MHB Show that limit is equal to zero

[evinda]

Hello! (Wave)Suppose that $B_{\epsilon}=\{ |x- \xi|< \epsilon\}$ and $E(|x- \xi|)=\left\{\begin{matrix}
\frac{1}{(2-n)w_n}|x-\xi|^{2-n} &, n \geq 3 \\ \\
\frac{1}{2 \pi} \ln{|x-\xi|} &, n=2
\end{matrix}\right.$

So when $ |x- \xi|=\epsilon $ then

$$E(|x- \xi|)=\left\{\begin{matrix}
\frac{1}{(2-n)w_n}\epsilon^{2-n} &, n \geq 3 \\ \\
\frac{1}{2 \pi} \ln{\epsilon} &, n=2
\end{matrix}\right.$$I want to show that $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x- \xi|) f(\xi)ds=0$ for any continuous and bounded $f$.

There is a hint that we should check the integral $\int_{|x|=\epsilon} E(|x|) ds=0$ by using spherical $(n \geq 3)$ or polar $(n=2)$ coordinates.So, suppose that $n \geq 3$.

Then I thought the following:

$\int_{|x|=\epsilon} E(|x|) ds=\int_{|x|=\epsilon} \frac{1}{(2-n) w_n} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} 2 \pi \epsilon^{n-1}=\frac{2 \pi \epsilon}{(2-n) w_n} \to 0 \text{ while } \epsilon \to 0$.

But I didn't use spherical coordinates. Have I done something wrong? (Thinking)

 

[I like Serena]

So, suppose that $n \geq 3$.

Then I thought the following:

$\int_{|x|=\epsilon} E(|x|) ds=\int_{|x|=\epsilon} \frac{1}{(2-n) w_n} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} 2 \pi \epsilon^{n-1}=\frac{2 \pi \epsilon}{(2-n) w_n} \to 0 \text{ while } \epsilon \to 0$.

But I didn't use spherical coordinates. Have I done something wrong? (Thinking)

Hey evinda!

Well... there is something wrong...

It should be:
$$\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} w_n \epsilon^{n-1}$$
Otherwise I think it's all good! (Happy)

 

[evinda]

I see... So like that we show that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$.

How do we use this in order to calculate the limit $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x-\xi|) f(\xi) ds$ ?

 

[I like Serena]

I see... So like that we show that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$.

How do we use this in order to calculate the limit $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x-\xi|) f(\xi) ds$ ?

We have:
$$0 \le \left| \int_{\partial{B_{\epsilon}(x)}} E(|x-\xi|) f(\xi) ds \right| \le \sup_{\xi\in \partial B_\epsilon(x)} |E(|x-\xi|) | \cdot \sup_{\xi\in \partial B_\epsilon(x)} |f(\xi)| \cdot \int_{\partial{B_{\epsilon}(x)}} ds
$$
don't we? (Wondering)

 

[evinda]

We have:
$$0 \le \left| \int_{\partial{B_{\epsilon}(x)}} E(|x-\xi|) f(\xi) ds \right| \le \sup_{\xi\in \partial B_\epsilon(x)} |E(|x-\xi|) | \cdot \sup_{\xi\in \partial B_\epsilon(x)} |f(\xi)| \cdot \int_{\partial{B_{\epsilon}(x)}} ds
$$
don't we? (Wondering)

Yes. So don't we use the fact that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$ ?

When $\xi \in \partial{B_{\epsilon}(x)}$ do we have that $|x-\xi|=\epsilon$ ?

If so then $\sup_{\xi \in \partial{B_{\epsilon}(x)}} |E(|x-\xi|)|=\frac{1}{(2-n) w_n} \epsilon^{2-n}$ and $\int_{\partial{B_{\epsilon}(x)}} ds=w_n \epsilon^{n-1}$ . Right?

 

[I like Serena]

Yes. So don't we use the fact that $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} E(|x|) ds=0$ ?

When $\xi \in \partial{B_{\epsilon}(x)}$ do we have that $|x-\xi|=\epsilon$ ?

If so then $\sup_{\xi \in \partial{B_{\epsilon}(x)}} |E(|x-\xi|)|=\frac{1}{(2-n) w_n} \epsilon^{2-n}$ and $\int_{\partial{B_{\epsilon}(x)}} ds=w_n \epsilon^{n-1}$ . Right?

Right! (Nod)

And $f$ is given to be bounded, so $\displaystyle\sup_{\xi \in \partial{B_{\epsilon}(x)}} |f(\xi)| = M$ for some $M$.

 

[evinda]

Nice... Thank you! (Smile)