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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.