Prove This Is Discontinuous At Zero

Nov 2009
59
0
Pocatello, ID
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
Nov 2009
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Berkeley, California
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.
 
Nov 2009
59
0
Pocatello, ID
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
Nov 2009
4,563
1,566
Berkeley, California
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?
 
Nov 2009
59
0
Pocatello, ID
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 :-(
 
Nov 2009
485
184
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.
 

Drexel28

MHF Hall of Honor
Nov 2009
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Berkeley, California
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?