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

Question

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?

Answer 1

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 $M = \sup |g(t)|$.