# Prove This Is Discontinuous At Zero

#### mathematicalbagpiper

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

#### Drexel28

MHF Hall of Honor
Let E be a normed vector space. Let $$\displaystyle A:E\longrightarrow E$$ be defined by $$\displaystyle A(p(x))=p'(x)$$, the derivative of $$\displaystyle p(x)$$. Show that the $$\displaystyle \epsilon, \delta$$ definition of a continuous function fails at zero.
This makes no sense. I assume that $$\displaystyle E$$ is some kind of polynomial space. What kind of norm? The sup norm? You need be much more specific.

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

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

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

I hope that's enough.

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle E=\mathbb{R}[x]$$.

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

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

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

#### mathematicalbagpiper

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

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

#### roninpro

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

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

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

• mathematicalbagpiper

#### Drexel28

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

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

• TheCoffeeMachine and mathematicalbagpiper

#### mathematicalbagpiper

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