Electronic – input capacitor for buck converter

The ESR value of your capacitor and inductor affect the ripple output voltage. The output ripple voltage can be estimated based on the inductor ripple current (delta IL) and the ESR of your capacitor.
The inductor shall be used to reduce the ripple current (delta IL).
delta IL = (1/F*L)Vout(1-Vout/Vin). Where F is the switching frequency. In the same time the capacitor is selected with the following calculation: C=Ip/(F*Vripple); where Ip is the peak output current and F your switching frequency. The Ripple voltage is based on requirements and performance expectation.

I'm not sure I'm following your line of thought, but it sounds like you are attacking the problem backwards from the usual way.

Normally, you'll first set a limit on ripple current (\$\Delta{}I_L\$), and choose an inductor to meet that requirement.

Then you will choose your output capacitor to achieve the ripple voltage that you want.

For idealized components, you will have

$$\Delta{}V_{out} = \Delta{}I_L \sqrt{R_{ESR}^2 + \left(\frac{1}{8 f_{sw} C_{out}}\right)^2}$$

So the lower the ESR and the higher the capacitance, the lower the ripple voltage will be.

This ripple voltage is ripple due the inherent transients in each cycle of the switching circuit, when operating with a constant load current.

Ripple or ring due to load transients is more due to the voltage control loop than due to these considerations. So to minimize that you need to model your control loop and ensure you have sufficient phase margin to avoid an unwanted ring response to load transients.

Is this amount of current (ripple current, Irms) identical with Inductor ripple current of regulator?

Yes. If the load current is constant (or near enough) then the deviations of inductor current from the average value (which is the load current) have nowhere to go but in and out of the output capacitor.