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?


1 Answer 1


I will answer my questions since I was able to find out what it means.

Yes, the functions $\phi$ are from the real line $\mathbb{R}$ into $\mathbb{R}$.

Finite means actually compact supported. This was pointed out to me in the chat on math.stackexchange.

Vanishing outside some interval is because the maps are assumed to be compactly supported i.e. the subset of the domain of all non-zero values of the function is assumed to be compact. Since compactness in metric spaces is equivalent to being closed and totally bounded, in particular, for $\mathbb{R}^{n}$ and the real line $\mathbb{R}$ compactness is equivalent to being closed and bounded, we have that every bounded set is contained in an interval and also the support. Meaning that outside of this, the function values "vanish" i.e. are zero.

More about the support of a function can be found here.

Remark: When Kolmogorov writes that the functions are continuously differentiable at all orders it means in modern terminology that the maps are smooth i.e. infinitely many times differentiable.

Still looking into showing that a regular generalized function converges.

  • $\begingroup$ The convergence isn't too difficult. You've deduced that there is an interval $[a,b]$ such that all of the $\phi_n$ vanish outside of $[a,b]$. This means that $\int_{-\infty}^\infty f(x) \phi_n(x) dx = \int_a^b f(x) \phi_n(x) dx$. If $\phi_n \to \phi$ in $K$ then $\phi_n \to \phi$ uniformly on $[a,b]$. Now $\bigg | \int_a^b f(x) (\phi_n(x) - \phi(x) ) dx \bigg | \leq \int_a^b |f(x)| |\phi_n(x) - \phi(x)| dx$. Use uniform convergence of $\phi_n$ and integrability of $f$ on $[a,b]$ to see that the right hand side goes to $0$ as $n \to \infty$. $\endgroup$ Feb 24, 2020 at 21:21

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