1. ## bounded operator

I have been set the following question.

Show that the operator L[f](x) = INT(0,x) f(s) dx , 0<= x <=1 , is bounded from C0(0,1) into itself.

where INT(0,x) is the integral between 0 and x, and C0(0,1) is the space of continuous functions with no continuous derivatives on (0,1).

I know that to show this I must show there exists c>0 such that
llLxll <= cllxll for all x.

However I am having trouble actually showing this exists. any ideas?

2. Originally Posted by johnbarkwith
I have been set the following question.

Show that the operator L[f](x) = INT(0,x) f(s) dx , 0<= x <=1 , is bounded from C0(0,1) into itself.

where INT(0,x) is the integral between 0 and x, and C0(0,1) is the space of continuous functions with no continuous derivatives on (0,1).

I know that to show this I must show there exists c>0 such that
llLxll <= cllxll for all x.

However I am having trouble actually showing this exists. any ideas?
To start with, you need to be clear about what the space C0(0,1) is, and what the norm on it is. I think you probably mean the space that I would call C[0,1], which consists of all continuous functions on the closed interval [0,1]. The norm of a function f in this space is $\displaystyle \|f\| = \max\{|f(x)|:0\leqslant x\leqslant1\}$, the maximum value of |f| on the interval.

The operator L takes a function to its integral: L(f) is the function whose value at x is $\displaystyle \int_0^x\!\!\!f(s)\,ds$. What you need to do is to see whether $\displaystyle \|L(f)\|$ is less than some multiple of $\displaystyle \|f\|$. So you want to estimate the maximum value of $\displaystyle \Bigl|\int_0^x\!\!\!f(s)\,ds\Bigr|$. This must be less then or equal to $\displaystyle \int_0^x\!\!\!|f(s)|\,ds$. But $\displaystyle |f(s)|\leqslant\|f\|$ for all s, by the definition of $\displaystyle \|f\|$.

Now see if you can take it from there and show that $\displaystyle \|L(f)\|\leqslant c\|f\|$ for some constant c.

3. You surely do NOT mean that "C0(0,1) is the space of continuous functions with no continuous derivatives on (0,1)"! functions in C0(0,1) are not required to be differentiable but they surely are not required not to! They you have it defined, C0(0,1) would not be a space because it is not closed under addition.

If F(x) = $\displaystyle \int_0^x f(t)dt$, the F'(x)= f(x) and F(0)= 0.

Apply the mean value theorem to [0,x]: [F(x)-F(0)]/x= F(x)/x= f'(c) where c is between 0 and x. Since f is continuous, it is bounded.

4. Originally Posted by HallsofIvy
You surely do NOT mean that "C0(0,1) is the space of continuous functions with no continuous derivatives on (0,1)"! functions in C0(0,1) are not required to be differentiable but they surely are not required not to! They you have it defined, C0(0,1) would not be a space because it is not closed under addition.
I was at a research seminar yesterday where the speaker referred to $\displaystyle C^{\,k}[0,1]$ as "the space of functions on the unit interval with k continuous derivatives". Of course, everyone in the audience understood him to mean "... at least k continuous derivatives". But somehow the English language doesn't admit that interpretation in the case k=0.

5. any ideas how i could further show that it is bounded from C0(0,1) to L2(0,1)...

where L2 is lebesgue space.