Real Analysis Folland, Proposition 6.13, The Dual of $L^p$

Question

Proposition 6.13 - Suppose $p$ and $q$ are conjugate exponents and $1\leq q < \infty$. If $g\in L^q$, then $$\|g\|_{q} = \|\phi_g\| = \sup\{\left|\int fg\right|: \|f\|_{p} = 1 \}$$ If $\mu$ is semifinite, this result also holds for $q = \infty$.

Part of Proof -

We know that for Holder's inequality that $\|\phi_g\| \leq \|g\|_{q}$, equality is trivial if $g = 0$ a.e. If $g\neq 0$ and $q < \infty$, let $$f = \frac{|g|^{q-1}\overline{sgn g}}{\|g\|_{q}^{q-1}}$$ Then, $$\|f\|_{p} = \frac{\int |g|^{(q-1)p}}{\|g\|_{q}^{(q-1)p}} = \frac{\int |g|^q}{\int |g|^q} = 1$$ So, $$\|\phi_g\| \geq \int fg $$

I don't understand why this is the case? Any insight or suggestions is greatly appreciated

Answer 1

Proposition 6.13 - Suppose $p$ and $q$ are conjugate exponents and $1\leq q < \infty$. If $g\in L^q$, then $$\|g\|_{q} = \|\phi_g\| = \sup\{\left|\int fg\right|: \|f\|_{p} = 1 \}$$ If $\mu$ is semifinite, this result also holds for $q = \infty$.

Proof -

Note that for $1\leq q \leq \infty$, we know, for Holder's inequality, that $\|\phi_g\| \leq \|g\|_{q}$, because for all $f \in L^p$ such that $\|f\|_p=1$, we have
$$\left|\int fg\right| \leq \|f\|_p\|g\|_{q}=\|g\|_{q}$$ So $$ \|\phi_g\| = \sup\left \{\left|\int fg\right|: \|f\|_{p} = 1 \right \} \leq \|g\|_{q} \tag{1} $$

Note that the equality is trivial if $\|g\|_{q} = 0$. Now, if $\|g\|_{q}\neq 0$ and $q < \infty$, let $$h = \frac{|g|^{q-1}\overline{\textrm{sgn} g}}{\|g\|_{q}^{q-1}}$$ Then, (and it is important) let us prove that such $h$ is in $L^p$ and $\|h\|_{p}=1$. In fact, we have $$\|h\|_{p}^p =\int | h |^p= \frac{\int |g|^{(q-1)p}}{\|g\|_{q}^{(q-1)p}} = \frac{\int |g|^q}{\int |g|^q} = 1$$

Note that $$\int hg = \frac{\int |g|^{(q-1)}(\overline{\textrm{sgn} g})g}{\|g\|_{q}^{(q-1)}} = \frac{\int |g|^{(q-1)}| g|}{\|g\|_{q}^{(q-1)}}= \frac{\int |g|^{q}}{\|g\|_{q}^{(q-1)}}= \frac{\|g\|_{q}^{q}}{\|g\|_{q}^{(q-1)}}=\|g\|_{q} $$

So, $$ \|g\|_{q} = \int hg = \left | \int hg \right | \leq \sup\left \{\left|\int fg\right|: \|f\|_{p} = 1 \right \} = \|\phi_g\|$$ So, for $q<\infty$, using $(1)$ we have $$\|g\|_{q} = \|\phi_g\| = \sup \left \{\left|\int fg\right|: \|f\|_{p} = 1 \right\}$$

Now suppose $q=\infty$ and $\mu$ is semifinite.

For any $\epsilon>0$, let $A =\{x: |g(x)| >\|g\|_\infty -\epsilon\}$. Then $\mu(A)>0$, so since $\mu$ is semifinite, there exists $B \subset A$ such that $0<\mu(B) <\infty$. Let $$h= \frac{\chi_B \overline{\textrm{sgn}g}}{\mu(B)}$$ Then $$\int |h| d\mu = \int \frac{\chi_B | \overline{\textrm{sgn}g} |}{\mu(B)}d\mu = \int \frac{\chi_B }{\mu(B)}d\mu =1$$ So $h\in L^1$ and $\|h\|_1=1$. And we have $$ \int h g d\mu = \frac{ \int\chi_B ( \overline{\textrm{sgn}g})g d\mu}{\mu(B)}= \frac{ \int\chi_B |g|d\mu}{\mu(B)}\geq \|g\|_\infty-\epsilon $$ So we have $$ \|g\|_\infty-\epsilon \leq \int h g d\mu \leq \left| \int h g d\mu \right| \leq \sup\left \{\left|\int fg\right|: \|f\|_{1} = 1 \right \} = \|\phi_g\|$$ So, using $(1)$ for $q=\infty$, we have $$\|g\|_{\infty} = \|\phi_g\| = \sup \left \{\left|\int fg\right|: \|f\|_{1} = 1 \right\}$$

Answer 2

In fact we always define $g^*$ as the conjugate function of $g \in L^q$ by \begin{equation*} g^* = \Vert g \Vert_q^{1-q} \cdot \textrm{sgn}g \cdot |g|^{q-1}. \end{equation*} It's easy to show that \begin{equation*} \int g\cdot g^* = \Vert g \Vert_q \text{ and } \Vert g^* \Vert_p = 1, \end{equation*} which shows that \begin{equation*} \sup\left\{\left|\int fg\right|:f \in L^p, \Vert f \Vert_p = 1\right\} \geqslant \Vert g \Vert_q. \end{equation*}