I think you have read my comment incorrectly.
I said it does not matter if the load is capacitive or resistive, the end result is heat is generated. This is true.
And I didn't disagree with you on this point; I said "Certainly, heat is generated...".
But you also said "..In fact the diode conducts for only a very small portion of the cycle and during this time the current flow is very high....", right after the previous statement without saying that you only intended this to apply to capacitive loads, which to me suggests that you apparently intended it to apply to both cases, capacitive load and resistive load; if you didn't, you should have said so. It isn't true for a resistive load.
Furthermore, if the load is resistive or capacitive, and the same wattage is being drawn from the supply, the heating of the diode will be slightly more in the case of a capacitive load due to the I squared R losses.
Using equal wattage drawn from the supply for comparison adds a complication because the current supplied to a resistive load is pulsating; a series of half sine waves, not smooth DC. Previously we had been discussing just the average current through the diode.
Also, the heating in the diode is more complicated to calculate than just the I^2*R losses. When the current in a diode is large, the voltage drop vs. current is no longer an exponential relationship. The intrinsic resistance of the silicon chip begins to dominate. For lower currents, the drop across the diode is nearly constant at the well-known .7 volts.
When the current in the diode is large, the resistive part of the diode voltage drop dissipates power as I^2*R, but for smaller currents, the voltage Vd across the diode is more nearly constant, and the loss is proportional to I*Vd. The total loss is a combination of these two, and is difficult to calculate theoretically. It's probably just as well to resort to measurement.
I set up a variable 12+ volt transformer secondary with a diode bridge and a 15 ohm load. Adjusting the transformer output voltage so that 25 watts were delivered to the load, without any capacitor, the measured diode loss (of one diode) was .605 watts, with an RMS diode current of 570 mA. With a 3000 uF cap across the load, and adjusting the transformer output to again deliver 25 watts to the load, the measured diode loss was .77 watts, with an RMS diode current of 640 mA.
The heating of the diode is more that just "slightly" more in the case of a capacitive load.
With a resistive load, the diode current is not the narrow pulse that it is with a capacitive load, and this results is substantially less diode heating.