Rate of convergence of mollified exponential on $[0, \infty)$

Question

Consider the function $$ f(t)=e^{-at} $$ with $a>0$, $t \in [0, \infty)$ and let $\rho$ a smooth function on $[0, \infty)$ such that $$ \int_0^\infty \rho(t)=1 $$ and let $\rho_\delta(t):= \frac{1}{\delta} \rho\left(\frac{x}{\delta}\right)$ so that $$ \lim_{\delta \to 0} f * \rho_\delta(t)= f(t) $$ I'm trying to obtain the rate of convergence of $(f *\rho_\delta, \phi)$, but if I explicitely write the convolution I obtain (after a change of variable) $$ \int_0^T \phi(t) f * \rho_\delta(t)dt= \int_0^T \phi(t) \int_{-\infty}^{t/\delta} e^{-a(t-\delta \tau)} \rho(\tau) d\tau dt \lesssim \int_0^T \phi(t) \frac{e^{-at}}{a\delta} dt $$ where I have used that $\rho_\delta$ is supported in $[0, \infty)$. As far as I can see this does not converges to $f$. What is my mistake? Is there any way to obtain the rate of convergence of the convolution in terms of $\delta$?

Answer 1

We have: $$ \begin{align} \int_0^T \phi(t) f * \rho_\delta(t)dt &= \int_0^T \phi(t) \int_{-\infty}^{t} e^{-a\tau} \rho_\delta(t-\tau) d\tau \,dt \\ &= \int_0^T \phi(t) \int_{-\infty}^{t} e^{-a\tau} \frac{\rho(\frac{t-\tau}{\delta})}{\delta} d\tau \,dt \end{align} $$

Let $u(\tau)=\frac{t-\tau}{\delta}$, then $\tau = t - \delta u$, $u(t) = 0$, $u(-\infty)=+\infty$ and $d\tau=-\delta du$ $$ \begin{align} \int_0^T \phi(t) f * \rho_\delta(t)dt &= \int_0^T \phi(t) \int_{0}^{+\infty} e^{-a(t-\delta u)} \rho(u)du\,dt\\ &= \int_0^T \phi(t) e^{-at}\int_{0}^{+\infty} e^{a\delta u} \rho(u)du\,dt \end{align} $$

When $\delta \rightarrow 0$, the integral reaches $$ \int_0^T \phi(t) e^{-at}\int_{0}^{+\infty} \rho(u)du\,dt = \int_0^T \phi(t) e^{-at}dt$$