On Mon, 13 Dec 2004 13:28:32 +0100, Han de Bruijn

Well of course you can't prove that because it's not true -
by now I know that you mean the limit of that as sigma -> 0,
but the fact that you mean that doesn't make the notation
correct. How to prove that

  int(-oo,+oo) f(t).g_sigma(t-T) dt -> f(T)

as sigma -> 0 is a standard thing in harmonic analysis -
look for "approximate identity" arguments.

Let's see. Of course we need some technical hypotheses
of f; let's assume that f is bounded and continuous.
Using the fact that the gaussian has integral 1 we need
to show that

  int(-oo,+oo) (f(t) - f(T)).g_sigma(t-T) dt -> 0.

Let's assume T = 0 just to save typing - we need to show
that

  int(-oo,+oo) (f(t) - f(0)).g_sigma(t) dt -> 0

as sigma -> 0. So it's more than enough to show that

  int(-oo,+oo) |f(t) - f(0)|.g_sigma(t) dt -> 0

Let epsilon > 0. Choose delta > 0 so that |f(t) - f(0)|
< epsilon whenever |t| < 0. Write the last integral
as the sum of two integrals, one where |t| < delta and
one where |t| > delta.

The integral over |t| < delta is less than epsilon
for all sigma, by our choice of delta and the fact
that the gaussian has integral 1. On the other
hand the fact that f is bounded and property (iii)
above shows that the integral over |t| > delta is
< epsilon if sigma is small enough. QED. (more details
if you want.)

************************

David C. Ullrich