Shorting the inductor temporarily converts some of the energy in the stored electric field of the input capacitor to "stored" magnetic field in the inductor (not quite, since it's only present when current is flowing, but close enough). Roughly the same amount of energy is eventually converted back into stored electric field in the output capacitor, minus the losses from parasitic capacitances/resistances and radiated emissions.
To keep things simple, imagine a lossless boost converter. In terms of power-in and power-out, a lossless converter has the same input and output power. If the converter output is doing real work (resistive load), the output power is equal to the output voltage times the load current. Therefore, at a lower input voltage, the converter sees a much higher input current than the load current - it has to, because power-in equals power-out. If it intuitively feels like you're pulling more energy out of the input capacitor than you're pushing into the output capacitor because of the temporary shorting of the inductor, remember - it's not lost to real work, just cleverly exploited by the transformation to magnetic field to losslessly overcome the difference in potential energy between the input charge at low voltage and the output charge at high voltage.
Energy in cap: E = 1/2 * C * V^2
C = Q / V
E = Q * V / 2
Energy in inductor: E = 1/2 * L * I^2
I = dQ/dt
I don't know how to write out the integral notation on HN, so you can fill in the blanks (sorry) - integrate the inductor current ramp up and ramp down portions, set equal to input charge pulled and output charge pushed, observe that energy is conserved with less charge at higher voltage on the output.
To keep things simple, imagine a lossless boost converter. In terms of power-in and power-out, a lossless converter has the same input and output power. If the converter output is doing real work (resistive load), the output power is equal to the output voltage times the load current. Therefore, at a lower input voltage, the converter sees a much higher input current than the load current - it has to, because power-in equals power-out. If it intuitively feels like you're pulling more energy out of the input capacitor than you're pushing into the output capacitor because of the temporary shorting of the inductor, remember - it's not lost to real work, just cleverly exploited by the transformation to magnetic field to losslessly overcome the difference in potential energy between the input charge at low voltage and the output charge at high voltage.
I don't know how to write out the integral notation on HN, so you can fill in the blanks (sorry) - integrate the inductor current ramp up and ramp down portions, set equal to input charge pulled and output charge pushed, observe that energy is conserved with less charge at higher voltage on the output.