# Show that T is continuos linear functional and find the norm of T

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.

## 2 Answers

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

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