Confused About Rudin's Opening Statements on his Chapter on Distributions

Question

I have a few doubts regarding Rudin's opening paragraphs in Chapter 6 of Functional Analysis. The passage (with small edits) reads:

A complex function $f$ is said to be locally integrable if $f$ is measurable and $\int_K |f| < \infty$ for every compact $K \subseteq R$. We let $\mathcal{D} = \mathcal{D}(\mathbb{R})$ be the vector space of all $\phi\in C^\infty(\mathbb{R})$ whose support is compact. Then

1. $\int f\phi$ exists for every locally integrable $f$ and for every $\phi\in\mathcal{D}$.
2. $\mathcal{D}$ is sufficiently large to assure that $f$ is determined (a.e.) by the integrals $\int \phi f$. (To see this, note that the uniform closure of $\mathcal{D}$ contains every continuous function with compact support.)
3. If $f$ happens to be continuously differentiable, then $$\int f'\phi = -\int f\phi' \ \ \ \ (\phi\in\mathcal{D}).$$
4. If $f\in C^\infty(\mathbb{R})$, then $$\int f^{(k)}\phi = (-1)^k\int f\phi^{(k)} \ \ \ \ (\phi\in\mathcal{D}).$$ The compactness of the support of $\phi$ was used in these integrations by parts.

Regarding 2: while it is true that the uniform closure of $\mathcal{D}$ contains all continuous functions with compact support, how does that imply $f$ is determined by the $\int \phi f$?

Regarding 3: it seems to me that the actual equation should be $$\int f'\phi = f\phi -\int f\phi' \ \ \ \ (\phi\in\mathcal{D}).$$
Why does Rudin omit the $f\phi$ term?

Regarding 4: again, it seems to me the actual equation should be $$\int f^{(k)}\phi = \sum_{i=0}^{k-1}(-1)^if^{(i)}\phi^{(k-i-1)} + (-1)^k\int f\phi^{(k)}.$$ Why does Rudin omit the second term?

Lastly: how is compactness of the support of $\phi$ used?

Answer 1

The second remark ($\mathcal{D}$ being sufficiently large to determine $f$ a.e. through its various integrals) alludes to the following fact: if $f$ and $g$ are locally integrable with respect to a $\sigma-$finite Borel measure $\mu$ of $\mathbb{R}$ and $$\int_\mathbb{R} f\phi \, d\mu = \int_\mathbb{R} g\phi \, d\mu \qquad \text{ for all } \phi \in \mathcal{D},$$ then $f = g$ a.e.[$\mu$].

To prove the above statement, it's enough to show that if $$\int_\mathbb{R} f\phi \, d\mu = 0 \qquad \text{ for all } \phi \in \mathcal{D}$$ then $f = 0$ a.e.[$\mu$]. Suppose there is an open interval $A \subset \mathbb{R}$ such that $f$ is positive a.e.[$\mu$] on $A.$ Take $K \subset A$ a (nondegenerate) proper compact subinterval (so that $\mu(A \smallsetminus K) > 0$) and pick a nonnegative $\phi \in \mathcal{D}$ such that $\operatorname{supp} \phi \subset A$ and $\phi = 1$ on $K$ (you can always construct these smooth bump functions using smooth cutoff functions). Then $$0 = \int_\mathbb{R} f\phi \, d\mu = \int_A f\phi \, d\mu = \int_K f\phi \, d\mu + \int_{A \smallsetminus K} f\phi \, d\mu \, > \, \int_K f \, d\mu.$$ This contradicts the positivity a.e.[$\mu$] of $f.$ A similar argument works for $f$ negative a.e.[$\mu$], and thus we conclude that $f$ must be zero a.e.[$\mu$] on any open interval, and also on $\mathbb{R}.$

The third and fourth remarks are explained by noting that the boundary terms in the integration-by-parts formula always vanish since every $\phi \in \mathcal{D}$ has compact support.

Lastly, you want the functions $\phi \in \mathcal{D}$ to have compact support so the integrals $\int_\mathbb{R} f\phi$ are all finite for every locally integrable $f$ (and because of the nice "topological/real analysis" properties that you get when you work with these function spaces).