Inner product and parallelogram

Question

Let $V$ be a normed vector space (over $\mathbb{R}$, say) with norm $||•||$. I have already proven that if $$||•|| = \sqrt{\langle•,•\rangle}$$ for some real inner product $\langle•,•\rangle$ then the parallelogram law $$||u+v||^2 + ||u-v||^2 = 2||u||^2 + 2||v||^2$$ holds for all pairs of $u,v \in V$.

I'm having difficulties with the converse. Assuming the parallelogram law, and defining the map
$$\langle u,v\rangle = (1/4)(||u+v||^2 - ||u-v||^2)$$ I'm able to show that
1) $\langle v,v\rangle = ||v||^2$,
2) $\langle u,v\rangle = \langle v,u\rangle$,
3) $\langle u, v+w\rangle = \langle u,v\rangle + \langle u,w\rangle$,

Note to 3), this is done by showing that
3.1) $\langle u, v+w\rangle = 2\langle u/2, v\rangle + 2\langle u/2, w\rangle$
3.2) $\langle u,v\rangle = 2\langle u/2, v\rangle$
so 3.1) is equivalent to 3)

Now to show the last part
4) $\langle au,v\rangle = a\langle u,v\rangle$ for a real number a, I have been able to show that
4.1) $\langle nu, v\rangle = n\langle u,v\rangle$ for natural n and zero
4.2) $\langle pu, v\rangle = p\langle u,v\rangle$ for rational p

But to show that this linearity holds for a real number, I have tried to extend the argument, and seem stuck. I searched the net and found this https://jlch3554.files.wordpress.com/2012/11/la-solution-2011-7.pdf (Page 157, exercise 27) it seems to be shown by using Cauchy-schwartz (which is not difficult to prove for this $\langle •,•\rangle$) and then use

For real $r$ and rational $p$ (using linearity of $p$)
5.1) $r\langle u,v\rangle - \langle ru,v\rangle = (r-p)\langle u,v\rangle - \langle(r-p)u,v\rangle$

Together with (Since $|\langle u,v\rangle| ≤ ||u|| ||v||$ by Cauchy Schwartz)
5.2) $-|r-p| ||u|| ||v|| ≤ (r-p) ||u|| ||v|| ≤ |r-p| ||u|| ||v||$

That $$|r\langle u,v\rangle - \langle ru,v\rangle| = |(r-p)\langle u,v\rangle - \langle (r-p)u,v\rangle| ≤ 2|r-p| ||u|| ||v||$$

But... I think that this is only true if both (r-p)\langle u,v\rangle and \langle (r-p)u, v\rangle are in the middle of the inequality of 5.2) for the same ends.

That is: WHY is $$|\langle (p-r)u, v\rangle| \leq |(p-r)| ||u||\cdot ||v||$$ I recognise the use of Cauchy Schwartz, but the linearity (that is; pulling out the factor $(p-r)$, seems to violate - or be exactly what we try to prove? Therefore, I do not see that this last step is true. If it is however, the proof seems to be done, as by squeezing the difference it must be zero so the parts are equal.

Answer 1

Let us assume you have proved for all $u, v, w \in V$ and all $p \in \mathbb Q $, $$\begin{align} &\langle u+v, w \rangle = \langle u,w \rangle + \langle v,w \rangle &\text{additivity} \tag 1\\ &\langle pu, v \rangle = p \langle u,v \rangle &\text{multiplication by rational} \tag{2} \\ &\left| \langle u,v \rangle \right | \leqslant || u || \cdot || v || &\text{Cauchy-Schwarz} \tag{3} \end{align}$$ Then the missing steps are:

Choose any $u,v \in V$. Let $ r $ be a real and $ p $ a rational that is close to $r$, so $|r-p| < \varepsilon $ for some freely chosen small $\varepsilon$.
$$\begin{align} \left| \langle ru,v \rangle - r\langle u,v \rangle \right| &= \left| \langle ru,v \rangle - \langle pu,v \rangle + \langle pu,v \rangle - r \langle u,v \rangle \right| \\ &=\left | \langle ru,v \rangle - \langle pu,v \rangle + p\langle u,v \rangle - r \langle u,v \rangle \right| &\text{multiplication by rational} \\ &\leqslant \left| \langle (r-p)u,v\rangle \right| + |p-r| \left| \langle u,v \rangle \right| &\text{additivity, triangle inequality for reals} \\ &\leqslant || (r-p)u ||\cdot ||v|| + |p-r| ~|| u || \cdot || v || &\text{Cauchy-Schwarz} \\ &\leqslant |r-p| ~ ||u|| \cdot || v || + |p-r| ~ || u|| \cdot || v|| &\text{property of norm, } ||\lambda u || = |\lambda|\cdot ||u|| \\ &=2|r-p|~ || u|| \cdot ||v|| \\ &\leqslant 2 \varepsilon \cdot ||u|| \cdot ||v|| \end{align}$$ Now $||u||, ||v||$ are fixed and $ \varepsilon $ arbitrary, so $\langle ru,v \rangle = r\langle u,v \rangle $.