Is a parallel capacitor needed in this application?
R1 and C1 in the low voltage section (it would be nice if your component designators were unique accross all the schematic sections you post) are pointless.

R1 puts unnecessary load on the transformer output and has no useful purpose.

C1 will filter some high frequencies that may come in from the high voltage feed and make it accross the transformer, but there is no point to this.  Only the positive peaks of the output of the secondary will conduct thru D1, which causes C2 to be charged up.  A little high freqency noise superimposed on the power waveform would only make the peaks slightly higher, which is harmless.  Furthermore, the inductance of the transformer will filter out much of the high frequency content anyway.

C1 in the high voltage circuit does make sense since it will reduce some of the high frequencies produced by this circuit and thereby fed back onto the power line. The half wave bridge rectifier will draw current in short spurts, which will contain significant harmonics.  The inductance of the transformer together with C1 will reduce the magnitude of the higher harmonics that will be coupled back onto the power line.

For such a low power (tranformer is rated for only 2 VA), a simple parallel capacitor like C1 may be sufficient.  For higher power levels a more deliberate filter may be required.  For even higher power levels, a certain minimum power factor must be met, depending on what jurisdiction the device will be used or sold in.

However, all this is noise compared to another glaring problem with this circuit.  The 9 V RMS output of the transformer will be half-wave rectified to produce about 12 V.  So far so good.  From the 2 VA rating of the transformer, we know that this supply is intended to supply less than 2 W.  2 W at 12 V would be 167 mA, so let's be generous and say it is intended to provide 100 mA only.  Note that C1 will only be charged up once per power line cycles.  Judging from the 230 V input and your location, we can assume this is at 50 Hz.  That means C2 will be charged up every 20 ms.

For simplicity for now, let's see how much the voltage on C2 will drop assuming it gets charged to 12 V instantaneously once every 20 ms.  (100 mA)(20 ms) / 220 &micro;F = 9 V, which means the "12 V" line would drop to 3 V, which is clearly no good for the input of a 7805 regulator.

We can work this backwards and see what current this supply can maintain.  Let's say the 7805 needs at least 7.5 V in to work properly, which mean C2 can drop no more than 4.5 V.  (4.5 V)(220 &micro;F) / 20 ms = 20 mA.  Of course C2 doesn't get charged up instantaneously, so the actual value is a little higher.  But still, this supply as shown is basically no good for output currents beyond 20 mA.

Added:

I just noticed that there is a bridge rectifier immediately on the output of the transformer.  Somehow my brain skipped over that originally since I saw D1 there, which appeared to be a half wave rectifier.

Having both the bridge rectifier and D1 is more silliness.  Because of the small value of C2, I'd keep the bridge and get rid of D1.  This means the output will be charged up twice per power line cycle, but with one more diode drop in series.  Let's say the total voltage drop of the bridge is 1.4 V.  If the transformer can be counted on to produce 9 V sine, then its peaks will be 12.7 V.  That minus the diode drops is 11.3 V.  That leaves 3.8 V headroom for the 7805 to work.  (3.8 V)(220 &micro;F) / 10 ms = 84 mA, which is the maximum this supply can be counted on to produce from the C2 voltage drop point of view.  (84 mA)(9 V) = 760 mW, so that should be within the capabilities of the 2 VA transformer.

If you need more than 80 mA or so from this supply, then you have to change something.  This is assuming already that D1 is replaced by a short.