Theorem: Tuning Fork Delta Sequence

thumb300pxThe graph of the tuning fork delta sequence. As $n$ grows, the rectangles becomes thinner and longer. The area of each shape is equal to $1$, where the area under the axis contributes negatively. Note that at $x = 0$ the graph is always negative, and its value here approaches $-\infty$. Only taking the entire graph we can ensure the proper behaviour for $n$ approaching $\infty$,
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:<nowiki>$\map {\delta_n} x := \begin {cases}
-n & : \size x < \dfrac 1 {2n} \\ \\
2n & : \dfrac 1 {2n} \le \size x \le \dfrac 1 n \\ \\
0 & : \size x > \dfrac 1 n \end {cases}$</nowiki>
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where:
:$\phi \in \map \DD \R$ is a test function
:$\delta$ is the Dirac delta distribution
:$\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map {\delta_n} x \rd x
      | r = 2n \int_{- \frac 1 n }^{- \frac 1 {2n} } \rd x - n \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \rd x + 2n \int_{\frac 1 {2n} }^{\frac 1 n} \rd x
}}
{{eqn | r =  n \paren {2 \paren {- \frac 1 {2n} - \paren {- \frac 1 n} } - \paren {\frac 1 {2n} - \paren {- \frac 1 {2n} } } + 2 \paren {\frac 1 n - \frac 1 {2n} } }
}}
{{eqn | r =  n \paren {\frac 2 {2n} - \frac 2 {2n} + \frac 2 {2n} }
}}
{{eqn | r =  1
}}
{{end-eqn}}
Let $\map g x = \map \phi x - \map \phi 0$
Then:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x
      | r = \map \phi 0 \int_{-\infty}^\infty \map {\delta_n} x \rd x + \int_{-\infty}^\infty \map {\delta_n} x \map g x \rd x
}}
{{eqn | r = \map \phi 0 + \int_{-\infty}^\infty \map {\delta_n} x \map g x \rd x
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x
      | r = 2n \int_{- \frac 1 n}^{- \frac 1 {2n} } \map g x \rd x - n \int_{- \frac 1 {2 n} }^{\frac 1 {2n} } \map g x \rd x + 2n \int_{\frac 1 {2 n} }^{\frac 1 n} \map g x \rd x
}}
{{end-eqn}}
We have that $\map g 0 = 0$.
By definition, $\phi$ is a smooth real function on $\R$.
By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$.
Then:
:$\ds \forall \epsilon' \in \R_{> 0} : \exists \delta \in \R_{> 0} : 0 < \size x < \delta \implies \size {\map g x} < \epsilon'$
Let $\ds \delta = \frac 1 N$ where $N \in \R_{> 0}$.
Let $\ds \epsilon' = \frac \epsilon 3$ where $\epsilon \in \R_{> 0}$.
Then
:$\ds \forall \epsilon > 0 : \exists N \in \R_{> 0} : \size x < \frac 1 N \implies \size {\map g x} < \frac \epsilon 3$
Suppose:
:$\ds n \in \N : \size x \le \frac 1 n < \frac 1 N$
That is, suppose:
:$\ds n \in \N : \frac 1 {\size {x} } \ge n > N$
Then:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x
      | o = \le
      | r = 2n \int_{- \frac 1 n}^{- \frac 1 {2 n} } \size {\map g x} \rd x + n \int_{- \frac 1 {2n} }^{\frac 1 {2 n} } \size {\map g x} \rd x + 2n \int_{\frac 1 {2 n} }^{\frac 1 n} \size {\map g x} \rd x
}}
{{eqn | o = \le
      | r = \frac \epsilon 3 \paren {2 n \paren {- \frac 1 {2 n} + \frac 1 n} + n \paren {\frac 1 {2 n} + \frac 1 {2 n} } + 2 n \paren {\frac 1 n - \frac 1 {2 n} } }
      | c = $\ds \frac 1 {2n} < \frac 1 n < \frac 1 N$
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
Since $n$ was arbitrary:
:$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$
By definition of the limit of a real sequence:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$
However:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x
      | r = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0 \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \rd x
      | c = Sum Rule for Limits of Real Functions
}}
{{eqn | r = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0
}}
{{eqn | r = 0
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x
      | r = \map \phi 0
}}
{{end-eqn}}
{{qed}}