I have suggested, in comments, that an "outlier" in this situation might be defined as a member of a "small" cluster centered at an "extreme" value. The meanings of the quoted terms need to be quantified, but apparently they can be: "small" would be a cluster of less than 10 values and "extreme" can be determined as outlying relative to the set of component means in the mixture model. In this case, outliers can be found with simple post-processing of any reasonable cluster analysis of the data.
Choices have to be made in fine-tuning this approach. These choices will depend on the nature of the data and therefore cannot be completely specified in a general answer like this. Instead, let's analyze some data. I use R due to its popularity on this site and succinctness (even compared to Python).
First, create some data as described in the question:
set.seed(17) # For reproducible results
centers <- rnorm(100, mean=100, sd=20)
x <- c(centers + rnorm(100*100, mean=0, sd=1),
rnorm(100, mean=250, sd=1),
rnorm(9, mean=300, sd=1))
This command specifies 102 components: 100 of them are situated like 100 independent draws from a normal(100, 20) distribution (and will therefore tend to lie between 50 and 150); one of them is centered at 250, and one is centered at 300. It then draws 100 values independently from each component (using a common standard deviation of 1) but, in the last component centered at 300, it draws only 9 values. According to the characterization of outliers, the 100 values centered at 250 do not constitute outliers: they should be viewed as a component of the mixture, albeit situated far from the others. However, one cluster of nine high values consists entirely of outliers. We need to detect these but no others.
Most omnibus univariate outlier-detection procedures would either not detect any of these 109 highest values or would indicate all 109 are outliers.
Suppose we have a good sense of the standard deviations of the components (obtained from prior information or from exploring the data). Use this to construct a kernel density estimate of the mixture:
d <- density(x, bw=1, n=1000)
plot(d, main="Kernel density")
The (almost invisible) blip at the extreme right qualifies as a set of outliers: its small area (less than 10/10109 = 0.001 of the total) indicates it consists of just a few values and its situation at one extreme of the x-axis earns it the appellation of "outlier" rather than "inlier." Checking these things is straightforward:
x0 <- d$x[d$y > 1000/length(x) * dnorm(5)]
gaps <- tail(x0, -1) - head(x0, -1)
histogram(gaps, main="Gap Counts")
The density estimate d is represented by a 1D grid of 1000 bins. These commands have retained all bins in which the density is sufficiently large. For "large" I chose a very small value, to make sure that even the density of a single isolated value is picked up, but not so small that obviously separated components are merged.
Evidently the gap distribution has two high outliers (which can automatically be detected using any simple procedure, even an ad hoc one). One characterization is that they both exceed 25 (in this example). Let's find the values associated with them:
large.gaps <- gaps > 25
ranges <- rbind(tail(x0,-1)[large.gaps], c(tail(head(x0,-1)[large.gaps], -1), max(x))
The output is
[,1] [,2]
[1,] 243.9937 295.7732
[2,] 256.3758 300.9340
Within the range of data (from 25 to 301) these gaps determine two potential outlying ranges, one from 244 to 256 (column 1) and another from 296 to 301 (column 2). Let's see how many values lie within these ranges:
lapply(apply(ranges, 2, function(r){x[r[1] <= x & x <= r[2]]}), length)
The result is
[[1]]
[1] 100
[[2]]
[1] 9
The 100 is too large to be unusual: that's one of the components of the mixture. But the 9 is small enough. It remains to see whether any of these components might be considered outlying (as opposed to inlying):
apply(ranges, 2, mean)
The result is
[1] 250.1848 298.3536
The center of the 100-point cluster is at 250 and the center of the 9-point cluster is at 298, far enough from the rest of the data to constitute a cluster of outliers. We conclude there are nine outliers. Specifically, these are the values determined by column 2 of ranges,
x[ranges[1,2] <= x & x <= ranges[2,2]]
In order, they are
299.0379 300.0376 300.2696 300.3892 300.4250 300.5659 300.7018 300.8436 300.9340
The objective of EM is to maximize the observed data log-likelihood,
$$l(\theta) = \sum_i \ln \left[ \sum_{z} p(x_i, z| \theta) \right] $$
Unfortunately, this tends to be difficult to optimize with respect to $\theta$. Instead, EM repeatedly forms and maximizes the auxiliary function
$$Q(\theta , \theta^t) = \mathbb{E}_{z|\theta^t} \left (\sum_i \ln p(x_i, z_i| \theta) \right)$$
If $\theta^{t+1}$ maximizes $Q(\theta, \theta^t)$, EM guarantees that
$$l(\theta^{t+1}) \geq Q(\theta^{t+1}, \theta^t) \geq Q(\theta^t, \theta^t) = l(\theta^t)$$
If you'd like to know exactly why this is the case, Section 11.4.7 of Murphy's Machine Learning: A Probabilistic Perspective gives a good explanation. If your implementation doesn't satisfy these inequalities, you've made a mistake somewhere. Saying things like
I have a close to perfect fit, indicating there are no programming errors
is dangerous. With a lot of optimization and learning algorithms, it's very easy to make mistakes yet still get correct-looking answers most of the time. An intuition I'm fond of is that these algorithms are intended to deal with messy data, so it's not surprising that they also deal well with bugs!
On to the other half of your question,
is there a conventional search heuristic or likewise to increase the likelihood of finding the global minimum (or maximum)
Random restarts is the easiest approach; next easiest is probably simulated annealing over the initial parameters. I've also heard of a variant of EM called deterministic annealing, but I haven't used it personally so can't tell you much about it.
Best Answer
To complement @matt's answer, you can also consider the beta distribution with $\alpha = .5$ and $\beta = .5$. It is illustrated by the red line in the figure below (copied from Wikipedia). As you can see, it is multimodal (viz., bimodal), but it isn't a mixture distribution: