Norm of functional on $L^4[0, 1]$

Question

I am trying to calculate the norm of the operator $$ \begin{align} f: L^4[0, 1] &\rightarrow \mathbb{R} \\ x &\mapsto \int_0^1 t^3x(t) dt \end{align} $$ I started off by estimating $$ ||fx|| = \left| \int_0^1 t^3x(t) dt \right| \le \int_0^1 |t^3x(t)| dt \stackrel{Hölder}{\le} \left( \int_0^1 t^{12} dt \right)^{\frac{1}{4}} ||x|| \le \frac{1}{\sqrt[4]{13}} ||x|| $$ So therefore I know that $||f|| \le \frac{1}{\sqrt[4]{13}}$. Now I need to find some $x \in L^4[0, 1]$ such that $$ ||f|| \ge \frac{||fx||}{||x||} = \frac{1}{\sqrt[4]{13}} $$ But I can't find any. Am I overseeing something simple?

Answer 1

Applying Holder's inequality with $p=4$ and $q=\frac 4 3$ we se that $|f(x)| \leq 5^{-3/4} \|x||$. Hence the norm is at most equal to $5^{-3/4}$. To see that equality holds just take $x(t)=t$. I will let you verify that $\frac {|f(x)|} {\|x\|} =5^{-3/4}$ in this case.

Note: The choice of $x(t)$ is dictated by the condition for equality in Holder's inequality.