Let $C([0,1],\mathbb{R})$ be the vector space of continuous real-valued functions in the unit interval $[0,1]$. The norm in that space is given as the following integral: $\|f\|=\int_0^1 |f(x)|\,dx$.
A function $I: C([0,1],\mathbb{R})\to\mathbb{R}$ is defined as:
$I(f)=\int_0^1 f(x)\,dx$.
I need to find the operator norm for $I$.
So far I have reached the following:
$\|I(f)\| = \int_0^1 \left|\int_0^1 f(x)\,dx\right|\,dx \leq \int_0^1 |f(x)|\,dx$.
The part $\int_0^1 |f(x)|\,dx$ is $\|f\|$. So we have that $\|I(f)\|\leq\|f\|$. The operator norm can be found as the infimum of $k$ such that: $\|I(f)\|\leq k\|f\|$.
But I cant really work further… someone who can help
Best Answer
You have proved that if $\|f\|=1$, then $|I(f)|\le 1$. Now try to find a function $f$ with $\|f\|=1$ and $|I(f)|=1.$ Is there such a function?
I have in mind the defition of the operator norm: $$\|I\|=\sup\{|I(f)|:\|f\|=1\}.$$
If there exists $f$, which realizes this supremum (i.e. if $\sup=\max$), then $\|I\|=I(f)$. Sometimes there are no $f$ fulfilling such a condition.