# Find the norm $||T||$, where $T f(x) = \int^ x_0f(s) ds$ and the norm $||f|| = max_{0≤x≤1}|f(x)|$. [duplicate]

Let $$C[0, 1]$$ be the linear space of all continuous functions on the interval $$[0, 1]$$ equipped with the norm $$||f|| = max_{0≤x≤1}|f(x)|$$. Define the operator $$T : C[0, 1] → C[0, 1]$$ by $$T f(x) = \int^ x_0f(s) ds$$. Show that $$T$$ is bounded and find its norm $$||T||$$.

Proof:

$$||Tf||=||\int_0^x f(s)ds||=max_{0≤x≤1}|\int_0^x f(s)ds|\leq max_{0≤x≤1}\int_0^x |f(s)|ds\leq\int_0^1 max_{0≤s≤1}|f(s)|ds = \int_0^1 ||f||ds=||f||$$

$$||T||=sup_{f\neq0}\frac{||Tf||}{||f||}=...$$

I am not sure if the part $$||Tf||\leq ||f||$$ if correct. If this is correct, how to pick $$f$$ to prove $$||Tf||\geq ||f||$$?

## 2 Answers

There are two details in the proof that are incorrect (but are easily fixed):

• In "$$\max_{0≤x≤1}|\int_0^x f(s)ds|=\max_{0≤x≤1}\int_0^x |f(s)|ds$$" you need to use $$\le$$ instead of $$=$$.
• On the other hand, $$\int_0^1 max_{0≤s≤1}|f(s)|ds\leq\int_0^1 ||f||$$ is correct, but there you can have equality, and one $$ds$$ is missing: $$\int_0^1 max_{0≤s≤1}|f(s)|ds=\int_0^1 ||f||ds$$ would be better.

For the example of $$||Tf||=||f||$$ with $$f\ne 0$$, take $$f(x)=1$$, with $$||f||=1$$. Then, $$(Tf)(x)=\int_0^x f(s)ds=\int_0^x ds=x$$ and so $$||Tf||=\max_{0\le x\le 1}|x|=1=||f||$$

• So for $f_1(x)=1$ $||T||=sup\frac{||Tf||}{||f||}\geq \frac{||Tf_1||}{||f_1||} = 1$. To prove $||T||=1$ I need the $||T||\leq 1$. Nov 16 '21 at 7:33
• @Mihai.Mehe Of course, but that would come out of what you have already proven (barring the two easily fixable points above), which is that $||Tf||\le||f||$, Nov 16 '21 at 13:26

Let $$f(s)=1$$, then $$T(f(s)) = \int_0^x ds = x$$ and $$\max_{x\in[0,1]} |x| = 1$$. Hence $$1=\|Tf(s)\|\le \|T\|\|f(s)\|=\|T\|$$