# Bounded linear operator

1. Nov 22, 2008

### dirk_mec1

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

http://img389.imageshack.us/img389/9272/33055553mf5.png [Broken]

3. The attempt at a solution

Via induction: for n=1 equality holds now assume that Vn=Jn.
I introduce a dummy variable b and the fundamental theorem of calculus and change order of integration:

$$V_{n+1}f(t) = \int_a^t \frac{(t-s)^n}{n!} f(s) \mbox{d}s = \int_a^t \frac{t-s}{n} \frac{(t-s)^{n-1} } {(n-1)!} \mbox{d}s = \int_a^t \int_s^t \frac{1}{n} \frac{(t-s)^{n-1} } {(n-1)!} f(s) \mbox{d}b\ \mbox{d}s = \int_a^t \frac{1}{n} \int_a^b \frac{(t-s)^{n-1} } {(n-1)!} f(s) \mbox{d}s\ \mbox{d}b = \int_a^t \frac{1}{n} J^n f(b) \mbox{d}b$$

$$= \frac{1}{n} J(Jf) = \frac{1}{n} J^{n+1}f$$

But how do I get rid of the 1/n?

Last edited by a moderator: May 3, 2017
2. Nov 22, 2008

### Palindrom

First of all I would use a different "dummy" variable as $$b$$ is the right end of the interval. Let's us $$x$$.

For the matter at hand, your fifth equality is wrong: you can't use the induction hypothesis here, since in order for it to work you have to have $$\left(x-s\right)^{n-1}$$ in the numerator and not $$\left(t-s\right)^{n-1}$$.

I would take a slightly different approach (although induction is obviously the way to go).

3. Nov 22, 2008

### fikus

you should use integration by parts to integrate
$$V_{n+1}f(t) = \int_a^t \frac{(t-s)^n}{n!} f(s) \mbox{d}s$$
try with
$$\frac{(t-s)^n}{n!} = u , f(s)= dv$$
It works for me.

4. Nov 22, 2008

### dirk_mec1

Yes you're right, I should have used for example x. I still don't see the mistake it shouldn't read b-a it should read t-s.

Which is?

Yes that worked, thanks!

5. Nov 23, 2008

### Palindrom

I don't see that I need to convince you anymore seeing as you solved it, but the induction hypothesis reads

$$J^{n}f\left(t\right)=\int_{a}^{t}\frac{\left(t-s\right)^{n-1}}{\left(n-1\right)!}f\left(s\right)ds$$

Or, since the argument is now x and not t,

$$J^{n}f\left(x\right)=\int_{a}^{x}\frac{\left(x-s\right)^{n-1}}{\left(n-1\right)!}f\left(s\right)ds$$

Right?

6. Nov 23, 2008

### dirk_mec1

Yes you're right I see my mistake now thanks.

So I found this:

$$V_{n+1}f = \int_a^t \frac{(t-s)^n}{n!}\ f(s)\ \mbox{d}s = \left[ Jf(s) \frac{(t-s)^n}{n!} \right] _a^t + \int_a^t \frac{(t-s)^{n-1}}{(n-1)!)}\ Jf(s) \mbox{d}s = V_n(Jf) = J^n (Jf) =J^{n+1}f$$

Last edited: Nov 23, 2008
7. Nov 23, 2008

### dirk_mec1

Last edited by a moderator: May 3, 2017
8. Nov 23, 2008

### Pere Callahan

Show that for any function f mit sup norm 1, you have $||J^nf||\leq||J^n\underline 1||$ where $\underline 1$ is the constant one function. Calculate $||J^n\underline 1||$ and conclude!

9. Nov 23, 2008

### dirk_mec1

Suppose $||J^nf||\leq||J^n\underline 1||$ then how can I calculate $||J^n\underline 1||$?

Is it this:

$$\int_a^b \int_a^t \left| \frac{(t-s)^{n-1}}{(n-1)!}\ \right| \mbox{d}s\ \mbox{d}t$$

10. Nov 23, 2008

### Pere Callahan

What is the norm you are using? Normally, C([a,b]) is endowed with the sup norm. Are you using the L_1 norm?

11. Nov 23, 2008

### dirk_mec1

Yes I'm using the L_1 norm but what do you mean by 'C([a,b]) is endowed by the sup norm' that's only valid if we use the infinity norm, right?

So you're saying that is should be:

$$\left| \left| \int_a^t \frac{(t-s)^{n-1}}{(n-1)!}\ \mbox{d}s\ \right| \right| _{\infty}$$

12. Nov 23, 2008

### Palindrom

In your assignment it says "calculate the norm in the Banach space...". The space C[a,b] with the L^1 norm is not a Banach space, and so if you take this norm the space BL(C[a,b]) is not a Banach space either. This should imply that they meant the supremum norm (and in fact, like Pere correctly noted, unless explicitly said otherwise the space C[a,b] is taken with the supremum norm).

13. Nov 23, 2008

### dirk_mec1

So the expression below is wrong?

$$\left| \left| \int_a^t \frac{(t-s)^{n-1}}{(n-1)!}\ \mbox{d}s\ \right| \right| _{\infty}$$

Then what should it be?

Last edited: Nov 23, 2008
14. Nov 23, 2008

### Pere Callahan

This is correct; you just need to evaluate the integral and take the $\sup_{t\in[a,b]}$

15. Nov 23, 2008

### dirk_mec1

Indeed the answer is deduced from this expression but how do you prove that:

$||J^nf||\leq||J^n\underline 1||$ ?

16. Nov 23, 2008

### Pere Callahan

$$||J^nf||=\sup_{t\in[a,b]}\left|\int_a^t{ds\frac{(t-s)^{n-1}}{(n-1)!}f(s}\right| \leq\int_a^b{ds\frac{(t-s)^{n-1}}{(n-1)!}|f(s)|}$$
For any f with ||f||<=1, you can bound ||f(s)|| by 1, and you get what I wrote.