proof of $L^p$-norm is dual to $L^q$

Let $(X,𝔐,\mu )$ be a $\sigma$-finite measure space and $p,q$ be Hölder conjugates. Then, we show that a measurable function $f:X\to ℝ$ has $L^p$-norm

$${\parallel f\parallel }_p=sup\{{\parallel fg\parallel }_1:g\in L^q,{\parallel g\parallel }_q=1\}.$$(1)

Furthermore, if either and or $p=1$ then $\mu$ is not required to be $\sigma$-finite.

If ${\parallel f\parallel }_p=0$ then $f$ is zero almost everywhere, and both sides of equality (1) are zero. So, we only need to consider the case where ${\parallel f\parallel }_p>0$.

Let $K$ be the right hand side of equality (1).For any $g\in L^q$ with ${\parallel g\parallel }_q=1$, the Hölder inequality gives ${\parallel fg\parallel }_1\le {\parallel f\parallel }_p$, so $K\le {\parallel f\parallel }_p$. Only the reverse inequality remains to be shown.

If and then, setting $g={|f|}^{p-1}$ gives

Therefore, $g\in L^q$ and,

$$K\ge {\parallel f(g/{\parallel g\parallel }_q)\parallel }_1={\parallel {|f|}^p\parallel }_1/{\parallel g\parallel }_q={\parallel f\parallel }_p^p/{\parallel f\parallel }_p^{p-1}={\parallel f\parallel }_p.$$

On the other hand, if $p=1$ so that $q=\mathrm{\infty }$, then setting $g=1$ gives ${\parallel g\parallel }_q=1$ and

$$K\ge {\parallel fg\parallel }_1={\parallel f\parallel }_1.$$

So, we have shown that $K={\parallel f\parallel }_p$ when and , and when $p=1$. From now on, it is assumed that the measure is $\sigma$-finite. Then there is a sequence $A_n\in 𝔐$ increasing to the whole of $X$ and such that .

Now consider the case where and ${\parallel f\parallel }_p=\mathrm{\infty }$. Let $f_n$ be the sequence of functions

$$f_n=1_{A_n}{1}_{|f|\le n}f$$

then, $|f_n|\le |f|$ and monotone convergence gives ${\parallel f_n\parallel }_p\to {\parallel f\parallel }_p=\mathrm{\infty }$. Therefore,

$$K\ge sup\{{\parallel f_{n}g\parallel }_1:g\in L^q,{\parallel g\parallel }_q=1\}={\parallel f_n\parallel }_p.$$

and letting $n$ go to infinity gives $K=\mathrm{\infty }$.

We finally consider $p=\mathrm{\infty }$. Then, for any there exists a set $A\in 𝔐$ with $\mu (A)>0$ such that $|f|\ge L$ on $A$.Also, monotone convergence gives $\mu (A\cap A_n)\to \mu (A)$ and, therefore, $\mu (A\cap A_n)>0$ eventually. Replacing $A$ by $A\cap A_n$ if necessary, we may suppose that . So, setting $g=1_A/\mu (A)$ gives ${\parallel g\parallel }_1=1$ and,

$$K\ge {\parallel fg\parallel }_1={\int }_A|f|𝑑\mu /\mu (A)\ge L.$$

Letting $L$ increase to ${\parallel f\parallel }_p$ gives $K\ge {\parallel f\parallel }_p$ as required.