# Thread: Prove This Is Discontinuous At Zero

1. ## Prove This Is Discontinuous At Zero

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.

2. 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.

3. 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.

4. 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?

5. 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 :-(

6. 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.

7. 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?

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