Why losslessness requires voltage and current must be both zero?

Question

In the textbook (pg. 260 Linear Circuit Analysis, by Artice Davis), losslessness is defined in such way:

Losslessness: If w(\$\infty\$) = 0 for any v(t) and i(t) waveforms supported by the element and having the property that v(\$ \infty\$) and i(\$ \infty\$) are both zero, then we say that the element is lossless ... we must require that both the voltage and current waveforms approach zero at t = \$ \infty\$ in order to test an element to see if it is lossless.

An element is lossless when there is no energy loss or energy exchanges. And for instance, if I start with the equation for the energy of a capacitor \$ w(t) = {1 \over 2} C v(t)^2 \$. So long if there's a voltage across the capacitor, energy cannot be zero (not lossless). BUT If I start the calculation with power, where \$ P = v(t) \cdot i(t) \$, then as long as either (and not necessarily both) current or voltage is zero, power has to be zero (so will its integration, or energy). Hence my questions are:

1. Mathematically, if I calculate energy by integrating power through all time, I get zero power and zero energy (lossless) if either voltage or current is zero. So why does the textbook claim that it is required for both current and voltage to be zero for an element to be considered lossless?

2. Conceptually, as long as a capacitor "holds" its charge eternally, then the capacitor has to be "lossless" because no charge is leaving the capacitor, and nothing disappears in form of energy away from the capacitor. So capacitor, ideally, can hold a voltage and still be lossless.

Answer 1

Indeed, if you charge an ideal capacitor and leave it charged forever, it hasn't lost any energy.

But according to this definition of losslessness, that's not a valid test. You must discharge the capacitor and the amount of energy you get back by discharging it must be equal to the amount of energy you spend to charge it. That's a valid test (according to this definition).

Why not just calculate the stored charge? Well, you could imagine a component like a capacitor where the capacitance somehow reduces after it's charged. The voltage stays the same, but the stored energy doesn't. The only way to test that is to take the energy back out.

Answer 2

I think the point is that we cannot measure the energy inside a component but we can only empty it and sum up the energy flow at its terminals.

From an electric point of view it's like a black box with an interface.

So, at the end of time we shall have:

Net energy zero, w=0

No energy inside,i.e. state variables at zero, so any v=0 and I=0