Obviously, the filter WOULD shift the (non-periodic) signal along the time axis

No it doesn't because this would be a contradiction to #1,#2,#3 of the proof.

How about doing the exercise I mentioned? The result *is* admittedly counter-intuitive. Go ahead and try.

Edit: You know what? I'm saving you the trouble: Install GNU Octave and try this:

Save this to the file "fddelay.m":

function [y,x] = fddelay(radius)

x = -floor(radius):radius;

f2 = 1;

y = x.*0;

while (f2>0.01) %% cutoff frequency

f1 = f2 * 0.95; %% narrow bands

fcenter = (f1+f2)/2;

delayinsamples = 2/fcenter;

y = y + bandpass(x-delayinsamples,f1,f2);

f2 = f1;

end;

return;

function y = bandpass(x,f1,f2)

y = lowpass(x,f2)-lowpass(x,f1);

return;

function y = lowpass(x,freq)

y = sinc(x.*freq).*freq;

return

In the Octave console do this:

[y,x] = fddelay(999);

plot(x,y);

It'll show you the impulse response of the filter's approximation.

This thing converges to the unit impulse when you further reduce the band pass filters' bandwidths and lower the cut off frequency...

I'm done here.

Cheers,

SG