To appreciate the Dirac delta function as the limit of a certain function that is more elementary, consider |x| < ε → gε(x) = 8|x| > ε. Plot the functions gε(x) and g(x0) for ε = 0.1, 0.01, and ε → 0. Show that lim gε(x) = 0 for ε → 0. Show that lim gε(0) = 0 for ε → 0. Show that ∫gε(ε0) dx = 1 for any ε0 and ∫f(c)g(x0) dx = f(ε0) for any ε0 where f is continuous.
The Correct Answer and Explanation is:
The Dirac delta function is often considered in the context of distributions, which are not functions in the traditional sense but rather generalized functions that act on test functions. A standard approach to defining the Dirac delta function, δ(x)\delta(x)δ(x), is through its limiting behavior as a sequence of functions.
The function gϵ(x)g_\epsilon(x)gϵ(x):
Given that you want to appreciate the Dirac delta function as the limit of a simpler function, consider gϵ(x)=8∣x∣ϵg_\epsilon(x) = \frac{8|x|}{\epsilon}gϵ(x)=ϵ8∣x∣ for ∣x∣<ϵ|x| < \epsilon∣x∣<ϵ and gϵ(x)=0g_\epsilon(x) = 0gϵ(x)=0 otherwise.
This function is a piecewise linear function that has a “spike” at x=0x = 0x=0, with a peak that becomes increasingly narrow as ϵ→0\epsilon \to 0ϵ→0, while its total area remains constant.
- Plotting gϵ(x)g_\epsilon(x)gϵ(x) for different values of ϵ\epsilonϵ:
- For ϵ=0.1\epsilon = 0.1ϵ=0.1, gϵ(x)g_\epsilon(x)gϵ(x) will be a triangle with base 2ϵ=0.22\epsilon = 0.22ϵ=0.2 and height 808080 at x=0x = 0x=0.
- For ϵ=0.01\epsilon = 0.01ϵ=0.01, the triangle becomes much narrower and taller, with a height of 800800800 at x=0x = 0x=0.
- As ϵ→0\epsilon \to 0ϵ→0, the function becomes increasingly concentrated around x=0x = 0x=0, approaching the Dirac delta function.
- Limit as ϵ→0\epsilon \to 0ϵ→0:
- For ϵ→0\epsilon \to 0ϵ→0, the function gϵ(x)g_\epsilon(x)gϵ(x) tends to zero for any x≠0x \neq 0x=0 because the area under the function remains constant while the base of the triangle narrows.
- Therefore, limϵ→0gϵ(x)=0\lim_{\epsilon \to 0} g_\epsilon(x) = 0limϵ→0gϵ(x)=0 for all x≠0x \neq 0x=0.
- Evaluating at x=0x = 0x=0:
- At x=0x = 0x=0, gϵ(0)=8∣0∣ϵ=0g_\epsilon(0) = \frac{8|0|}{\epsilon} = 0gϵ(0)=ϵ8∣0∣=0 for all ϵ\epsilonϵ. Hence, limϵ→0gϵ(0)=0\lim_{\epsilon \to 0} g_\epsilon(0) = 0limϵ→0gϵ(0)=0.
- Integral of gϵ(x)g_\epsilon(x)gϵ(x):
- The integral of gϵ(x)g_\epsilon(x)gϵ(x) over R\mathbb{R}R is given by: ∫−∞∞gϵ(x) dx=∫−ϵϵ8∣x∣ϵ dx=1.\int_{-\infty}^{\infty} g_\epsilon(x) \, dx = \int_{-\epsilon}^{\epsilon} \frac{8|x|}{\epsilon} \, dx = 1.∫−∞∞gϵ(x)dx=∫−ϵϵϵ8∣x∣dx=1. The area under the triangle is always 1, regardless of the value of ϵ\epsilonϵ.
- Action of gϵ(x)g_\epsilon(x)gϵ(x) on a continuous function:
- For any continuous function f(x)f(x)f(x), the action of gϵ(x)g_\epsilon(x)gϵ(x) is given by: ∫−∞∞f(x)gϵ(x) dx=f(0).\int_{-\infty}^{\infty} f(x) g_\epsilon(x) \, dx = f(0).∫−∞∞f(x)gϵ(x)dx=f(0). This is because the function gϵ(x)g_\epsilon(x)gϵ(x) is nonzero only in a narrow region around x=0x = 0x=0, and as ϵ\epsilonϵ becomes smaller, the integral becomes more concentrated at x=0x = 0x=0, thus approaching f(0)f(0)f(0).
Conclusion:
The function gϵ(x)g_\epsilon(x)gϵ(x) approaches the Dirac delta function δ(x)\delta(x)δ(x) as ϵ→0\epsilon \to 0ϵ→0, because:
- limϵ→0gϵ(x)=0\lim_{\epsilon \to 0} g_\epsilon(x) = 0limϵ→0gϵ(x)=0 for x≠0x \neq 0x=0,
- gϵ(0)=0g_\epsilon(0) = 0gϵ(0)=0 for all ϵ\epsilonϵ,
- ∫−∞∞gϵ(x) dx=1\int_{-\infty}^{\infty} g_\epsilon(x) \, dx = 1∫−∞∞gϵ(x)dx=1 for any ϵ\epsilonϵ,
- ∫−∞∞f(x)gϵ(x) dx\int_{-\infty}^{\infty} f(x) g_\epsilon(x) \, dx∫−∞∞f(x)gϵ(x)dx approaches f(0)f(0)f(0) for continuous functions f(x)f(x)f(x).
This demonstrates that the Dirac delta function is the limiting form of gϵ(x)g_\epsilon(x)gϵ(x) as ϵ→0\epsilon \to 0ϵ→0.
