This is false statement. I'd refer to microchip application note AN905.

I disagree. Both the current rise against induction, and peak current constrained by resistance, are proportional to current. Dissipated power is proportional to the integral of voltage and current over time. If you use PWM the current will be less than if you use a constant voltage, but with or without PWM if you double the voltage you will get roughly double the current and roughly four times the dissipated power.

It depends on the PWM frequency and the inductance of the motor. In all cases, the power dissipation in the motor will be the integral of I^2 * R over time, where I is the motor current and R is the motor resistance. You can't calculate the power from the motor voltage so easily because the motor has inductance and back emf as well as resistance

With low frequency/low inductance, the current through the inductor will settle near V/R over most of the active part of the PWM cycle, where V is the applied voltage (12V less the voltage drop of the switching device) and R is the resistance of the motor. The flyback diode(s) will conduct for a very small part of the cycle. Torque will be proportional to the average current, and at 50% PWM power dissipation will be roughly twice as much as if the motor were powered continuously at half the voltage (i.e. 4x the power dissipation for half of the time).

With high frequency/high inductance, the motor inductance together with the flyback diode(s) will smooth out the motor current so that it is roughly constant throughout the cycle. In this case, there will be no increase in power dissipation in the motor compared to using a lower voltage and no PWM.

That's why the application note referred to "PWM of sufficient frequency".