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{,} &amp; x&lt;-\epsilon\text{,}\\<br /> \frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} &amp; \epsilon\leq x\leq\epsilon\text{,}\\<br /> 0\text{,} &amp; \epsilon&lt;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&gt;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&gt;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{,} &amp; x&lt;-\epsilon\text{,}\\<br /> \frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} &amp; \epsilon\leq x\leq\epsilon\text{,}\\<br /> 0\text{,} &amp; \epsilon&lt;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&gt;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&gt;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