Is it possible to have an estimator that is unbiased and bounded? I have a parameter $\theta$ which lies between $[0,1]$. Let us say that I can run an experiment and obtain
$\hat{\theta} = \theta + w$, where $w$ is a standard Gaussian. What I need is an 
estimate of $\theta$ which is 1) unbiased 2) almost surely bounded. Requirement (2) is crucial for me. 
The natural think to do is to construct a new estimator setting $\hat{\theta}$ to $1$ if it is above $1$ and 
to $0$ if it is below $0$. But then the estimator will not be unbiased. So what should I do?
Formally, the question is whether there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\hat \theta)$ satisfies (1) and (2) above. Moreover, would the situation be different if I drew more than a single sample?
 A: I will present conditions under which an unbiased estimator remains unbiased, even after it is bounded. But I am not sure that they amount to something interesting or useful.  
Let an estimator $\hat \theta$ of the unknown parameter $\theta$ of a continuous distribution, and $E(\hat \theta) =\theta$.
Assume that for some reasons, under repeated sampling we want the estimator to produce estimates that range in $[\delta_l,\delta_u]$. We assume that $\theta \in [\delta_l,\delta_u]$ and so we can write when convenient the interval as $[\theta-a,\theta+b]$ with $\{a,b\}$ positive numbers but of course unknown.  
Then the constrained estimator is 
$$\hat \theta_c = \left\{\begin{matrix}  \delta_l & \hat \theta <\delta_l\\\hat \theta &\delta_l \leq \hat \theta \leq \delta_u \\\delta_u & \delta_u < \hat \theta \end{matrix} \right\}$$
and its expected value is
$$\begin{align} E(\hat \theta_c) &= \delta_l\cdot P[\hat \theta \leq \delta_l] \\&+ E(\hat \theta \mid \delta_l \leq\hat \theta \leq\delta_u )\cdot P[\delta_l \leq\hat \theta \leq \delta_u] \\
&+\delta_u\cdot P[\hat \theta > \delta_u]\end{align}$$
Define now the indicator functions
$$I_l = I(\hat \theta \leq \delta_l),\;\; I_m = I(\delta_l\leq \hat \theta \leq \delta_l),\;\; I_u = I(\hat \theta > \delta_u)$$
and note that 
$$I_l + I_u = 1- I_m   \tag{1}$$
using these indicator functions, and integrals, we can write the expected value of the constrained estimator as ($f(\hat \theta)$ is the density function of $\hat \theta$),
$$E(\hat \theta_c) = \int_{-\infty}^{\infty}\delta_lf(\hat \theta)I_ld\hat \theta + \int_{-\infty}^{\infty}\hat \theta f(\hat \theta)I_md\hat \theta + \int_{-\infty}^{\infty} \delta_uf(\hat \theta)I_ud\hat \theta$$
$$=\int_{-\infty}^{\infty}f(\hat \theta)\Big[\delta_lI_l + \hat \theta I_m + \delta_uI_u\Big]d\hat \theta$$
$$=E\Big[\delta_lI_l + \hat \theta I_m + \delta_uI_u\Big] \tag{2}$$
Decomposing the upper and lower bound, we have
$$E(\hat \theta_c)  = E\Big[(\theta-a)I_l + \hat \theta I_m + (\theta+b)I_u\Big]$$
$$=E\Big[\theta\cdot(I_l+I_u) + \hat \theta I_m\Big] -aE(I_l)+bE(I_u) $$
and using $(1)$,
$$ = E\Big[\theta\cdot(1-I_m) + \hat \theta I_m\Big] -aE(I_l)+bE(I_u) $$
$$\Rightarrow E(\hat \theta_c) = \theta +E\big[(\hat \theta -\theta)I_m\big]-aE(I_l)+bE(I_u)  \tag {3}$$
Now, since $E(\hat \theta) = \theta$ we have
$$E\big[(\hat \theta -\theta)I_m\big] = E\big(\hat \theta I_m\big) - E(\hat \theta)E(I_m)$$ 
But
$$E\big(\hat \theta I_m\big) = E\big(\hat \theta I_m\mid I_m=1\big)E(I_m) = E\big(\hat \theta \big)E(I_m)$$
Hence, $E\big[(\hat \theta -\theta)I_m\big] =0$ and so
$$\begin{align} E(\hat \theta_c) &= \theta -aE(I_l)+bE(I_u) \\
&= \theta -aP(\hat \theta \leq \delta_l)+bP(\hat \theta > \delta_u)\end{align}\tag {4}$$
or alternatively
$$ E(\hat \theta_c) = \theta -(\theta-\delta_l)P(\hat \theta \leq \delta_l)+(\delta_u-\theta)P(\hat \theta > \delta_u)\tag {4a}$$
Therefore from $(4)$, we see that for the constrained estimator to also be unbiased, we must have
$$aP(\hat \theta \leq \delta_l) = bP(\hat \theta > \delta_u) \tag {5}$$
What is the problem with condition $(5)$? It involves the unknown numbers $\{a,b\}$, so in practice we will not be able to actually determine an interval to bound the estimator and keep it unbiased. 
But let's say this is some controlled simulation experiment, where we want to investigate other properties of estimators, given unbiasedness. Then we can "neutralize" $a$ and $b$ by setting $a=b$, which essentially creates a symmetric interval around the value of $\theta$... In this case, to achieve unbiasedness, we must more over have $P(\hat \theta \leq \delta_l) = P(\hat \theta > \delta_u)$, i.e. we must have that the probability mass of the unconstrained estimator is equal to the left and to the right of the (symmetric around $\theta$) interval...  
...and so we learn that (as sufficient conditions), if the distribution of the unconstrained estimator is symmetric around the true value, then the estimator constrained in an interval symmetric around the true value will also be unbiased... but this is almost trivially evident or intuitive, isn't it?  
It becomes a little more interesting, if we realize that the necessary and sufficient  condition (given a symmetric interval) a) does not require a symmetric distribution, only equal probability mass "in the tails" (and this in turn does not imply that the distribution of the mass in each tail has to be identical) and b)  permits that inside the interval, the estimator's density can have any non-symmetric shape that is consistent with maintaining unbiasedness -it will still make the constrained estimator unbiased. 
APPLICATION: The OP's case
Our estimator is $\hat \theta = \theta + w,\;\; w \sim N(0,1)$ and so $\hat \theta \sim N(\theta,1)$.
Then, using $(4)$ while writing $a,b$ in terms of $\theta, \delta$, we have, for bounding interval $[0,1]$,
$$E[\hat \theta_c] = \theta -\theta P(\hat \theta \leq 0) +(1-\theta)P(\hat \theta > 1)$$
The distribution is symmetric around $\theta$. Transforming ($\Phi()$ is the standard normal CDF)
$$E[\hat \theta_c] = \theta -\theta P(\hat \theta-\theta \leq -\theta) +(1-\theta)P(\hat \theta -\theta > 1-\theta)$$
$$=\theta -\theta \Phi(-\theta) +(1-\theta)[1-\Phi(1-\theta)]$$
One can verify that the additional terms cancel off only if $\theta =1/2$, namely, only if the bounding interval is also symmetric around $\theta$.
