Compute norm of linear functional

Question

I was trying to calculate norm of functional $F: L^p([0;1]) \to {\rm I\!R}$, given by formula:
$$
F(f) = \int_{[0;1]} x f(x) \ d\lambda(x),
$$
for $p \in [1; \infty)$ and $\lambda$ – Lebesgue measure on $[0;1]$.
I have seen the solution for $p = 1$ here, but I am not sure if for $p > 1$ it can be done the same way.
If we take $f \in L^p([0;1])$, such that $\Vert f \Vert_{L^p} = 1$, we can apply Hölder inequality:
$$
|F(f)| \le \int_{[0;1]} x |f(x)| \ d\lambda(x) \le \int_{[0;1]} |f(x)| \ d\lambda(x) \le \Big(\int_{[0;1]} |f(x)|^p \ d\lambda(x)\Big)^{\frac{1}{p}}\Big(\int_{[0;1]} |1|^{1 – \frac{1}{p}} \ d\lambda(x)\Big)^{1 – \frac{1}{p}} = \Vert f \Vert_{L^p} \cdot \lambda([0;1]) = 1.
$$
So $|F(f)| \le 1$.
Unfortunatelly I fail at showing that $|F(f)| \ge 1$. I was trying to construct similar sequence as here for every $p > 1$. How to show second inequality? Thank you in advance for any tips.

Answer 1

It is a bad idea to use $x \leq 1$ in getting an upper bound for $\|F\|$ . Let $q$ be the index conjugate to $p$ (i.e. $q=\frac p {p-1}$). Then $|F(f)| \leq \|f\|_p (\int x^{q})^{1/q}$ by Holder's inequality. Hence $\|F\| \leq (\int_o^{1} x^{q}dx)^{1/q}=\frac 1 {(q+1)^{1/q}}$. This is actually an equality. To see this you have to use the condition for equality in Holder's inequality. Recall that for $f,g \geq 0$ the conidtion for $\int fg =(\int f^{p})^{1/p} (\int g^{q})^{1/q}$ is $f^{p}= cg^{q}$. So take $f(x)=x^{q/p}$. Compute $\|f\|$ and check that $|F(\frac f {\|f\|})| =\frac 1 {(q+1)^{1/q}}$. This proves that $\|F\| =\frac 1 {(q+1)^{1/q}}$.

PS

This is precisely the argument used to prove that the dual of $L^{p}$ is $L^{q}$. So we have the following general fact: If we define $F(f)=\int_0^{1} f(x)g(x)d\lambda (x)$ for all $f \in L^{p}$ then the norm of $F$ is equal to $(\int_0^{1}|g(x)|^{q}d\lambda(x))^{1/q}$.