How to prove a given statistical test has the greatest power It is very conventional in evaluating the null hypothesis to consider the distribution of the mean of a sample of some size if the null hypothesis were true and to compare the mean of your own sample to that distribution.  (At least if the underlying distribution is thought to be normal) But why, even given the normality assumption, do statisticians generally compare the distribution of the mean to the sample mean as opposed to comparing the distribution of some other statistic, say the median or the minimum, to the median (or minimum) of the sample.
I believe the answer to my question is that the mean is a powerful statistic.  If the sample indeed did not come from the null hypothesis distribution, then the mean test is likely to pick it up.  It might take a much greater effect for the median or min test to obtain an adequately low p-value?  But, how do we KNOW that there are not better statistics than the mean?  Is there a proof out there in the literature?  Or how would one possibly go about proving it?
 A: You seem to be interlacing two questions: 1) Does this sample come from a distribution with mean $\mu$? and 2) Does this sample come from a specific distribution (e.g. a normal distribution with mean $\mu$ and standard deviation $\sigma$). If you want to test whether your sample comes from a particular distribution, or test whether two samples come from the same (unknown) distribution, there are more powerful tests than comparison of means, like Kolmogorov-Smirnov test that you can look up online (that can reject the null that a sample comes from a distribution even if the means are equal). Such a test basically compare all quantiles of the two distributions, not just the mean or median or min or max.
A: If it were actually known somehow (so this is not fully realistic, to say the least) that the population is exactly normally distributed, but the only information about the mean $\mu$ and the S.D. $\sigma$ are estimates based on an i.i.d. sample, then the optimality of the sample mean as opposed to the median, or the average of the max and the min, or any other statistic, as a way of estimating location, comes down to the one-to-one nature of the two-sided Laplace transform (analysts, please write an appendix to this answer explaining how to prove that; I'll show how the reduction is done.)
Let $X_1,\ldots,X_n$ be the i.i.d. sample. Let $\bar X=\dfrac{X_1+\cdots+X_n}n$ be the sample mean, and $S^2=\dfrac 1 n \sum_{i=1}^n (X_i-\bar X)^2$ be the MLE of the variance.
Lemma: The conditional distribution of the sample $X_1,\ldots,X_n$ given $\bar X$ and $S^2$ does not depend on $\mu$ and $\sigma^2$.
Hint on how to prove this: Let (lower-case) $x_1,\ldots,x_n$ be the arguments to the density function and let lower-case $s^2$ and $\bar x$ bear the same relation to those as capital $S^2$ and $\bar X$ do to $X_1,\ldots,X_n$.  Then the density is proportional to
$$
\frac{1}{\sigma^n} \exp\left( -\frac 1 2 \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2 \right) = \frac 1 {\sigma^n} \exp\left( \frac{-n}{\sigma^2} (s^2 + (\bar x-\mu)^2) \right).
$$
Use that to prove the lemma.  (If in addition we had had that times a factor that depends on $x_1,\ldots,x_n$ but not on $\mu$ and $\sigma$, we would still have had enough to prove the lemma, and that happens with some families of distributions.)
It follows that if we take any other unbiased estimator $T$ of $\mu$ (i.e. any function of $X_1,\ldots,X_n$ that does not depend on $\mu$ and $\sigma$ and whose expected value remains equal to $\mu$ if $\mu$ and $\sigma$ change) then the expected value of that other estimator $T$ given $\bar X$ and $S^2$ is a function only of $X_1,\ldots,X_n$ that does not depend on $\mu$ and $\sigma$, i.e. it is a statistic.  But since it's an expected value given $\bar X$, it is determined by $\bar X$.  Call it $U(\bar X)$.
The Rao--Blackwell theorem (google that) tells us that therefore $U(\bar X)$ has a mean squared error no bigger than that of $T$.
Conclusion so far: To minimize the mean square error of estimators of $\mu$, we need not consider any that are not functions of $\bar X$.
But now $\mathbb E(U(\bar X)-\bar X)$ is $0$ regardless of the values of $\mu$ and $\sigma$.  It is an "unbiased estimator of $0$".  If we can show that this family of distributions admits no nonzero unbiased estimators of $0$, then we've got $U(\bar X)=\bar X$ and we're done.
An unbiased estimator of $0$ is a function $g(\bar X)$ whose expected value remains $0$ as $\mu$ and $\sigma$ change.  So we have
$$
\int_{\mathbb R} g(x) \exp\left(\frac{-1}{\sigma^2/n} (x-\mu)^2) \right)\,dx =0\text{ for all }\mu\in\mathbb R,\ \sigma>0.
$$
(Since $\bar X\sim N(0,\sigma^2/n)$.)
Do a bit of algebra and find that this says the two-sided Laplace transform
$$
\mu\mapsto\int_{\mathbb R} g(x) e^{-\mu x}\,dx
$$
is $0$ for all values of $\mu$.
Hence $g\equiv 0$, and so there are no nontrivial unbiased estimators of $0$ for this distribution.
