Is the Functional C_n on Polynomials with Supremum Norm Discontinuous?

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Homework Help Overview

The discussion revolves around the continuity of the functional \( C_n \) defined on the space of polynomials over the interval [0,1], specifically examining its behavior with respect to the supremum norm. The original poster is tasked with demonstrating that this functional is discontinuous.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the continuity of the functional by considering specific polynomial sequences and their supremum norms. They express uncertainty regarding their calculations and the implications of their findings.
  • Some participants question the validity of the original poster's calculations and suggest alternative reasoning based on limits and properties of the functional.
  • Others propose examining different polynomial forms to illustrate the unbounded nature of the functional as \( k \) approaches infinity.

Discussion Status

Contextual Notes

Participants note that the functional's continuity is tied to the behavior of the polynomial sequences as \( k \) increases, and there is a focus on the supremum norm's properties. The original poster expresses confusion about the necessity of demonstrating infinite behavior for discontinuity, indicating a potential gap in understanding the underlying concepts.

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Homework Statement



I have to show that the functional C_n on the space of polynomials on the interval [0,1], that takes the n'th coefficient ie

C_n\left( \sum_{j=0}^m a_j t^j \right) = a_n

is discontinuous with respect to the supremum norm \|p\|_{\infty} = \sup_{t\in[0,1]}|p(t)|.

Homework Equations



A functional F on a normed space is continuous if and only if

\sup_{\|x\|\leq 1} |F(x)| < \infty


The Attempt at a Solution



Our normed space in the problem is the polynomials with the supremum norm.

As a hint it says to consider

p_k(t) = \frac{(1-t)^k}{bin(k,n)}

bin(k,n) is the binomial coefficient.

The supremum of |p_k(t)| in the interval is attained when t = 0 so

\|p_k\|_{\infty} = \frac{1}{bin(k,n)}

and I get by the binomial formula for (1-t)^k, that

C_n(p_k) = (-1)^n \|p_k\|_{\infty} bin(k,n)

but then I get

\sup_{\|p_k\|\leq 1} | \|p_k\|_{\infty} bin(k,n) | = bin(k,n)

but I think the point with hint was that this was supposed to be infinite so that C_n would be shown to be discontinuous. What am I doing wrong?
 
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I don't think there is anything wrong. You have pk->0 in your supremum norm as k->infinity (at least for n>0). If Cn were continuous this should mean Cn(pk)->0. But you have |Cn(pk)|=1. That's enough to show Cn is discontinuous without any explicit infinities anyplace, right?
 
That actually sounds reasonable. But then my calculation of the functional norm ie

\sup_{\|p_k\|\leq 1} | C_n(p_k) |

is wrong, since this has to be infinite if C_n has to be discontinuous, though I can't see the mistake.
Maybe I'll just stick to argument that lim |C_n(p_k)| = 1 but lim p_k = 0, as you said. This shows |C_n| to be discont and therefore C_n is discont, because |.| is a cont map.
 
P3X-018 said:
That actually sounds reasonable. But then my calculation of the functional norm ie

\sup_{\|p_k\|\leq 1} | C_n(p_k) |

is wrong, since this has to be infinite if C_n has to be discontinuous, though I can't see the mistake.
Maybe I'll just stick to argument that lim |C_n(p_k)| = 1 but lim p_k = 0, as you said. This shows |C_n| to be discont and therefore C_n is discont, because |.| is a cont map.

You can do it that way as well. But then change pk to just be (1-t)^k. Now |pk|<=1. But Cn(pk)=bin(k,n) which is unbounded (for n>0).
 

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