parallelogram equation in euclidean spaces

Question

I have to show that normed space is euclidean space iff for any $x$ and $y$ perform parallelogram equation, i.e $||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$. I proved this one way. in the opposite direction, I take $(x,y)=\frac{1}{4}(||x+y||^2-||x-y||^2)$ and I proved the linearity, but I have some problem with proving that $(\lambda x,y)=\lambda(x,y)$. Firsly we can easly prove that for $-1$ and by induction that this is done $\forall \lambda\in\mathbb Z$ and therefore $\forall \lambda\in\mathbb N$.(This follows from linearity). I'm also already proof that $\forall \lambda\in\mathbb Q$, but I have some problems to prove this for irrational $\lambda$

Answer 1

For fixed $x,y$ the function $\lambda \mapsto \langle \lambda x,y\rangle- \lambda \langle x,y \rangle$ is continuous, and you've proved it's $0$ on a dense set.

Answer 2

Hint: For any real number $r$, take a sequence $q_n\in\mathbb{Q}$ of rationals converging to $r$ and use continuity.