Question concerning specific part of proof that the dual of $L^p$ is $L^q$ where $p$ and $q$ are Hölder conjugates and $p<2$.

Question

I want to show that if $p<2$ then the dual of $L^p([0,1])$ is $L^q([0,1])$ where $\frac{1}{p}+\frac{1}{q} = 1$ without referring to the fact that $L^p([0,1])$ is uniformly convex. My question concern a particular step in the proof. Since $2/p$ and $2/(2-p)$ are Hölder conjugates we see that

$$\|f\|_p^p \leq \|1\|_{2/(2-p)}\|f^p\|_{2/p} = \|f\|_2^p\Rightarrow L^2([0,1])\subseteq L^p([0,1]).$$

Now let $l$ be any bounded linear functional on $L^p([0,1])$ we want to show that we can identify with $l$ an element of $L^q([0,1])$. Since

$$|l(f)|\leq c\|f\|_p\leq c\|f\|_2$$

we find that $l\vert_{L^2}$ is a bounded linear functional on $L^2$ and therefore by Riesz representation theorem there exists some $u\in L^2([0,1])$ such that

$$l(f)=\int_{[0,1]}f(x)u(x)\ d\lambda(x).$$

Now I know how to show that $u$ in fact belongs to $L^q([0,1])$ what I don't know how to do is to show that this implies that $l$ is actually given by

$$l(f)=\int_{[0,1]}f(x)u(x)\ d\lambda(x)$$

for any $f$ which belongs to $L^p$. Suppose $l_1$ is the linear functional defined by

$$l_1(f)=\int_{[0,1]}f(x)u(x)\ d\lambda(x)$$

then certainly $l_1$ is bounded for every $f\in L^p([0,1])$ and $l-l_1$ vanishes on $L^2$ but can we say that $l-l_1$ is zero everywhere? In order for the proof to be complete I believe that this is what we want to show since then any linear functional $l$ over $L^p$ then can be associated with a $u\in L^q$ by a mapping which is an isometry and therefore the dual of $L^p$ is isometrically isomorphic to $L^q$ but I can't see how this should be true in general?

Answer 1

Let $f \in L^p$ and for each $n \ge 1$ define the truncation $$f_n(x) = \left\{ \begin{array}{cc} n & f(x) \ge n \\ f(x) & -n \le f(x) \le n \\ -n & f(x) \le -n \end{array} \right.$$ Then $f_n \in L^p \cap L^2$ for all $n$ so that $$l(f_n) = \int f_n(x) u(x) \, d \lambda (x)$$ for all $n$. Since $f_n \to f$ and $|f - f_n|^p \le |f|^p$, the dominated convergence theorem gives you $$\|f_n - f\|_p = \left( \int |f_n - f|^p \, d\lambda(x) \right)^{1/p} \to 0.$$ Since $l$ is bounded on $L^p$ you get $|l(f) - l(f_n)| = |l(f-f_n)| \le C \|f - f_n\|_p$ so that $$l(f_n) \to l(f).$$

On the other hand, Holder's inequality gives you $$\left| \int f_n(x) u(x) \, d \lambda (x) - \int f(x) u(x) \, d \lambda (x) \right| \le \int |f - f_n| |u(x)| \, dx \le \|f - f_n\|_p \|u\|_q $$ so that $$\int f_n(x) u(x) \, d \lambda (x) \to \int f(x) u(x) \, d \lambda (x).$$

Since limits are unique you conclude $$l(f) = \int f(x) u(x) \, d \lambda (x).$$

Answer 2

I have a suggestion for an answer but I'm very uncertain. Suppose that $f$ is positive and belongs to $L^p([0,1])$ and let $f_k$ be a sequence of simple functions such that $f_k\leq f_{k+1}$ and $\lim_kf_k(x) = f(x)$ for every $x\in [0,1]$ and $|f_k(x)|\leq \max\{f(x),k\}$ for every $x\in [0,1]$. Then $f_k\in L^2([0,1])$ and therefore

$$\left|\int_{[0,1]}f_k(x) u(x)\ d\lambda-l(f)\right|=|l(f_k)-l(f)|\leq c\|f-f_k\|_p$$

By Lebesue's dominated convergence theorem the right-hand side tends to zero and therefore

$$ 0 = \lim_{k\rightarrow \infty}\left|\int_{[0,1]}f_k(x)u(x)\ d\lambda(x)-l(f)\right| = \left|\int_{[0,1]}f(x)u(x)\ d\lambda(x)-l(f)\right|$$