I am not sure why the question you link to does not provide an answer. I’ll try to answer it but it is really similar to what has already been said there. Bottom line is: if the value $K$ is reachable by the underlying asset $S$, that is $K$ belongs to the domain of process $S$, then the butterfly should be strictly positive.
First note that the butterfly is actually an approximation of the second derivative w.r.t. to strike:
$$\lim_{h \rightarrow 0}\frac{C(t,K+h)-2C(t,K)+C(t,K-h)}{h^2}=\frac{\partial^2C}{\partial K^2}(t,K)$$
where obviously $h^2>0$. Yet by the Breeden-Litzenberger formula, we know that:
$$\frac{\partial^2C}{\partial K^2}(T,K)=e^{-rt}q(t,K)\geq 0$$
where $q$ is the risk-neutral density of the underlying $S$ and $r$ the risk-free rate. You now see that if $K$ is a value which $S$ can reach, that is $K$ belongs to the domain of $S$, then the density of $S$ at $K$ must be strictly positive, that is:
$$C(t,K+h)-2C(t,K)+C(t,K-h)\approx h^2e^{-rt}q(t,K)>0$$
Going further, let us introduce the Dirac delta function $\delta$, which is characterized by the following property for any real-valued function $f$:
$$\int_{-\infty}^{+\infty}\delta(x)f(x)dx=f(0)$$
Hence the density can be expressed as:
$$q(t,K)=\int_{-\infty}^{+\infty}\delta(s-K)q(t,s)ds=E^Q\left(\delta(S_t-K)\right)$$
That is, the risk-neutral density corresponds to the price of a payoff which is non-negative everywhere and strictly positive for one state the world, i.e. if $S_t=K$ $-$ informally the payoff would be infinite if $S_t=K$, see the definition of the Dirac delta. Hence to avoid arbitrage the price of this claim, $e^{-rt}q(t,K)$, must be strictly positive.
