Help understanding AC signal across a charged (coupling) capacitor

Question

(Assuming ideal elements) My friend is trying to convince me that the voltage across \$R_L\$ varies between 4 and 6 volts, says the average value doesn't decrease and stays fixed at \$5V\$ forever.
I'm not getting how this is possible:

^{simulate this circuit – Schematic created using CircuitLab}

The amplitude of the input ac signal varies between -1 and +1 volts, always less than the capacitor voltage of \$5V\$. Wouldn't the capacitor see this low voltage and start discharging?
How can the 1V input AC signal put more charge into the capacitor so that the capacitor's voltage increases, even slightly?
Does not make much sense to me.
Highly appreciate any help. Thanks!

Answer 1

Yes, There is a DC path around the cap its going to discharge. DC response follows RC time-constant.

Answer 2

How about this, from Signal Wave Explorer: 1MHz DC-block (high pass filter), with 10MHz sin input 1volt peakpeak with 2.5 volt offset

Answer 3

You can replace the sinewave with the square wave and now the situation will look like this:

As you can see the capacitor will have a path for a discharge current to flow. And after some time the capacitor reach steady state and in steady state condition, the average capacitor current is \$0A\$ over a cycle (I_charge = -I_discharge).
AC Circuit Having Only Capacitor

EDIT

In real amp life the situation will look like this:

At DC the Cin capacitor is charged to \$2.6V\$

For positive half-cycle (+1V peak):

As you can see \$5V\$ source provide \$3.5μA\$ of current, but the base current needed to be equal to \$9μA\$, so this additional current will have to come from the AC signal source.

And the Cin capacitor is discharging (I_discharge = -5.5μA)

As I show in the diagram.

For Negative half-cycle:

This time the base need only \$3μA\$, but our \$5V\$ bias source provides \$8.5μA\$.

This implies that our AC signal source needs to "sink" this excess current \$5.5μA\$.

And Cin this time is charging (I_charge = +5.5μA)

As you can see the average capacitor current is 0A over a cycle, hence the average (DC) voltage across capacitor stay unchanged (2.6V).