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Consider a sequence:

g_n(x)= { \sqrt{ n \over 2\pi}} e^{ -x^2 n^2/2}

When $n\rightarrow \infty$, the sequence converges to the so-called $\delta$-function (which, by the way, is not a function but a new object: The distribution). $\delta$-function equals zero if $x\neq 0$, is infinitely large if $x=0 $, and, additionally, keeps the area under its graph equal to one. This last extra requirement differs $\delta$-function from ``normal'' functions. It comes from the constancy of the integrals

\int_{- \infty}^{\infty} g_n(x) \, dx =1 \quad \forall n

\par\Huge {Animated figure
to be inserted here}

Problem: Prove the basic propety of the $\delta$-function

\int_{- \infty}^{\infty} g_n(x) \,f(x)\, dx \rightarrow f(0)
\quad \mbox{when } \rightarrow \infty.

for all smooth functions $f$.

Andre Cherkaev