3.7 V to 12 V boost converter - effect on cell

Using a booster can have an effect on the battery.

Say you need to get 1 ampere at 12 volts from your 3.7 volt battery.. Your booster has to put out 12 watts continuosly (1A x 12V = 12 watts.)

To put out those 12 watts continuosly, the converter has to take in 12 watts continuosly. Given 12 watts and 3.7 volts, the converter will draw 3.24 amperes (12 watts/3.7 volts = 3.24 amperes.) Your battery must be capable of supplying over 3 amperes of current to the converter so it can put out 12 volts at 1 ampere.

Batteries have a capacity rating (ampere hours) and a current rating (amperes.)

If you draw more than the rated current from your battery, you may permanently damage it. That damage may show as reduced capacity, or it may out right destroy the battery (effectively no capacity at all.)

If you are using cells with over current protection, the protection might kick in and shut things off before (much) damage occurs.

Summary:

You have to consider the current rating of the battery along with the current the booster will need to deliver the required power to the load.

Basically your capacity will reduce in accordance with the efficiency of the boost converter and the voltage transformation ratio of the boost converter (if you are measuring the current at the 12V output. Assuming the efficiency of the converter is 80%, then 2100mAh will be effectively reduced to: $$ 2100 text{mAh}times 0.8 times frac{3.7}{12} = 518 text{mAh}$$

You won't get as many mAh from the 12V supply as you do from the battery, but you'll get almost as many mWh.

A boost supply can't provide free energy, so when it generates a higher voltage than at the input, it must draw more current from the input than it puts out. In the ideal case, Po=Pi, meaning Vo * Io = Vi * Ii. In real life, the power supply will have an efficiency lower than 100%, so Vo * Io = Vi * Ii * Eff.