Gaussian 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:

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

Then:

It follows that:


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

Then:

Analogously:

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 $A$ be such that $A < \delta$.

Then:

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: