# A discontinuous Functional

1. Jan 8, 2008

### P3X-018

1. The problem statement, all variables and given/known data

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 $\|p\|_{\infty} = \sup_{t\in[0,1]}|p(t)|$.

2. Relevant equations

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

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

3. 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?

2. Jan 8, 2008

### Dick

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. Jan 8, 2008

### P3X-018

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.

4. Jan 8, 2008

### Dick

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).