Let T : $L^{3}$ [0,1] $\to$ R

T(f) = $\int_{0}^1$ $t^{2}$ f(t) dt

1- Show that T is continuos linear functional

2- Find the norm of T

My solution :

1- first I proved that $t^{2}$ $\in$ $L^{3/2}$

Now we have f $\in$ $L^{3}$ and $t^{2}$ $\in$ $L^{3/2}$

And since p, q are conjugate then T is bounded linear functional Then so it's continuous

Is it true?

2- I know the norm of T is : || T || = sup || Tt || where || t || = 1

But how can I apply on this?

Thanks a lot.


$\|T\| \leq 4^{-2/3}$ because $|\int f(t)t^{2}\, dt| \leq (\int |f|^{3})^{1/3} (\int t^{3})^{2/3}=4^{-2/3} \|f\|_3$. Now take $f(t)=4^{1/3}t$ and verify that $\|f\|_3=1$ and that $Tf=4^{-2/3}$. Hence $\|T\|=4^{-2/3}$.

  • $\begingroup$ Thank you very much.. But how the " <= " became “ = “ ? $\endgroup$ – Duaa Hamzeh Dec 18 '18 at 10:30
  • 1
    $\begingroup$ By definition of operator norm, $\|T|| \geq \|f\|$ if $\|f\|=1$. I have given a specific $f$ such that $\|f\|=1$ and $Tf=4^{-2/3}$. $\endgroup$ – Kavi Rama Murthy Dec 18 '18 at 10:33
  • $\begingroup$ For the record I meant $\|T\| \geq |Tf|$ if $\|f\|=1$. $\endgroup$ – Kavi Rama Murthy Dec 18 '18 at 12:02
  • $\begingroup$ It's help me very much.. Grateful 🌸 $\endgroup$ – Duaa Hamzeh Dec 18 '18 at 12:15
  • $\begingroup$ Also.. I have seen a theorem now from Royden say that || T || = ||$t^{2}$ || on $L^{3/2}$ space since it is bounded $\endgroup$ – Duaa Hamzeh Dec 18 '18 at 12:19

With Hölder we get

$|T(f)| \le (\int_0^1t^3 dt)^{2/3} (\int_0^1 |f(t)|^3)^{1/3} = c ||f||_3$,

where $c:= (\int_0^1t^3 dt)^{2/3}$.

This shows that $T$ is continuous.

It is your turn to determine $||T||$.


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.