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This will depend on the definition of "return on the long run". If we define the annualized return on the long run by $\frac{1}{T}\ln \frac{S_T}{S_0}$ for a certain time $T$ in the future, then \begin{align*} E\left( \frac{1}{T}\ln \frac{S_T}{S_0} \right) = \mu-\frac{1}{2}\sigma^2, \end{align*} as claimed. Note that $\mu$ is the instant, or instantaneous, ...
Trying to shed some light here: What we also see using this here, is that if returns are log-normally distributed, ie. $$1 + r = \exp(\mu + \sigma Z),$$ with $Z$ standard-normal, then $$E[1+r] = \exp(\mu + \frac 12 \sigma^2)$$ holds. But the geometric mean $GM$ is given by $\exp(\mu)$ and we have $$\log(GM) = \mu = \log(E[1+r]) - \sigma^2 /2$$ and ...