A question about "Coin Cells And Peak Current Draw" App Note
The load draws a high current spike during 1ms, which makes the voltage drop due to the high internal resistance of the battery. The chip shuts down at some undervoltage threshold, so it will shut down while there is still some juice left in the battery.
If a capacitor is used, and it has much lower internal resistance than the battery, then it will provide the energy for the current spike, and slowly recharge from the battery afterwards. So the voltage will not drop as much, and the chip will not shut down while the battery is still usable.
If this is a low average current application, the capacitor needs to be low leakage. Some electrolytics leakage in the hundreds of nA, but some are much worse, so make sure to select the right one.
If there is no size constraint on the device, two AA batteries instead of a coin cell also solves the problem.