Question

Prove that the test space \(\boldsymbol{K}\) of all infinitely differentiable finite functions has "sufficiently many" functions in the sense that, given any two distinct continuous functions \(f_{1}\) and \(f_{2}\), there exists a function \(\varphi \in \mathrm{K}\) such that $$ \int_{-\infty}^{\infty} f_{1}(x) \varphi(x) d x \neq \int_{-\infty}^{\infty} f_{2}(x) \varphi(x) d x $$ Hint. Since \(\mathrm{f}(\mathrm{x})=f_{1}(x)-f_{2}(x) \neq 0\), there is a point \(x_{0}\) such that \(\mathrm{f}\left(x_{0}\right) \\# 0\), and hence an interval \([a, \beta]\) in whichf \((\mathrm{x})\) does not change sign. Let $$ \varphi(x)= \begin{cases}e^{-1 /(x-\alpha)^{2}} e^{-1 /(x-\beta)^{2}} & \text { if } \alpha

Short answer

The function \( \varphi(x) \) has compact support and distinguishes \( f_1 \) from \( f_2 \), proving sufficient existence in \( \boldsymbol{K} \).

Step-by-step solution

Understand the Problem

We need to prove that there exists a function in the test space \( \boldsymbol{K} \), denoted as \( \varphi \), such that it can distinguish between any two distinct continuous functions \( f_1 \) and \( f_2 \). This means finding a \( \varphi \) for which the integrals \( \int_{-\infty}^{\infty} f_1(x) \varphi(x) \, dx \) and \( \int_{-\infty}^{\infty} f_2(x) \varphi(x) \, dx \) are not equal.

Define the Difference Function

Define \( f(x) = f_1(x) - f_2(x) \). Given \( f_1 \) and \( f_2 \) are distinct, \( f(x) eq 0 \) for at least some \( x \). Locate a point \( x_0 \) such that \( f(x_0) eq 0 \), implying there exists an interval \( [\alpha, \beta] \) where \( f(x) \) does not change sign.

Define the Test Function

Define the function \( \varphi(x) \) as follows: \[ \varphi(x) = \begin{cases} e^{-1/(x-\alpha)^2} e^{-1/(x-\beta)^2} & \alpha < x < \beta \ 0 & \text{otherwise} \end{cases} \] The function \( \varphi(x) \) has compact support in the interval \( [\alpha, \beta] \) and is infinitely differentiable. Therefore, \( \varphi(x) \in \boldsymbol{K} \).

Evaluate the Integral of Difference

Since \( \varphi(x) = 0 \) outside \( (\alpha, \beta) \), the integral simplifies to: \[ \int_{-\infty}^{\infty} f(x) \varphi(x) \, dx = \int_{\alpha}^{\beta} f(x) \varphi(x) \, dx \] Since \( f(x) \) does not change sign on this interval, and \( \varphi(x) > 0 \) for \( x \in (\alpha, \beta) \), this integral is not zero. Hence, the integrals for \( f_1 \) and \( f_2 \) with \( \varphi(x) \) are different.

Conclude the Proof

The existence of such a \( \varphi(x) \) proves that for any two distinct continuous functions \( f_1 \) and \( f_2 \), there is a function in the test space \( \boldsymbol{K} \) that results in different integral values when integrated against \( f_1 \) and \( f_2 \). This confirms the test space \( \boldsymbol{K} \) has sufficiently many functions as required.

Key concepts

Infinitely Differentiable Functions

Infinitely differentiable functions are an important concept in mathematical analysis. These are functions that can be differentiated an unlimited number of times. This means that the derivative can be taken repeatedly without encountering any points where the function is not defined or not smooth.

The beauty of infinitely differentiable functions lies in their smoothness and the predictability of their derivatives. When a function is infinitely differentiable, it behaves very nicely at and between every point in its domain. This makes them a perfect candidate for various advanced mathematical applications.

In our context, the test function \( \varphi(x) = \begin{cases} e^{-1/(x-\alpha)^2} e^{-1/(x-\beta)^2} & \text{if}\ \alpha < x < \beta \ 0 & \text{otherwise}\ \end{cases} \)is an example of an infinitely differentiable function. Although it looks complex, its true power is its infinite differentiability in the open interval \((\alpha, \beta)\).

• Each derivative of the function within this interval is continuous, ensuring seamless transitions across points.
• The infinite series of derivatives help in accurately capturing the behavior of functions it's multiplied with—critical while operating in integral calculus.

Integral Calculus

Integral calculus is all about measuring the area under a curve. It helps us in determining various characteristics of functions by integrating them over specific intervals. In our problem, integral calculus plays a crucial role in separating the behaviors of two distinct continuous functions using a test function.

By integrating the product of a difference function and a test function, we aim to find out whether the integral values are distinct for different functions. Choosing a test function that leads to different integral results verifies the divergence of the two initial functions. This happens without requiring the point values of the functions themselves.

• The integral \( \int_{\alpha}^{\beta} f(x) \varphi(x) \, dx \) is evaluated over a compact interval where \( f(x) \) remains sign-consistent.
• Therefore, the test function \( \varphi(x) \), when integrated, clearly reveals the disparity as it magnifies any difference present within this specific range.

Compact Support

The term "compact support" in mathematics refers to functions that are non-zero only within a closed and bounded interval and zero everywhere else. This is a significant property in both real and complex analysis, allowing mathematicians to control where the function is effective.

In our problem, the test function \( \varphi(x) \) has compact support over the interval \([\alpha, \beta]\), where its effects are concentrated. Having compact support is highly valuable as it confines its influence within a finite region, which makes computations manageable.

• Functions with compact support can be differentiated repeatedly without leaving the interval where they are defined, making them excellent for representing local behaviors.
• Their smoothness and delineated influence make them ideal for use in advanced techniques like integrating by parts or in convolution operations.

By ensuring that the function \( \varphi(x) \) is zero outside \([\alpha, \beta]\), we avoid troublesome edge effects and focus strictly on the core region of interest defined by continuous functions like \( f_1 \) and \( f_2 \). This is crucial for accurate integral evaluations that reflect true differences.