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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

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    HINT: Is there a concrete function for wich the equality holds ($\|I(f)\|=\|f\|$)? – 2017-01-19
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    It holds for zero ? – 2017-01-19
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    Any other than zero? By the way you calculated $|I(f)|$ wrong. $I(f)$ is an element of $\mathbb R$. It happens to be the same result but that is a coincidence. – 2017-01-19
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    Where is the error ? – 2017-01-19
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    You don't need an additional integral in $\| I(f) \|$ as $\| I(f) \| = | I(f) |$ . – 2017-01-19

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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.

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    I dont really get why i have showed that ?.. I mean, i did it generally ? – 2017-01-19
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    You demonstrated that $$|I(f)|\le\int_0^1 |f(t)|\text{d}t.$$ Indeed, now take any $f$ with $\|f\|=1$. – 2017-01-19
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    That could be f(x)=1..? And what can I use that for ?.. I think i am bit confused... – 2017-01-19
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    Yes, of course, $f\equiv 1$ realizes this supremum, so $\|I\|=1$. – 2017-01-19
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    I really appreciate your help... But I just want to understand it 100%. Where does supremum have a role ? – 2017-01-19
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    You have proved that $\|I\|\le 1$. Because $|I(1)|=1$, then the supremum is attained. So, $\|I\|=1$. – 2017-01-19