Dirac Delta/Mean Value Theorem Problem

[EdMel]

Homework Statement

Consider the function $\delta_{\epsilon}(x)$ defined by
$$\delta_{\epsilon}(x)=\begin{cases}<br /> 0\text{,} & x<-\epsilon\text{,}\\<br /> \frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} & \epsilon\leq x\leq\epsilon\text{,}\\<br /> 0\text{,} & \epsilon<x\text{,}<br /> \end{cases}$$
(b) Consider a function $f$, defined on $\mathbb{R}$. which is continuous. Use
the Mean Value Theorem of integral calculus to show that
$$\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)$$
where $-\epsilon\leq\theta\leq\epsilon$.

Homework Equations

I guess the definition of $\delta_{\epsilon}(x)$ implies $\epsilon>0$, but I will assume this for my answer anyway.

I will refer to the Mean Value Theorem as MVT.

The Attempt at a Solution

Let,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=\overset{\epsilon}{\underset{-\epsilon}{\intop}}f(x)\delta_{\epsilon}(x)dx\quad\text{, as }\delta_{\epsilon}(x)=0\quad\forall\quad\vert x\vert\geq\epsilon\text{,}$$
$=(\epsilon-(-\epsilon))f(\theta)\delta_{\epsilon}(\theta)\text{, }$ $-\epsilon\leq\theta\leq\epsilon$, by Mean Value Theorem,
$$=2\epsilon f(\theta)\frac{3}{4\epsilon^{3}}(\epsilon^{2}-\theta^{2})\\=f(\theta)\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})$$
Then I am not sure how to proceed. I thought maybe there would be another application of MVT. The best argument I can come up with continues as ...
as $\epsilon>0$ is arbitrary let $\epsilon=\sqrt{3}\vert\theta\vert$ ($-\epsilon\leq\theta\leq\epsilon$ is still satisfied), then
$$\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})=\frac{3}{2(\sqrt{3}\vert\theta\vert)^{3}}((\sqrt{3}\vert \theta\vert)^{2}-\theta^{2})=1$$
so,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)\text{,}\quad \epsilon=\sqrt{3}\vert\theta\vert\text{.}$$
The problem I have now is the $\theta$ we know exists by MVT is dependent on $\epsilon$ we choose. Is it then valid to set $\epsilon$ as a function of $\theta$?

Thanks in advance.

 

[Dick]

Homework Statement

Consider the function $\delta_{\epsilon}(x)$ defined by
$$\delta_{\epsilon}(x)=\begin{cases}<br /> 0\text{,} & x<-\epsilon\text{,}\\<br /> \frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} & \epsilon\leq x\leq\epsilon\text{,}\\<br /> 0\text{,} & \epsilon<x\text{,}<br /> \end{cases}$$
(b) Consider a function $f$, defined on $\mathbb{R}$. which is continuous. Use
the Mean Value Theorem of integral calculus to show that
$$\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)$$
where $-\epsilon\leq\theta\leq\epsilon$.

Homework Equations

I guess the definition of $\delta_{\epsilon}(x)$ implies $\epsilon>0$, but I will assume this for my answer anyway.

I will refer to the Mean Value Theorem as MVT.

The Attempt at a Solution

Let,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=\overset{\epsilon}{\underset{-\epsilon}{\intop}}f(x)\delta_{\epsilon}(x)dx\quad\text{, as }\delta_{\epsilon}(x)=0\quad\forall\quad\vert x\vert\geq\epsilon\text{,}$$
$=(\epsilon-(-\epsilon))f(\theta)\delta_{\epsilon}(\theta)\text{, }$ $-\epsilon\leq\theta\leq\epsilon$, by Mean Value Theorem,
$$=2\epsilon f(\theta)\frac{3}{4\epsilon^{3}}(\epsilon^{2}-\theta^{2})\\=f(\theta)\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})$$
Then I am not sure how to proceed. I thought maybe there would be another application of MVT. The best argument I can come up with continues as ...
as $\epsilon>0$ is arbitrary let $\epsilon=\sqrt{3}\vert\theta\vert$ ($-\epsilon\leq\theta\leq\epsilon$ is still satisfied), then
$$\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})=\frac{3}{2(\sqrt{3}\vert\theta\vert)^{3}}((\sqrt{3}\vert \theta\vert)^{2}-\theta^{2})=1$$
so,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)\text{,}\quad \epsilon=\sqrt{3}\vert\theta\vert\text{.}$$
The problem I have now is the $\theta$ we know exists by MVT is dependent on $\epsilon$ we choose. Is it then valid to set $\epsilon$ as a function of $\theta$?

Thanks in advance.

I think you aren't using the version of the MVT best suited to this problem. See http://en.wikipedia.org/wiki/Mean_value_theorem#First_mean_value_theorem_for_integration