Trying to understand
For (b), I understand that there are two claims are proposed in respective the cases $[1,\infty]$ and $[1,\infty)$.
i. For $p\in[1,\infty]$, one has $$\|\rho_k * f\|_p\le\|f\|_p,\qquad (i.e.\quad \rho_k * f\in L^p(\mathbb{R})\quad)$$ ii. For $p\in[1,\infty)$, one has $$\|\rho_k * f-f\|_p\to 0,\qquad (k\to\infty)$$ (convergence in $p-norm$)
However, I have a rather vague idea of the difference of the two, and I cannot explain to myself the meaning/implication of the two equations and how they differ. 1) Could anyone describe to me the meaning of the two equations, the significance of including $\infty$, in words/simple mathematical language?
2) In the case $p=\infty$, equation in $(ii)$ no longer holds, why? And equation in $(ii)$ always implies equation in $(i)$, right?
Notice that $(i)$ is not only saying that $\rho_k * f \in L^p$. The information encoded in the inequality is that the convolution with one of the $\rho_k$ does not increase the $L^p$-norm of $f$. This is true for every $p$, including $\infty$, and for every $k$.
On the other hand $(ii)$ exploits the fact that $\rho_k$ is a $\delta$-sequence to recover $f$ when $k$ approaches $\infty$. One of the main differences with $(i)$ is that it is a statement about the limit as $k$ goes to infinity, so there is no information about "small" values of $k$. Hence it is not correct to say that $(ii)$ implies $(i)$.
Moreover, the convergence in $(ii)$ implies that for $k$ large $\rho_k * f$ is close to $f$ in an $L^p$ sense, but there is no obvious inequality between the norms of these two $L^p$ functions.
This is important: the reason why $p = \infty$ is not included in $(ii)$ is that the corresponding result is false. Indeed, assume by contradiction that the result is true, i.e. $$\|\rho_k * f - f\|_{\infty} \to 0. \tag 1$$ Notice that $\rho_k * f$ is continuous since $\rho_k$ is assumed to be continuous. Then by $(1)$ we have that the continuous sequence of functions $\{\rho_k * f\}$ that converges uniformly to $f$. This would imply that $f$ is continuous, hence yielding a contradiction (not every $L^{\infty}$ function is continuous!).
The following is a function in $L^{\infty}(\mathbb{R})$ that cannot be made into a continuous function by redefining it on a null a set.
Consider $\chi(x) = \chi_{[0,1]}(x)$ and assume by contradiction that $f$ is a continuous function such that $f(x) = \chi(x)$ almost everywhere. Say that the set where they differ is $E$. Let $x_0$ be the smallest point such that $f(x_0) = \chi(x_0) = 1$. Consider intervals of the form $(x_0 - 1/n,x_0)$: since they all have positive measure we can find $x_n \in (x_0 - 1/n,x_0) \cap E^c$. By definition of $x_0$ we can also assume that $\chi(x_n) = 0$, indeed $x_0$ is only a set of measure $0$ apart from $0$. Then $x_n \to x_0$, so by continuity of $f$ we also have $f(x_n) \to 0$. This gives a contradiction because $$0 = \lim_n 0 = \lim_n \chi(x_n) = \lim_n f(x_n) = f(x_0) = 1.$$