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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 $||Tf||\geq ||f||$?

Mihai.Mehe
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2 Answers2

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

Just a user
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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 $||T||\leq 1$. – Mihai.Mehe Nov 16 '21 at 07:33
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    @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