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I have been following the course in Introduction to Probability, Statistics and Random Processes, by Hossien Pishro-Nik, and I was having some troubles understanding proof in chapter 4.3.2 Using the Delta Function. The lemma states that $$\int_{-\infty}^{\infty} g(x) \delta(x-x_0) dx = g(x_0).$$

The proof states that, let $I$ bet the value of the above integral. Then, we have $$I= \lim_{\alpha \rightarrow 0} \bigg[ \int_{-\infty}^{\infty} g(x) \delta_{\alpha} (x-x_0) dx \bigg]$$ $$=\lim_{\alpha \rightarrow 0} \bigg[ \int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx \bigg].$$ By the mean value theorem in calculus, for any $α>0$, we have $$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx=\alpha \frac{g(x_{\alpha})}{\alpha}=g(x_{\alpha}),$$ for some $x_{\alpha} \in (x_0-\frac{\alpha}{2},x_0+\frac{\alpha}{2}).$ Thus, we have $$I = \lim_{\alpha \rightarrow 0} g(x_{\alpha})=g(x_0).$$

Now, the lemma makes sense intuitively but I am unable to understand why $$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx=\alpha \frac{g(x_{\alpha})}{\alpha}=g(x_{\alpha}).$$ From my understanding of the mean value theorem, shouldn’t $$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}} \frac{g(x)}{\alpha} dx=\frac{1}{\alpha}\frac{G(x_{0}+\frac{\alpha}{2})-G(x_{0}-\frac{\alpha}{2})}{(x_{0}+\frac{\alpha}{2})-(x_{0}-\frac{\alpha}{2})}$$ $$=\frac{1}{\alpha}\frac{G(x_{0}+\frac{\alpha}{2})-G(x_{0}-\frac{\alpha}{2})}{\alpha}=\frac{G(x_{\alpha})}{\alpha^2},$$ where $G(x_{\alpha})$ is $G(x)$ defined in some range $x_{\alpha} \in (x_0-\frac{\alpha}{2},x_0+\frac{\alpha}{2}).$

As you can see I am unable to understand how the author managed to achieve the result he stated. Can you please help me understand how the author achieved the result?

BlockFace
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  • He's using generalized mean value theorem for integrals. – PNDas Oct 26 '20 at 04:33
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    See https://en.wikipedia.org/wiki/Mean_value_theorem, and search for First mean value theorem for definite integrals – PNDas Oct 26 '20 at 04:35
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    If the lemma states that $\int_{-\infty}^{\infty} g(x) \delta(x-x_0) dx = g(x_0),$ what is the definition of $\delta$? That integral is normally taken to be the definition of $\delta.$ – md2perpe Oct 26 '20 at 07:29
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    @PNDas Ah, of course. Thank you for mentioning the topic and sending the link with the relevant key terms. It makes sense now. I would upvote your comment but unfortunately I do not have enough rep points to be able to do that now. I am happy to write an answer and mention your help, in case someone else in the future may run into the same problem. – BlockFace Oct 26 '20 at 14:33
  • @md2perpe like I mentioned in the question, the lemma makes sense and I am able to see a picture in my head as to why it does. I guess my issue was being able to apply the mean value theorem in one part of the proof. I guess I was not able to apply the mean value theorem correctly in this case which is why I was confused. – BlockFace Oct 26 '20 at 14:38
  • @BlockFace. But what definition of $\delta$ do you have? – md2perpe Oct 26 '20 at 16:05
  • @md2perpe ${\delta}(x)$ is zero everywhere, except at $x=0$ where it is $\infty$. ${\delta}_{\alpha}(x) = \frac{1}{\alpha},{\space} |x|<\frac{\alpha}{2}$, 0 otherwise. To see how these were established please follow https://www.probabilitycourse.com/chapter4/4_3_2_delta_function.php – BlockFace Oct 26 '20 at 16:59
  • @BlockFace. That article does not define $\delta(x)$ to be zero everywhere except at $x=0$. It defines $\delta$ as the functional limit of $\delta_\alpha$ as $\alpha\to0.$ In distribution theory, the formula in the lemma is taken as definition of $\delta.$ It can then be shown that $\delta_\alpha \to \delta$ in the space of distributions. – md2perpe Oct 26 '20 at 17:45
  • @md2perpe sorry for the confusion, you are correct. I am new to all of this. I have made the edits to the question. Hopefully the question states the problem more clearly now. – BlockFace Oct 26 '20 at 18:51

1 Answers1

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Full credit goes to PNDas for his comment.

The mean value theorem states that if a function $f(x)$ satisfies the following conditions

  1. $f(x)$ is continuous on the closed interval $[a,b]$
  2. $f(x)$ is differentiable on the open interval $(a,b)$

then there is a number c such that $a<c<b$ and $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ $${\implies}{f(b)-f(a)} = {f'(c)}{(b-a)},$$ where $f'$ is the differential of $f$.

Since $$\int_{a}^{b}{f'(x)dx} = {f(b)-f(a)},$$ then application of the mean value theorem would imply that $$\frac{1}{b-a}\int_{a}^{b}{f'(x)dx} = {f'(c)}$$ $${\implies}\int_{a}^{b}{f'(x)dx} = {f'(c)}{(b-a)},$$ where $a<c<b$.

Applying this to the problem in the question would produce $$\int_{x_0-\frac{\alpha}{2}}^{x_0+\frac{\alpha}{2}}{\frac{g(x)}{\alpha}dx} = {\frac{g(x_{\alpha})}{\alpha}}{[(x_0+\frac{\alpha}{2})-(x_0-\frac{\alpha}{2})]}$$ $$= {\alpha}{\frac{g(x_{\alpha})}{\alpha}} = g(x_{\alpha}),$$ where $x_{\alpha}{\space}{\in}{\space}({x_0-\frac{\alpha}{2}}, {x_0+\frac{\alpha}{2}}).$

The definition of the mean value theorem was taken from Paul's Online Notes.

BlockFace
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