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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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  • $\begingroup$ HINT: Is there a concrete function for wich the equality holds ($\|I(f)\|=\|f\|$)? $\endgroup$ – Tito Eliatron Jan 19 '17 at 16:40
  • $\begingroup$ It holds for zero ? $\endgroup$ – Elias S. Jan 19 '17 at 16:43
  • $\begingroup$ 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. $\endgroup$ – Tim B. Jan 19 '17 at 16:45
  • $\begingroup$ Where is the error ? $\endgroup$ – Elias S. Jan 19 '17 at 16:50
  • $\begingroup$ You don't need an additional integral in $\| I(f) \|$ as $\| I(f) \| = | I(f) |$ . $\endgroup$ – Nik Pronko Jan 19 '17 at 17:38
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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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  • $\begingroup$ I dont really get why i have showed that ?.. I mean, i did it generally ? $\endgroup$ – Elias S. Jan 19 '17 at 22:12
  • $\begingroup$ You demonstrated that $$|I(f)|\le\int_0^1 |f(t)|\text{d}t.$$ Indeed, now take any $f$ with $\|f\|=1$. $\endgroup$ – szw1710 Jan 19 '17 at 22:14
  • $\begingroup$ That could be f(x)=1..? And what can I use that for ?.. I think i am bit confused... $\endgroup$ – Elias S. Jan 19 '17 at 22:18
  • $\begingroup$ Yes, of course, $f\equiv 1$ realizes this supremum, so $\|I\|=1$. $\endgroup$ – szw1710 Jan 19 '17 at 22:20
  • $\begingroup$ I really appreciate your help... But I just want to understand it 100%. Where does supremum have a role ? $\endgroup$ – Elias S. Jan 19 '17 at 22:25

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