The buck OUTPUT current is NEVER discontinuous in a synchronous converter.If the inductor current is exhausted, the switching node will no longer be negative and will track the output voltage (via the inductor. However, if the lower switch is still on, it will start discharging the output capacitor through the inductor.
The current in off state can only be sourced by stored magnetic energy in the inductor. Once exhausted, the load gets fed by the output capacitor. To summarize- the Buck input current is discontinuous. During on time if feed the load via the inductor (while storing there energy), during off time it stops (the load is fed by the inductor and output cap. The Buck output current is continuous (either from the input or from the inductor and cap). The Boost topology has it other way around - continuous input current, choppy output current. This makes the Buck an input power quality liability - the choppy input current creates choppy voltage drops in the input loop-affecting the input power source.
Let's assume that the input is 12V and the output is 6V.
The lower FET is on. The voltage across the inductor is -6V. The current in the inductor therefore reduces by dI/dt=V/L.
But what happens when the inductor current reaches zero?
The lower FET is STILL on. The voltage across the inductor is still -6V. The current in the inductor therefore continues to become more negative. Current from the output capacitor flows though the inductor to the negative supply, and will continue to do so until the lower FET switches off.