Mollifiers: Asymptotic Convergence vs. Mean Convergence

Question

Problem

Does asymptotic convergence imply mean convergence: $$\varphi\in\mathcal{L}_\text{loc}(\mathbb{R}_+):\quad\varphi(T)\stackrel{T\to\infty}{\to}\varphi_\infty\implies\frac{1}{T}\int_0^T\varphi(s)\mathrm{d}s\stackrel{T\to\infty}{\to}\varphi_\infty$$

Remark

Three important classes fall under local integrability: $\mathcal{L}(\mathbb{R}_+),\mathcal{C}(\mathbb{R}_+),\mathcal{B}(\mathbb{R}_+)\subseteq\mathcal{L}_\text{loc}(\mathbb{R}_+)$

Answer 1

Splitting the integral gives: $$\left|\frac{1}{T}\int_0^T\varphi(s)\mathrm{d}s-\varphi_\infty\right|\leq\frac{1}{T}\int_0^{T_\varphi}|\varphi(s)|\mathrm{d}s+\frac{1}{T}\int_{T_\varphi}^T|\varphi(s)-\varphi_\infty|\mathrm{d}s+\frac{1}{T}\int_0^{T_\varphi}|\varphi_\infty|\\\leq\delta_T\left(\int_0^\infty|\varphi(s)|\mathrm{d}s\right)+1\cdot\delta_\varphi+\delta_T\cdot T_\varphi|\varphi_\infty|=\varepsilon\quad(T\geq T_\infty\geq\delta_T^{-1}T_\varphi)$$

(The first and third integral were bounded while the second integral was approximated.)