Prove This Is Discontinuous At Zero

**mathematicalbagpiper** · May 23rd 2010, 07:02 PM

Let E be a normed vector space. Let $A:E\longrightarrow E$ be defined by $A(p(x))=p'(x)$ , the derivative of $p(x)$ . Show that the $\epsilon, \delta$ definition of a continuous function fails at zero.

**Drexel28** · May 23rd 2010, 08:06 PM

Originally Posted by mathematicalbagpiper

Let E be a normed vector space. Let $A:E\longrightarrow E$ be defined by $A(p(x))=p'(x)$ , the derivative of $p(x)$ . Show that the $\epsilon, \delta$ definition of a continuous function fails at zero.

This makes no sense. I assume that $E$ is some kind of polynomial space. What kind of norm? The sup norm? You need be much more specific.

**mathematicalbagpiper** · May 23rd 2010, 08:15 PM

I just looked up the reference in the textbook (he references an earlier remark in the book). Here's a little more relevant information:

Let $E=\mathbb{R}[x]$ .

So $p(x)\in\mathbb{R}[x]$ .

Further, the norm on this space he defined as $||p||=max\{|a_0|,......,|a_n|\}$

I hope that's enough.

**Drexel28** · May 23rd 2010, 08:27 PM

Originally Posted by mathematicalbagpiper

I just looked up the reference in the textbook (he references an earlier remark in the book). Here's a little more relevant information:

Let $E=\mathbb{R}[x]$ .

So $p(x)\in\mathbb{R}[x]$ .

Further, the norm on this space he defined as $||p||=max\{|a_0|,......,|a_n|\}$

I hope that's enough.

Ok, so what have you tried? So when you say zero I assume you mean $0(x)=0+0+\cdots$ , right? So, you need to find some $\varepsilon>0$ such that given any $\delta>0$ there is some $p(x)\in\mathbb{R}[x]$ such that $\|p(x)\|<\delta$ but $\|p'(x)\|\geqslant \varepsilon$ . So what do you think?

**mathematicalbagpiper** · May 24th 2010, 04:54 PM

Honestly, I haven't a bloody clue, and just out of curiosity, why is it just that $||p(x)||<\delta$ but $||p'(x)||\geq\epsilon$ instead of $||p(x)-p(a)||<\delta$ but $||p'(x)-p'(a)||\geq\epsilon$ ?

Sorry, I know I should know this, but I barely scraped by analysis in undergrad :-(

**roninpro** · May 24th 2010, 04:59 PM

You should try writing down the definition of continuity for your map $A$ :

Given $\varepsilon>0$ there is $\delta>0$ so that $||p(x)-0||<\delta$ implies $||A(p(x))-A(0)||<\varepsilon$ .

When you clean this up, you get what Drexel suggested.

**Drexel28** · May 24th 2010, 05:01 PM

Originally Posted by mathematicalbagpiper

Honestly, I haven't a bloody clue, and just out of curiosity, why is it just that $||p(x)||<\delta$ but $||p'(x)||\geq\epsilon$ instead of $||p(x)-p(a)||<\delta$ but $||p'(x)-p'(a)||\geq\epsilon$ ?

Sorry, I know I should know this, but I barely scraped by analysis in undergrad :-(

Forget $\varepsilon$ , we can do this for $1$ . So ,if $A:\mathbb{R}[x]\to\mathbb{R}[x]$ were continuous at $0(x)$ then we should be able to find some $\delta>0$ such that $\|p(x)-0(x)\|=\|p(x)\|<\delta\implies \|p'(x)-0'(x)\|=\|p'(x)\|<1$ , right? But, for example if you claimed to me that you have found a $2\delta_0$ which works I can show you that you're wrong. To see this we know by the Archimedean principle I can find some $N\in\mathbb{N}$ such that $N\delta_0>1$ and $N\geqslant2$ , right? So, consider $p(x)=\delta_0x^N$ then clearly $\|p(x)\|=\max\{\delta_0\}=\delta_0<2\delta_0$ but $p'(x)=N\delta_0x^{N-1}$ , and $\|p'(x)\|=\|N\delta_0x^{N-1}\|=N\delta_0>1$ and so the $\delta=2\delta_0$ you claimed work didn't, and so you can't find any $\delta$ which works. So, it isn't continuous. Make sense?

**mathematicalbagpiper** · May 24th 2010, 05:01 PM

Now that makes sense. I think I have an idea where to go now.

Thread: Prove This Is Discontinuous At Zero

LinkBack

Thread Tools

Display

Prove This Is Discontinuous At Zero

Similar Math Help Forum Discussions

Where the function is discontinuous.?

function discontinuous only on (0,1)?

Discontinuous Functions

Discontinuous limits

Prove Function is Discontinuous Using Sequences

Search Tags