Using Hahn Banach for switching between $L^p$ spaces

Question

I want to understand the proof (or under which conditions a proof holds) of the following statement:

Let $f$ be a function in $L^p$ and let $q$ be such that $\frac{1}{p}+\frac{1}{q}=1$. Then we can find a function $g$ in $L^q$ such that:

1. $\| g \|_{q}=1$
2. $fg \in L^1$
3. $\|fg\|_{1} = \|f\|_{p}$

For context: This is used in Kenneth Hoffman "Banach Spaces and Analytic Functions" to prove that if $f \in L^p(-\pi, \pi)$ then the Césaro means of the Fourier series for $f$ converge to $f$ in the $L^p$ norm. In the book, the author merely states that this is a consequence of the Hahn-Banach theorem.

I know the Hahn-Banach version for extending linear functionals and for separating convex sets and I can't wrap my head around on how to use it here.

In addition to it, I am not entirely sure if what I stated is entirely correct, maybe we need the additional assumption of finite measure spaces like $(-\pi, \pi)$ as used in the book.

All in all, this seems like a very cool trick for going back and forth between $L^p$-spaces to establish inequalities. In addition to an explanation of this result, I would also like to know about other related tricks and applications of the Hahn-Banach to this type of questions.

Answer 1

Recall that the dual of $L^p$ is $L^q$, which is valid for arbitrary measure spaces when $1 < p < \infty$. (It is also valid for $p=1$ when the space is $\sigma$-finite, but for $p=1$ the statement at hand here is trivial; simply take $g=1$.)

Suppose without loss of generality $\|f\|_p = 1$. Consider the 1-dimensional subspace of $L^p$ spanned by $f$, and define a linear functional $\ell_0$ on this space by $\ell_0(f)=1$. Clearly $\|\ell_0\| = 1$. Use Hahn-Banach to extend $\ell_0$ to a bounded linear functional $\ell$ on all of $L^p$, whose norm is still 1. Since the dual of $L^p$ is $L^q$, there exists $g \in L^q$ such that $\int gh = \ell(h)$ for all $h \in L^p$; in particular, $\int fg = \ell(f) = 1$. Moreover, $\|g\|_q = \|\ell\| = 1$, so (1) and (2) are established.

For (3), by the triangle inequality and Hölder's inequality we have $$1 = \left|\int fg \right| \le \|fg\|_1 \le \|f\|_p \|g\|_q = 1$$ so that $\|fg\|_1 = 1$ as desired.

Answer 2

An answer involving Hahn-Banach has already been given. Let me suggest an alternative not involving Hahn-Banach. Because, while one can invoke Hahn-Banach, it seems like cracking a nut with a tank.

For the rest of this post $1<p,q<\infty$ with $1/p+1/q=1$. We can choose $$g(x)=\mathrm{sign}(f(x))\vert f(x)\vert^{p-1}/\Vert \vert f \vert^{p-1}\Vert_q.$$ As $1/p+1/q=1$, we get $1/q=1-1/p=(p-1)/p$ and therefore $$p=(p-1)q.$$ Thus, $\vert f\vert^{p-1}$ is in $L^q$ and we get $\Vert g\Vert_q=1$.

Next we see $$f(x)g(x)=\vert f(x)\vert^p/\Vert \vert f\vert^{p-1}\Vert_q$$ and hence $fg\in L^1$ and $$ \Vert fg \Vert_1=\Vert f \Vert_p^p / \Vert \vert f\vert^{p-1}\vert \Vert_q.$$ However, $$\Vert \vert f\vert^{p-1}\vert \Vert_q = \left( \int \vert f(x)\vert^p dx \right)^{1/q}.$$ Thus, $$ \Vert fg \Vert_1=\left( \int \vert f(x)\vert^p dx\right)^{1-1/q}=\Vert f\Vert_p.$$