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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.

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$\|T\| \leq 4^{-2/3}$ 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 $\|T\|=4^{-2/3}$.

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  • $\begingroup$ Thank you very much.. But how the " <= " became “ = “ ? $\endgroup$ – Duaa Hamzeh Dec 18 '18 at 10:30
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    $\begingroup$ 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
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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 $||T||$.

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