I recently heard that many american members of LG, says that fluorescents on their portable lanterns or emergency lighting feeds by DC current, because of a blackening on only one end and especially because of strong mercury migration after prolonged use of the lamp in the emergency lighting. These things don't occurs with our fluorescent emergency lighting at all (Especially mercury migration).
According to what Ash said me, there no inverter can produce a DC current (The inverter can only invert the DC current of the battery to HF AC current and nothing else), and so fluorescents on emergency lighting in Israel and outside the US operates on HF AC.
I noted that Ash itself think that what that i heard from the american members of LG about DC operated fluorescents in emergency lighting is unreal.
So it is true that the fluorescents in american emergency lighting, really recieves DC and not HF AC as the rest of the world?
One thing that i already thinks that happens in the american fluorescent emergency lighting/portable lanterns (If they indeed feeds their fluorescent lamps with DC current instead of HF AC), is that they are inverterless (Don't have an inverter at all), meaning that the lamp recieve the DC current from the batteries directly, with only some minor resistance to limit their current.

Update: I asked Ash again about this, and he said me that the resaon for only one end blackening and mercury migration of fluorescents in american emergency lighting/portable lanterns, aren't caused because the lamp recives DC current (Because there are no ways to feed fluorescent lamps with pure DC current from batteries, because of their too low supply voltage to the lamp), but because of the following reasons:
1. Half rapidstart: The inverter heats only one electrode, so the lamp operate half hot cathode and half cold cathode. In this case, a blackening will appear in the end, where the non heated electrode located, and the mercury will migrate from the cold electrode to the hot one.
2. Asymmetric AC: I don't remember what Ash said me about this, but this can cause most of the current on the lamp, to flow mostly in one direction, resulting in a mercury migration.

All single transistor inverter ballasts feed the lamps by the current with strong DC component, mainly when the lamp is fed directly from the transformer secondary. And it is really one polarity pulsed DC current, if that ballast does not provide heating supply, what is mostly the case with low power, battery operated ballasts. Feeding the lamp with DC have one huge advantage for low power fluorescent lighting: You have to keep only one electrode hot (the one, what is the cathode), what mean you have more power remaining to actually generate light. With short and not so narrow tubes operated at low current the DC component could be tolerated, as the diffusion is powerful enough to maintain the mercury vapor spread across the whole tube length, it is able to overcome the electrolysis (migrating the mercury towards one end). With high current density the electrolysis would be too strong, so the gas fill would separate, so for higher power and/or narrow tubes more complex, DC component free inverters are necessary.

You heavily mixed up voltages and currents. In nonlinear circuit (what is the circuit with a discharge) each may have way different shape, include the presence (or absence) of the DC component.
It is true, then the transformer can not generate the DC component in it's VOLTAGE, but nothing limit any DC component in the current.

The inverter itself work, indeed, with (HF)AC voltage. But unless the inverter generate really symmetrical waveform (from the simplest circuits only Royer oscillator does so), the lamp nonlinear behavior rectify the output of the inverter, so you end up with strong DC component.
So let's take the single transistor blocking oscillator inverter. When the transistor is ON, the battery voltage is connected to the primary winding, what force the flux in the core rise.
 This changing flux induce the voltage in all other windings, include the feedback (generate base drive for the transstor) and the secondary generate rectangular voltage pulse with the voltage given by the battery voltage and the turn ratio.
This is usually designed to be about the tube arc voltage with hot electrodes, but way insufficient to force any current trough the discharge, when the cathode (corresponding to this polarity) is cold.
Now as the flux rise, so does the magnetizing current, what is proportional to the flux and magnetic reluctance, so linearly rise as well. The primary current would be then a sum of the magnetizing and transformed secondary current, so it will rise too. When this current exceed the collector current capability of the transistor (with given base drive), the transistor will limit it, so prevent further rising. That mean the flux would stop to rise too, what mean all induced voltages drop. That mean the base loose it's drive, so the transistor switch OFF.
As there is no current path trough the transistor anymore, the flux start to drop, so generate voltage of the opposite polarity. And this voltage would rise so high, it find a current path trough any winding, capable to support the flux in the core.
As the flux fall, the voltage have the opposite polarity then before (with the transistor ON) and at the same time the voltage is limited only by the things connected to any of the winding to take over the magnetizing current - what is the lamp, in whatever state (with hot cathode it would mean quite low voltage, so long time, with cold electrodes high voltage, so short time).
Now during the ON time, given amount of energy was stored in the magnetic circuit and this is released during the OFF state into the lamp (faster or slower, depend on the voltage drop). It is only the state with the transistor OFF, when the inverter could adopt the voltage to the load, with the transistor ON all the voltages are related to the battery voltage (via transformers turn ratio)

Now it depend, how is connected the lamp circuit:
- If the lamp is connected directly to the secondary transformer winding, the VOLTAGE would have no DC component (as that is indeed not possible with transformers - each period the flux have to return exactly to the starting value), but the current would flow in pulses of only single direction, so it would be pulsed DC, so cause the electrolytic effects. In the transformer winding is nothing to prevent DC current to flow, transformer winding act as a short circuit for the DC. With this configuration there is no power dissipation in the lamp to heat up the cathode for the transistor ON state, while the current in the OFF state dissipate quite a lot on the coresponding cathode, only one cathode warm up, so the lamp become very asymmetrical load - in one direction (transistor OFF) the voltage drop would be about 9V (hot cathode drop) + 60V (anode column) = ~70V, in the other direction the voltage necessary to cause any current would be about 150V (cold cathode fall) + 60V (anode column) = ~210V, what is way above the inverter generate for this polarity (~70V). So even when the voltage become nearly sumetrical after the cathode warm up, at the same time the lamp become highly asymmetrical, so it would still maintain only the DC lamp current (pulsed, but no opposite polarity), unless the other electrode is heated by some independent means (auxiliary winding,...), so the lamp become more symmetrical and the DC current component is reduced (but never eliminated in this circuit).

- If a capacitor is in series with the lamp, this capacitor will prevent the CURRENT to have any DC component. Now as the lamp is a nonlinear load, the DC voltage develop across the capacitor, so across the lamp may have DC component so, the lamp would pass the same amount of charge in each direction. But the secondary winding of the transformer have still no DC voltage component - it is separated from the lamp by the capacitor... This usually ensure, then both electrodes would finally warm up, so the voltage drop across the tube fall, so the voltage become more symmetrical.