next up previous
Next: Distributions and weak limits: Up: ``Strange'' limits Previous: ``Strange'' limits

Delta-sequence

Consider a sequence:

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

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

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

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

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

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

for all smooth functions $f$.



Andre Cherkaev
2001-11-16