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Math Help - Prove This Is Discontinuous At Zero

  1. #1
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    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathematicalbagpiper View Post
    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.
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  3. #3
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    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.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathematicalbagpiper View Post
    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?
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  5. #5
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    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 :-(
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  6. #6
    Senior Member roninpro's Avatar
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    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.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathematicalbagpiper View Post
    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?
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  8. #8
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    Now that makes sense. I think I have an idea where to go now.
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