Have they solved this exercise correct?(banach space, function space).

Please look at this exercise: It is the last question I have a problem with

Here is the 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?

• Do you realize that it's difficult to read the images? May 29, 2014 at 7:38
• @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 May 29, 2014 at 7:41
• @user119615: The image is too small to read even in the link. Can you somehow made it bigger?
– user99914
May 29, 2014 at 8:10
• @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 May 29, 2014 at 8:22

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 $M = \sup |g(t)|$.
• 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? May 29, 2014 at 8:37
• 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 $||I|| \leq \int_0^1 |g(t)|\,dt$. May 29, 2014 at 9:45