Is a non-negative function which satisfies the parallelogram identity continuous? 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 $k \in \mathbb{N}$.
 A: 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.
