Mollifiers: Approximation Problem
Given a mollifier: $\varphi\in\mathcal{L}(\mathbb{R})$

Then it acts as an approximate identity:
  $$f\in\mathcal{C}(\mathbb{R}):\quad\int_{-\infty}^\infty n\varphi(nx)f(x)dx\to f(0)\cdot\int_{-\infty}^\infty\varphi(x)dx$$
  How to prove this under reasonable assumptions?

Example
As an example regard the Gaussian:
$$f\in\mathcal{C}(\mathbb{R}):\quad\frac{n}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-(nx)^2}f(x)\mathrm{d}x\to f(0)$$
(This is a useful technique when studying operator semigroups.)
 A: You can always approximate $j(x)$ by bounded continuous functions, for which the result holds by dominated convergence and the usual Riemann  substitution rule. 
A: Now, I got a proof for the case...
Theorem
Given a mollifier: $\varphi\in\mathcal{L}(\mathbb{R})$

For bounded functions one has:
  $$f\in\mathcal{C}(\mathbb{R}):\quad\int_{-\infty}^{\infty}n\varphi(nx)f(x)\mathrm{d}x\to f(0)\cdot\int_{-\infty}^{\infty}\varphi(x)\mathrm{d}x\quad(\|f\|_\infty<\infty)$$
  (In fact, this result extends to the Bochner integral!)

Proof
Split the integral into outer vanishing parts and the inner convergent part:
$$\left|\int_{-\infty}^{\infty}\varphi(\hat{x})f(\tfrac{1}{n}\hat{x})\mathrm{d}\hat{x}-f(0)\cdot\int_{-\infty}^{\infty}\varphi(\hat{x})\mathrm{d}\hat{x}\right|\\\leq\int_{-\infty}^{-R}|\varphi(\hat{x})|\cdot2\|f\|_\infty\mathrm{d}\hat{x}+\int_{-R}^{R}\|\varphi\|_\infty\cdot\left|f\left(\tfrac{1}{n}\hat{x}\right)-f(0)\right|\mathrm{d}\hat{x}+\int_{R}^{\infty}|\varphi(\hat{x})|\cdot2\|f\|_\infty\mathrm{d}\hat{x}\\\leq2\|f\|_\infty\delta_R+\left(\int_{-\infty}^{\infty}\varphi(x)\mathrm{d}x\right)\delta_N+2\|f\|_\infty\delta_R=\varepsilon\quad(n\geq N_R(\varepsilon))$$
That proofs the assertion.
Outview
Does it apply to unbounded functions too:
$$f\in\mathcal{C}(\mathbb{R}):\quad\int_{-\infty}^{\infty}n\varphi(nx)f(x)\mathrm{d}x\to f(0)\cdot\int_{-\infty}^{\infty}\varphi(x)\mathrm{d}x$$
(Intuition and examples suggest it does.)
