Is the Functional C_n on Polynomials with Supremum Norm Discontinuous?

  • Thread starter P3X-018
  • Start date
  • Tags
    Functional
In summary: So Cn is discontinuous.In summary, the functional C_n on the space of polynomials on the interval [0,1] that takes the n'th coefficient is discontinuous with respect to the supremum norm. This can be shown using the fact that a functional is continuous if and only if the supremum of its values on the unit ball is finite. By considering the polynomial p_k = (1-t)^k and using the binomial formula, it can be shown that the supremum of |p_k(t)| in the interval is attained when t = 0, resulting in \|p_k\|_{\infty} = \frac{1}{bin(k,n)}. However, C_n(p_k) = (-1
  • #1
P3X-018
144
0

Homework Statement



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

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

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

Homework Equations



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

[tex] \sup_{\|x\|\leq 1} |F(x)| < \infty [/tex]


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

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

bin(k,n) is the binomial coefficient.

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

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

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

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

but then I get

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

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?
 
Physics news on Phys.org
  • #2
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?
 
  • #3
That actually sounds reasonable. But then my calculation of the functional norm ie

[tex] \sup_{\|p_k\|\leq 1} | C_n(p_k) | [/tex]

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.
 
  • #4
P3X-018 said:
That actually sounds reasonable. But then my calculation of the functional norm ie

[tex] \sup_{\|p_k\|\leq 1} | C_n(p_k) | [/tex]

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).
 

1. What is a discontinuous functional?

A discontinuous functional is a mathematical function that is not continuous at one or more points in its domain. This means that there are abrupt changes or breaks in the function's graph, where the value of the function suddenly jumps or is undefined.

2. What causes a function to be discontinuous?

There are several possible causes for a function to be discontinuous, including: a removable discontinuity, where the function has a hole or gap in its graph that can be filled to make the function continuous; a jump discontinuity, where the left and right limits of the function at a point are different; and an infinite discontinuity, where the function approaches positive or negative infinity at a point.

3. How do you determine if a function is discontinuous?

To determine if a function is discontinuous, you can graph the function and look for any abrupt changes or breaks in the graph. You can also check for any points where the function is undefined or where the left and right limits are different. Additionally, you can use the formal definition of continuity which states that a function is continuous if the limit of the function at a point is equal to the value of the function at that point.

4. Can a discontinuous function still be useful?

Yes, a discontinuous function can still be useful in certain situations. For example, a discontinuous function may be used to model real-world phenomena that have sudden changes or jumps, such as the population of a species or the stock market. Discontinuous functions are also commonly used in physics and engineering to describe systems with sudden changes or breaks in behavior.

5. How can we deal with discontinuities in a function?

There are several techniques that can be used to deal with discontinuities in a function, depending on the type of discontinuity. For removable discontinuities, the function can be modified or redefined at the point of discontinuity to make it continuous. For jump discontinuities, the function can be split into two separate functions that approach the same value from each side of the discontinuity. And for infinite discontinuities, limits can be used to describe the behavior of the function near the point of discontinuity.

Similar threads

Replies
1
Views
564
  • Calculus and Beyond Homework Help
Replies
3
Views
397
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
542
  • Calculus and Beyond Homework Help
Replies
1
Views
585
Replies
4
Views
349
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
2
Views
1K
Back
Top