\right) \\ & = \frac{1}{4}\left(\Vert x + z \Vert^2 - \Vert x - z \Vert^2\right) + \frac{1}{4}\left(\Vert y + z \Vert^2 - \Vert y - z \Vert^2\right) \\ & = \langle x, z \rangle + \langle y, z \rangle \end{align*}$$ as desired.
Step 3. $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$.
This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x' = \dfrac{x}{q}$ that $$q \langle \lambda x, y \rangle = q\langle p x', y \rangle = p \langle q x', y \rangle = p\langle x,y \rangle,$$ so dividing this by $q$ gives $$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$ We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we're done.
Step 4. The complex case.
Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$).
Addendum. In fact we can weaken requirements of Jordan von Neumann theorem to $$ 2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2 $$ Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get $$ \Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2 $$ which together with previous inequality gives the equality.
Solution 2:
It's not immediate or trivial, so I wouldn't feel too bad for having trouble. This is an exercise in Friedberg, Insel, and Spence's Linear Algebra, 4th Edition, which has an extensive 8 part "Hint." Here's an edited sequence of hints, following theirs:
First, prove that the result holds for $\lambda = 2$, that is, $\langle 2u,v\rangle = 2\langle u,v\rangle$.
Then, prove that the inner product is additive in the first component: $\langle x+u,v\rangle = \langle x,v\rangle + \langle u,v\rangle$.
Then, prove the result holds for $\lambda$ any positive integer. Then for the reciprocal $\frac{1}{m}$ of any positive integer. Then for any rational number.
Then prove that $|\langle u,v\rangle|\leq ||u||\,||v||$
Then prove that for every $\lambda\in\mathbb{R}$, every $r\in\mathbb{Q}$, you have $$|\lambda\langle u,v\rangle - \langle \lambda u,v\rangle | = |(\lambda-r)\langle u,v\rangle - \langle(\lambda-r)u,v\rangle|\leq 2|\lambda-r|\,||u||\,||v||.$$
Finally, use that to prove homogeneity: for every $\lambda\in\mathbb{R}$, $\langle\lambda u,v\rangle = \lambda\langle u,v\rangle$.
Solution 3:
One should also convince oneself that: $$\langle\cdot,\cdot\rangle\to\|\cdot\|\to\langle\cdot,\cdot\rangle$$ $$\|\cdot\|\to\langle\cdot,\cdot\rangle\to\|\cdot\|$$ (Otherwise really bad things could happen...)
Luckily, this can be checked rather easily: $$\|x\|'=\sqrt{\frac{1}{4}\left(\|x+x\|^2-\|x-x\|^2\right)}=\|x\|$$ $$\langle x,y\rangle'=\frac{1}{4}\left(\sqrt{\langle x+y,x+y\rangle}^2-\sqrt{\langle x-y,x-y\rangle}^2\right)=\langle x,y\rangle$$
Solution 4:
(sorry, I'm not familiar with LaTex, hope you have no difficutly reading my answer).
I have another method to prove this result for $\ x,y \in \mathbb{R^n}$, $\lambda \in \mathbb{R}$.
step.1. prove that $\langle \lambda x,\lambda y \rangle = \lambda^2 \langle x, y \rangle$, use polarisation identity to expand inner product, it's easy to prove it.
step.2. prove that $\langle \lambda x, y \rangle = \langle x, \lambda y \rangle$, i.e.prove $\langle \lambda x, y \rangle - \langle x, \lambda y \rangle=0$, the proof is similar to proving $\langle x+y, z \rangle = \langle x, z \rangle+\langle y, z \rangle$, you may use $\langle x+y, z \rangle = \langle x, z \rangle+\langle y, z \rangle$ and $\langle -x, y \rangle = -\langle x, y \rangle=\langle x,-y \rangle$ in your proof and the latter is easy to prove.
step.3. use the result of step.2., we have $\langle \lambda x,\lambda y \rangle = \langle \lambda^2 x, y \rangle$, compare to the result in step.1., we get that $\langle \lambda^2 x , y \rangle = \lambda^2 \langle x, y \rangle$, for $\lambda^2 >0$.
step.4 when $\lambda <0$, $\langle \lambda x, y \rangle = \langle-(-\lambda x), y \rangle=-\langle(-\lambda x), y \rangle$, use the result in step.3., we get $\langle(-\lambda x), y \rangle=-\lambda\langle x, y \rangle$, which means the result holds for $\lambda <0$, and it's easy to prove when $\lambda =0$.
It's a bit more complex to prove for $\ x,y \in \mathbb{C^n}, \lambda \in \mathbb{C}$, while the method is general, you just have to seperate it into two parts, the real and the imaginary.