My question is motivated by this Mathoverflow post about inner products, where it's written in one of the answers that
...In fact, one can derive continuity using only the inequality |u|2≥0 and the parallelogram law.) Also, an algebraic argument must work over any field on characteristic 0.
My question: Let $G$ be an abelian group, and $\phi:G \to \mathbb{R}$ a non-negative function. If $\phi$ satisfies the parallelogram identity $$ 2 \phi(x) + 2 \phi(y) = \phi(x+y) + \phi(x-y) $$ then is it the following also true? $$ \lim_{\phi(h) \to 0} \phi(x + h) = \phi(x) $$
I have not been able to easily find any proofs online, but expected there to be some.
Thanks for any answers or references.
Edit: I think I came up with a proof which uses the parallelogram law a couple of times to show $$\phi(0) = 0$$ $$\phi(-x) = \phi(x)$$ $$\phi(kx) = k^2\phi(x) , \hspace{8mm}k \in \mathbb{N}$$ \begin{align} \Bigg| \phi(x + h) - \phi(x) \Bigg| &= \Bigg| \frac{1}{2 k} \Big\{ \phi(x + kh) - \phi(x - kh) \Big\} + \phi(h) \Bigg| \\ & \le \Bigg| \frac{1}{2 k} \Big\{ \phi(x + kh) + \phi(x - kh) \Big\} \Bigg| + \phi(h) \\ &= \frac{1}{k} \Big\{ \phi(x) + \phi(kh) \Big\} + \phi(h) \\ &= \frac{1}{k} \Big\{ \phi(x) + k^2 \phi(h) \Big\} + \phi(h) \\ &= \frac{1}{k} \Big\{ \phi(x) \Big\} + (k + 1) \phi(h) \\ \end{align} when $k \in \mathbb{N}$.
If $\phi(x)=0$, then the parallelogram law gives $$\phi(x+h)=2\phi(h)-\phi(x-h)\leq 2\phi(h)$$ and so your limit condition holds. Let us now assume $\phi(x)\neq 0$; we may scale $\phi$ to assume that $\phi(x)=1$. I now claim that $$\phi(x+h)\geq 1-4\sqrt{\phi(h)}$$ for all $h$ such that $\phi(h)$ is sufficiently small. The desired limit condition easily follows (we can get an upper bound on $\phi(x+h)=2+2\phi(h)-\phi(x-h)$ using the lower bound on $\phi(x-h)$).
To prove the claim, suppose we have $h$ such that $\phi(x+h)<1-4\sqrt{\phi(h)}$. Let $C=1-\phi(x+h)$, so $C>4\sqrt{\phi(h)}$. Note that the parallelogram law can be rearranged to $$\phi(x)-\phi(x+y)=\phi(x-y)-\phi(x)-2\phi(y).$$ We then see by induction that $$\phi(x+kh)-\phi(x+(k+1)h)= C-2k\phi(h)$$ for all $k\in\mathbb{Z}$ (the base case $k=0$ being the definition of $C$). It follows that $$\phi(x+kh)=1-kC+k(k-1)\phi(h)$$ for all $k\in\mathbb{Z}$. The idea now is that because $C$ is large, we can choose $k$ appropriately to make $\phi(x+kh)$ negative and reach a contradiction. If $\phi(h)=0$, we see immediately that $\phi(x+kh)<0$ for $k$ sufficiently large since $C>0$. So let us suppose $\phi(h)\neq 0$; to minimize $\phi(x+kh)$ we complete the square in $k$ and write $$\phi(x+kh)=\phi(h)(k-a)^2+1-\phi(h)a^2$$ where $a=\frac{\phi(h)+C}{2\phi(h)}>\frac{C}{2\phi(h)}$. Since $C>4\sqrt{\phi(h)}$, $\phi(h)a^2>4$. We can now choose an integer $k$ such that $|k-a|<1$ and find that $$\phi(x+kh)<\phi(h)-3$$ which is negative for $\phi(h)$ sufficiently small. This contradiction completes the proof.