Is the mean (Bayesian) posterior estimate of $\theta$ a (Frequentist) unbiased estimator of $\theta$? I am wondering about the different ways that Bayesian and Frequentist statistic connect with each other.
I recalled that the Maximum Likelihood estimate of a parameter $\theta$ is not necessarily an unbiased estimator of that parameter.
That made me wonder: Is the Mean Posterior estimate of $\theta$ an unbiased estimator?
That is,

Does $\phi(x)=E(\theta\mid x)$, imply $E(\phi(x)\mid\theta)=\theta$?

Note that this is indeed a meaningful question, since $\phi(x)$, while it is a Bayesian estimator, is simply a function from the data to the real line and so can also be seen as a classical frequentist estimator.
If this question cannot be answered in general, please assume the prior is uniform.
If not, is there some other Bayesian estimator (i.e. a function from the posterior to $\mathbb R$) that is always an unbiased estimator in the frequentist sense?
 A: This is a meaningful question which answer is well-known: when using a proper prior $\pi$ on $\theta$, the posterior mean $\delta^\pi(x) = \mathbb{E}^\pi[\theta|x]$ cannot be unbiased. As otherwise the integrated Bayes risk would be zero:
\begin{align*}
r(\pi; \delta^\pi) &= \overbrace{\mathbb{E}^\pi\{\underbrace{\mathbb{E}^X[(\delta^\pi(X)-\theta)^2|\theta]}_{\text{exp. under likelihood}}\}}^{\text{expectation under prior}}\\
&= \mathbb{E}^\pi\{\mathbb{E}^X[\delta^\pi(X)^2+\theta^2-2\delta^\pi(X)\theta|\theta]\}\\
&= \mathbb{E}^\pi\{\mathbb{E}^X[\delta^\pi(X)^2+\theta^2]|\theta\}-
\mathbb{E}^\pi\{\theta \mathbb{E}^X[\delta^\pi(X)|\theta]\}-\overbrace{\mathbb{E}^X\{\mathbb{E}^\pi[\theta|X]\delta^\pi(X)\}}^{\text{exp. under marginal}}\\
&= \mathbb{E}^\pi[\theta^2]+\underbrace{\mathbb{E}^X[\delta^\pi(X)^2]}_{\text{exp. under marginal}}
-\mathbb{E}^\pi[\theta^2]-\mathbb{E}^X[\delta^\pi(X)^2]\\
& = 0
\end{align*}
[Notations:  $\mathbb{E}^X$ means that $X$ is the random variable to be integrated in this expectation, either under likelihood (conditional on $\theta$) or marginal (integrating out $\theta$) while $^π$ considers $θ$ to be the random variable to be integrated. Note that $\mathbb{E}^X[\delta^\pi(X)]$ is an integral wrt to the marginal, while  $\mathbb{E}^X[\delta^\pi(X)|\theta]$ is an integral wrt to the sampling distribution.]
The argument does not extend to improper priors like the flat prior (which is not uniform!) since the integrated Bayes risk is infinite. Hence, some generalised Bayes estimators may turn out to be unbiased, as for instance the MLE in the Normal mean problem which is also a Bayes posterior expectation under the flat prior. (But there is no general property of unbiasedness for improper priors!)

A side property of interest is that $\delta^\pi(x) =
  \mathbb{E}^\pi[\theta|x]$ is sufficient in a Bayesian sense, as
$$\mathbb{E}^\pi\{\theta|\mathbb{E}^\pi[\theta|x]\}=\mathbb{E}^\pi[\theta|x]$$Conditioning
upon $\mathbb{E}^\pi[\theta|x]$ is the same as conditioning on $x$ for
estimating $\theta$.

