Equality in Hardy's inequality via Hölder's

I'm working on Exercise 3.14 in Rudin's Real and Complex Analysis. I was able to answer part (a): that for real $p$ satisfying $1<p<\infty$, for every function $f$ in $L^p(0,\infty)$, when $F$ is defined by $$ F(x)=\frac{1}{x}\int_{0}^{x} f(t)~dt, $$ then $$ \lVert F\rVert_p\leq\frac{p}{p-1}\lVert f\rVert_p. $$ My proof uses Rudin's suggestion to show the inequality first for nonnegative compactly supported continuous functions, and then the general case. For the nonnegative compactly supported continuous case, I used Hölder's inequality on $$ \int_{0}^{\infty} F^{p-1}(x)f(x)~dx, $$ and I used Fatou's lemma and a density argument for the general case. What I'm stuck on is part (b): showing equality holds in Hardy's inequality only when $f$ vanishes almost everywhere. I was able to do this with the additional assumption that $f$ is nonnegative, compactly supported, and continuous, by using the necessary and sufficient condition for equality in Hölder's. I wasn't able to extend this to the general case. Any suggestions?

(I saw a related question here which uses the Fubini theorem, but Rudin doesn't cover Fubini until Chapter 8.)

Solution 1:

Here's the approach that I used on this exercise. I feel like there's probably a simpler way, but I didn't see one without using Fubini's Theorem.

First, use Hölder's inequality to show that $f_n \in L^p$, $f \in L^p$, and $f_n \to f$ in $L^p$ guarantee that $F_n \to F$ pointwise.
In addition, show that $f_1 \le f_2$ pointwise guarantees $F_1 \le F_2$ pointwise.
Verify using Hölder's inequality that $F(x)$ is continuous in $x$ for $x \gt 0$.

The hint which Rudin gives a hint for part a) is to assume the continuous, non-negative compact support case for $f$ to show

$$ \int_0^{\infty}F^p\ dx = -p\int_0^{\infty}F^{p-1}(f-F) \ dx $$

Rewrite:

$$ \label{a}\tag{*} \int_0^{\infty}F^p\ dx = \frac{p}{p-1}\int_0^{\infty}F^{p-1}f \ dx $$

Use Lebesgue Monotone Convergence to show ($\ref{a}$) holds when $f$ is the characteristic function of an open set with finite measure.

Show that ($\ref{a}$) holds for non-negative simple functions $s$ which are nonzero only on a set of finite measure as follows: Let $U$ be an open set of finite measure such that $s = 0$ outside of $U$. Use Lusin's theorem to get a sequence $g_n$ of continuous, compactly supported, non-negative functions whose supports are contained in $U$, and such that $g_n \to s$ pointwise a.e., $g_n \to s$ in $L^p$, and with $0 \le g_n \le K\chi_{U}$ for some $K > 0$. Apply Lebesgue Dominated Convergence to show that ($\ref{a}$) holds for $s$.

Show that ($\ref{a}$) holds for any non-negative $f \in L^p$ using Monotone Convergence on a increasing sequence of non-negative simple functions which converges pointwise to $f$.

Now let $f$ be an arbitrary non-negative function in $L^p$.

Suppose (for a contradiction) that $||F||_p =\frac{p}{p-1}||f||_p$ but that it is not the case that $f = 0$ almost everywhere.

Then $||f||_p \gt 0$, and since we are in the equality situation, we also have $||F||_p \gt 0$

Using Hölder's inequality, show that $$ ||F||_p^p = \int_0^{\infty}F^p\ dx = \frac{p}{p-1}\int_0^{\infty}F^{p-1}f \ dx \\ \le \frac{p}{p-1}\left\{ \int_0^{\infty}F^p \ dx \right\}^{1 - \frac{1}{p}} ||f||_p \\ = \frac{p}{p-1} ||F||_{p-1}||f||_p $$

Thus Hölder's inequality must be an equality, and so there must be a non-negative constant $\alpha$ such that $F = \alpha f$ a.e. or $f = \alpha F$ a.e..
Verify that $\alpha$ must in fact be strictly positive.

The next part is what I wrestled with for some time.

We now have that WLOG $f = \alpha F$ a.e. for some $\alpha \gt 0$. Verify using ($\ref{a}$) that $\alpha = \frac{p-1}{p}$. From this we may conclude that $$ F(x) = \frac{1}{x} \int_0^x \alpha F(t) \ dt $$ Since $F(t)$ is continuous, it must be the case that $F(x)$ is differentiable for $x \gt 0$. Differentiating gives $$ xF'(x) = (\alpha - 1)F(x) = -\frac{1}{p}F(x) $$ Since $||F||_p > 0$, there must be a point $a$ with $F(a) > 0$.
But, the above equation shows that if $F(a) > 0$, then $F(x) > 0$ for $0 < x < a$.
So on $(0, a)$, have $\frac{F'}{F} = -\frac{1}{px}$ and hence $\log F(x) = C -\frac{1}{p} \log(x)$.

So, near zero, we have $F(x) = C x^{-\frac{1}{p}}$ for some $C > 0$. But this is a contradiction since then $F$ is not in $L^p$.

The case where $f$ is an arbitrary (complex-valued) function in $L^p$ follows from the case where $f$ is non-negative.

Let $g = |f|$.

If $||F||_p =\frac{p}{p-1}||f||_p$, then $$ \frac{p}{p-1}||f||_p = ||F||_p \le ||G||_p \le \frac{p}{p-1}||g||_p = \frac{p}{p-1}||f||_p $$ So $g = 0$ almost everywhere by the non-negative case, and the same is true for $f$.