N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence/Proof 2

Proof
Let $\map g x = \map \phi x - \map \phi 0$.

Then:


 * $\ds \int_{- \infty}^\infty \map \phi x \map {\delta_n} x \rd x = \map \phi 0 + \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$

Let $A \in \R_{> 0}$.

Then:

Let:


 * $\ds \max_{x \mathop \in \closedint {-A} A} \size {\map g x} := \map M A$

Then:

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 $x = 0$.

Furthermore:

By definition of the limit of a real function:


 * $\forall \epsilon' \in \R_{>0} : \exists \delta \in \R_{>0}: \forall A \in \R_{> 0}: 0 < A < \delta \implies \map M A < \epsilon'$

Let $\ds \epsilon' = \frac \epsilon 2$.

It follows that:


 * $\ds \forall \epsilon \in \R_{> 0} : \exists A \in \R_{> 0} : I_3 \le \map M A < \frac \epsilon 2$

Suppose $A$ is such that the above inequality holds.

By definition, $\map \phi x$ is bounded.

Then:

It follows that:


 * $\exists b \in \R_{> 0} : \forall x \in \R : \size {\map g x} < b$

Then:

With the number $A$ fixed:


 * $\ds \lim_{n \mathop \to \infty} \frac 2 \pi \map \arctan {n A} = 1$.

By Squeeze Theorem and for a fixed $A$ we have:


 * $\ds \lim_{n \mathop \to \infty} \size {I_1 + I_2} = 0$

By definition of the limit of a real sequence:


 * $\ds \forall \overline \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} \le b \size {1 - \frac 2 \pi \map \arctan {n A} } < \overline \epsilon$.

Let $\ds \overline \epsilon = \frac \epsilon 2$.

Then:


 * $\ds \forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} < \frac \epsilon 2$.

Let $A$ and $N$ be such that the above inequalities for $I_3$ and $I_1 + I_2$ hold.

Then:

To sum up:


 * $\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : \forall n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$

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: