Tuning Fork Delta Sequence/Proof 2

Proof
We have that:

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

Then:

Furthermore:

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:

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: