I am currently studying section 21.2 The test space and test functions. Generalized functions. from Kolmogorov and Fomin Introductory Real Analysis and I have some perhaps elementary questions regarding the material. I will highlight my questions like this.
The test space and the test functions are defined the following way:
Let $K$ be the set of all finite functions $\phi$ on $(- \infty, \infty)$ with continuous derivatives of all order, or equivalently, the set of all infinitely differentiable functions, where every function $\phi \in K$, being finite, vanishes outside some interval depending on the choice of $\phi$.
My question here is what does it mean "function $\phi$ on $(- \infty, \infty)$"? Does it mean the domain and the co-domain of the function are $(- \infty, \infty)$, i.e $\phi : (- \infty, \infty) \rightarrow (- \infty, \infty)$?
What does a finite function mean? Does it mean that the values are all numbers except $- \infty$ and $\infty$, i.e. $- \infty < f(x) < \infty$ for each $x$?
What does "every function $\phi \in K$, being finite vanishes outside some interval depending on the choice of $\phi$" mean? What does it mean for a function to vanish outside some interval? I.e. if the domain of the function would be $[1, 3]$, then the function vanishes outside of this interval means what? That its values are zero everywhere else?
The space $K$ is then equipped with addition and multiplication of scalar such that
$$+ : K \times K \rightarrow K, (\phi, \psi) \mapsto +(\phi, \psi) := (\phi + \psi)(x) := \phi(x) + \psi(x)$$
$$\cdot : \mathbb{K} \times K \rightarrow K, (\alpha, \phi) \mapsto \cdot (\alpha, \phi) := (\alpha \cdot \phi)(x) := (\alpha \phi)(x) := \alpha \phi(x)$$
Where $\mathbb{K}$ is a field and usually $\mathbb{R}$ or $\mathbb{C}$.
With these two operations, the space $K$ then becomes a linear space.
In the next step, a notion of convergence is introduced on the space $K$, namely:
A sequence $(\phi_{n})$ of functions in $K$ is said to converge to a function $\phi \in K$ iff.
(i) there exists an interval outside which all the functions $\phi_{n}$ vanish; and
(ii) The sequence $(\phi_{n}^{(k)})$ of the kth derivative converges uniformly on this interval to $\phi^{(k)}$ for every $k = 0, 1, 2, \ldots$ .
The linear space $K$ equipped with this notion of convergence is called the test space and its elements are called test functions.
Then a generalized function is defined by
The functional $T : K \rightarrow \mathbb{K}$ is a generalized function on $(- \infty, \infty)$ iff. $T$ is linear and continuous in the sense that $\phi_{n} \rightarrow \phi$ in $K$ implies $T(\phi_{n}) \rightarrow T(\phi)$.
I am stuck on showing that a regular generalized function is continuous. A regular generalized function is defined by:
Let $f(x)$ be a locally integrable function i.e. on every finite interval. Then, $f(x)$ generates a generalized function via the expression
$$T_{f}(\phi) = \langle f, \phi \rangle = \int^{\infty}_{- \infty} f(x)\phi(x)dx$$
Showing linearity of the above expression was simple by showing $T_{f}(\phi_{1} + \phi_{2}) = T_{f}(\phi_{1}) + T_{f}(\phi_{2})$ and $T_{f}(\alpha \phi) = \alpha T_{f}(\phi)$.
To show continuity I assumed $\phi_{n} \rightarrow \phi$ in the sense of the above definition of convergence. Then, I have to show $T(\phi_{n}) \rightarrow T(\phi)$.
So $$T(\phi_{n}) = \int^{\infty}_{- \infty} f(x)\phi_{n}(x)dx = \int^{b}_{a} f(x)\phi_{n}(x)dx$$
since all $\phi_{n}$ vanish outside some interval, I assumed that the integral will have some finite upper and lower bounds of integration and be zero everywhere else. And after this I am stuck. I was trying to use the second condition of the definition of convergence above, namely, that the kth derivative of $\phi_{n}$ converge uniformly but I can't seem to move forward. How do I show convergence of the regular generalized function?