Let $f \in C^1(\mathbb{R})$ and $f\in L^1([0,\infty))$ compute limits, reminiscent of reimann Lebesgue lemma
I am not sure how the fact that $f \in C^1$ helps here. I know that such functions are dense in $L^1$ so I would assume thorough a limiting procedure we could extend this result to all $f \in L^1$. It is easy to show that the limit is $\|f\|_1$ for a) when $f$ is a simple funciton. However, I do not know how to extend it to all $f\in L^1$. The problem i encountered is that when $\epsilon<0$ then $e^{-\epsilon t ^2}$ is not bounded and I cannot apply DCT or Holders. I have that same issue with all the parts. Hints or solutions are welcome.
For (c), by integration by parts
$$ \frac{1}{\varepsilon}\int_0^1 f(t)\; t\; e^{-t^2/\varepsilon}\; dt = \frac{f(0)}{2} - \frac{f(1)}{2} e^{-1/\varepsilon} + \frac{1}{2} \int_0^1 e^{-t^2/\varepsilon} f'(t)\; dt \to \frac{f(0)}{2}\ \text{as}\ \varepsilon \to 0+$$ For $t \ge 1 \ge \varepsilon > 0$, $$ \frac{t}{\varepsilon} e^{-t^2/\varepsilon} \le \frac{t}{\varepsilon} e^{-t/\varepsilon} $$ Note that $g(s) = s e^{-s}$ is decreasing for $s > 1$, with $g(s) \to 0$ as $s \to \infty$. Thus $$ \left|\dfrac{1}{\varepsilon} \int_1^\infty f(t)\; t\; e^{-t^2/\varepsilon}\; dt \right| \le \int_1^\infty g(t/\varepsilon) |f(t)|\; dt \le g(1/\varepsilon) \int_1^\infty |f(t)|\; dt \to 0 \ \text{as}\ \varepsilon \to 0+$$