- #1

EdMel

- 13

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## Homework Statement

Consider the function [itex]\delta_{\epsilon}(x)[/itex] defined by

[tex]\delta_{\epsilon}(x)=\begin{cases}

0\text{,} & x<-\epsilon\text{,}\\

\frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} & \epsilon\leq x\leq\epsilon\text{,}\\

0\text{,} & \epsilon<x\text{,}

\end{cases}[/tex]

(b) Consider a function [itex]f[/itex], defined on [itex]\mathbb{R}[/itex]. which is continuous. Use

the Mean Value Theorem of integral calculus to show that

[tex]\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)[/tex]

where [itex]-\epsilon\leq\theta\leq\epsilon[/itex].

## Homework Equations

I guess the definition of [itex]\delta_{\epsilon}(x)[/itex] implies [itex]\epsilon>0[/itex], 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,

[tex]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{,}[/tex]

[itex]=(\epsilon-(-\epsilon))f(\theta)\delta_{\epsilon}(\theta)\text{, }[/itex] [itex]-\epsilon\leq\theta\leq\epsilon[/itex], by Mean Value Theorem,

[tex]=2\epsilon f(\theta)\frac{3}{4\epsilon^{3}}(\epsilon^{2}-\theta^{2})\\=f(\theta)\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})[/tex]

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 [itex]\epsilon>0[/itex] is arbitrary let [itex]\epsilon=\sqrt{3}\vert\theta\vert[/itex] ([itex]-\epsilon\leq\theta\leq\epsilon[/itex] is still satisfied), then

[tex]\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[/tex]

so,

[tex]I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)\text{,}\quad \epsilon=\sqrt{3}\vert\theta\vert\text{.}[/tex]

The problem I have now is the [itex]\theta[/itex] we know exists by MVT is dependent on [itex]\epsilon[/itex] we choose. Is it then valid to set [itex]\epsilon[/itex] as a function of [itex]\theta[/itex]?

Thanks in advance.

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