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discontinuity of the derivative map

oblixps

Member
May 20, 2012
38
Let Z be the set of all polynomials in C([0,1]) with real coefficients. Let [tex] D: Z \rightarrow Z [/tex] be the derivative map where D(p) = p' and let [tex] ||p||_u = sup_{0 \leq t \leq 1}|p(t)| [/tex]. Show that D is discontinous at the zero polynomial 0. Furthermore, show that D is discontinuous at every point p in Z.

i am having trouble working through this. from the definitions, i know that for D to be continuous at 0, we need that ||D(p) - D(0)||u approaches 0 as ||p||u approaches 0. This is also equivalent to finding a sequence of polynomials pn in Z such that ||pn|| approaches 0 implies ||D(pn)|| approaches 0. So to show discontinuity, i need to find a sequence pn where ||pn|| approaches 0 but ||D(pn)|| does not. however i am having trouble coming up with such a sequence.

can someone give me some hints on how to approach this problem?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let Z be the set of all polynomials in C([0,1]) with real coefficients. Let [tex] D: Z \rightarrow Z [/tex] be the derivative map where D(p) = p' and let [tex] ||p||_u = sup_{0 \leq t \leq 1}|p(t)| [/tex]. Show that D is discontinous at the zero polynomial 0. Furthermore, show that D is discontinuous at every point p in Z.

i am having trouble working through this. from the definitions, i know that for D to be continuous at 0, we need that ||D(p) - D(0)||u approaches 0 as ||p||u approaches 0. This is also equivalent to finding a sequence of polynomials pn in Z such that ||pn|| approaches 0 implies ||D(pn)|| approaches 0. So to show discontinuity, i need to find a sequence pn where ||pn|| approaches 0 but ||D(pn)|| does not. however i am having trouble coming up with such a sequence.

can someone give me some hints on how to approach this problem?
As you have correctly said, you need to find a sequence of polynomials that are small (uniformly over the unit interval), but their derivatives are not small. You might start by noticing that the function $x^n$ has derivative $nx^{n-1}$. The derivative has a factor $n$ in it, which makes it a lot bigger than the original function. But $x^n$ is not uniformly small on the unit interval (because it takes the value 1 when $x=1$). So can you shrink it by a factor that makes it small, but not so small as to make the derivative also small?
 

oblixps

Member
May 20, 2012
38
ah i see. if i pick [tex] p_n = \frac{1}{n}x^n [/tex], then ||pn|| = 1/n and ||D(pn)|| = 1 so ||pn|| gets very small while ||D(pn)|| does not.

I am now trying to apply similar reasoning to the 2nd part of the problem. Now for some fixed p0 i want to find a sequence such that ||pn - p0|| goes to 0 while ||D(pn) - D(p0)|| does not.

If p0 = [tex] \sum b_k t^k [/tex], then i tried taking the polynomial sequence [tex] p_n = \sum (b_k + \frac{1}{n})t^k [/tex] so that as n approaches infinite the difference approaches 0. However, when i differentiate this, i get [tex] D(p_n) = \sum (kb_k + \frac{k}{n}) t^{k-1} [/tex] so in the difference i am left with the sum over k of k/n which still goes to 0 as n approaches infinite. i haven't been able to think of another way to make the difference ||pn - p0|| go to 0.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
ah i see. if i pick [tex] p_n = \frac{1}{n}x^n [/tex], then ||pn|| = 1/n and ||D(pn)|| = 1 so ||pn|| gets very small while ||D(pn)|| does not.

I am now trying to apply similar reasoning to the 2nd part of the problem. Now for some fixed p0 i want to find a sequence such that ||pn - p0|| goes to 0 while ||D(pn) - D(p0)|| does not.

If p0 = [tex] \sum b_k t^k [/tex], then i tried taking the polynomial sequence [tex] p_n = \sum (b_k + \frac{1}{n})t^k [/tex] so that as n approaches infinite the difference approaches 0. However, when i differentiate this, i get [tex] D(p_n) = \sum (kb_k + \frac{k}{n}) t^{k-1} [/tex] so in the difference i am left with the sum over k of k/n which still goes to 0 as n approaches infinite. i haven't been able to think of another way to make the difference ||pn - p0|| go to 0.
I think you're trying to make this harder than it need be. You already have the sequence of monomials $\frac1nx^n$ that make the mapping discontinuous at 0. Given a polynomial $p_0$, just take $p_n(x) = p_0(x) + \frac1nx^n.$
 

oblixps

Member
May 20, 2012
38
yes it seems i was too "locked into" that mode of thinking that i wasn't able to come up with anything else. now that i look at it, it seems so obvious to make that choice for pn(x). thank you for taking the time to reply to my questions!