2
$\begingroup$

Please look at this exercise: It is the last question I have a problem with Banach space of continuous functions

Here is the solution:

Solution

They say that $\|I(1)\|=\sup|g(t)|$. But isn't $\|I(1)\|=\int_0^1g(t)dt$? If so, what is the correct answer?

$\endgroup$
  • $\begingroup$ Do you realize that it's difficult to read the images? $\endgroup$ – user21820 May 29 '14 at 7:38
  • 1
    $\begingroup$ @user21820 I understand, I have no problem reading it, I really don't understand why it may be viewed differently for you. Here are the links directly to the images:, can you see them clear here: ? image 1: s27.postimg.org/y8d4c57wz/image.png , image 2: s1.postimg.org/am6ii8anz/image.png $\endgroup$ – user119615 May 29 '14 at 7:41
  • 1
    $\begingroup$ @user119615: The image is too small to read even in the link. Can you somehow made it bigger? $\endgroup$ – user99914 May 29 '14 at 8:10
  • $\begingroup$ @John Does it work if you open them as .jpg? Try these links: s27.postimg.org/y8d4c57wz/image.jpg and s1.postimg.org/am6ii8anz/image.jpg , if that works maybe it is a problem with the .png images $\endgroup$ – user119615 May 29 '14 at 8:22
2
$\begingroup$

You are right. Since $g(t) \geq 0$, it follows that $I(1)$ is a non-decreasing function on $[0,1]$, and hence, $||I(1)|| = (I(1))(1) =\int_0^1 g(t) dt$. In fact, one can use the estimate $$ \int_{0}^1 |f(t)g(t)|\, dt \leq \int_0^1 \left(\sup|f(t)|\right)|g(t)|\, dt = ||f||\int_0^1 |g(t)|\, dt $$ to show that one can take $M = \int_0^1 |g(t)|\, dt$ for all $g \in C$, which is better than $M = \sup |g(t)|$.

$\endgroup$
  • $\begingroup$ Thanks, and it is also the fact that $\int_0^1g(t)dt$, is the least of all possible M's you can chose, making that the operator-norm? $\endgroup$ – user119615 May 29 '14 at 8:37
  • $\begingroup$ Yes. For $g$ positive, we have $\int_0^1 g(t)\, dt = ||I(1)|| \leq ||I||\cdot||1|| = ||I||$. On the other hand, the inequality in my answer shows that for all $g \in C$, we have $||I|| \leq \int_0^1 |g(t)|\,dt$. $\endgroup$ – ivanpenev May 29 '14 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.